Algebraic Processes — NBT

Q1Basic

Topic: Common factor + difference of squares

Factorise fully: $3x^2 - 12$

A) $(3x-6)(x+2)$

B) $3(x-2)(x+2)$

C) $3(x-4)(x+1)$

D) $(x-2)(3x+6)$

Show the worked solution

Answer: B

Explanation

Step 1 — Take out the common factor. Both terms divide by 3. $3x^2 - 12 = 3(x^2-4)$

Step 2 — Factorise what is left. $x^2-4$ is a difference of two squares, $a^2-b^2=(a-b)(a+b)$ . $3x^2-12 = 3(x-2)(x+2)$

Why the others are wrong:

A and D still hide a 3 inside a bracket: $(3x-6)=3(x-2)$ . Not fully factorised.
C — expand it: $(x-4)(x+1)=x^2-3x-4$ . That is not $x^2-4$ .

Takeaway: "Factorise fully" means check every bracket again. Can anything still be divided down?

Q2Basic

Topic: Trinomial factorisation

Factorise: $x^2 + 5x + 6$

A) $(x+1)(x+6)$

B) $(x+6)(x-1)$

C) $(x-2)(x-3)$

D) $(x+2)(x+3)$

Show the worked solution

Answer: D

Explanation

For $x^2+bx+c$ , find two numbers that multiply to $c$ and add to $b$ . Here $c=6$ , $b=5$ .

Factor pairs of 6: $(1,6)$ and $(2,3)$ . Which adds to 5? $2+3=5$ ✓

$x^2+5x+6 = (x+2)(x+3)$

Check: $(x+2)(x+3) = x^2+5x+6$ ✓

Why the others are wrong:

A uses $(1,6)$ — multiplies to 6, but adds to 7, not 5.
C uses $(-2,-3)$ — multiplies to $+6$ , but adds to $-5$ . That gives $x^2-5x+6$ .
B uses $(6,-1)$ — multiplies to $-6$ , not $+6$ .

Takeaway: If $b$ and $c$ are both positive, both factors are positive. Checking the signs first cuts your options in half.

Q3Basic

Topic: Difference of squares with coefficients

Factorise fully: $9x^2 - 16y^2$

A) $(3x-4y)(3x+4y)$

B) $(9x-16y)(x+y)$

C) $(3x-4y)^2$

D) $(9x+4y)(x-4y)$

Show the worked solution

Answer: A

Explanation

Find the squares: $9x^2=(3x)^2$ and $16y^2=(4y)^2$ . So $a=3x$ and $b=4y$ , and $a^2-b^2=(a-b)(a+b)$ :

$9x^2-16y^2 = (3x-4y)(3x+4y)$

Check: $(3x-4y)(3x+4y) = 9x^2+12xy-12xy-16y^2 = 9x^2-16y^2$ ✓ The middle terms cancel — that always happens with a difference of squares.

Why the others are wrong:

B and D — expanding either leaves an $xy$ term. The original has no $xy$ term.
C is a perfect square, not a difference of squares: $(3x-4y)^2 = 9x^2-24xy+16y^2$ .

Takeaway: Difference of squares needs two terms, a minus between them, and both terms perfect squares. A plus between two squares ( $9x^2+16y^2$ ) cannot be factorised at all.

Q4Basic

Topic: Factorising by grouping (four terms)

Factorise fully: $ax + ay + bx + by$

A) $abxy$

B) $(a+x)(b+y)$

C) $(a+b)(x+y)$

D) $(a-b)(x-y)$

Show the worked solution

Answer: C

Explanation

Four terms, no overall common factor — so group them in pairs that do share a factor.

$ax+ay+bx+by = a(x+y) + b(x+y)$

Now $(x+y)$ is common to both halves. Take it out:

$= (a+b)(x+y)$

Check: $(a+b)(x+y) = ax+ay+bx+by$ ✓

Why the others are wrong:

B pairs the wrong terms. Expand it: $ab+ay+bx+xy$ — nothing like the original.
D has the right shape but wrong signs. It answers $ax-ay-bx+by$ .
A just multiplies everything together. That is not factorising.

Takeaway: With four terms, split them into two pairs, factorise each pair, then take out the bracket they share.

Q5Intermediate

Topic: Sum of cubes

Factorise fully: $x^3 + 27$

A) $(x+3)^3$

B) $(x+3)(x^2+3x+9)$

C) $(x+3)(x^2-3x+9)$

D) $(x+3)(x^2-6x+9)$

Show the worked solution

Answer: C

Explanation

$27 = 3^3$ , so this is a sum of cubes with $a=x$ and $b=3$ :

$a^3+b^3 = (a+b)(a^2-ab+b^2)$ $x^3+27 = (x+3)(x^2-3x+9)$

Watch the signs: the bracket $(x+3)$ keeps the original sign, but the middle term of the trinomial flips to the opposite sign, $-3x$ .

Why the others are wrong:

B forgot the middle sign flips. It expands to $x^3+6x^2+18x+27$ .
D uses $-6x$ , which belongs to $(x-3)^2$ , a different expansion.
A confuses "cube of a sum" with "sum of cubes". $(x+3)^3 = x^3+9x^2+27x+27$ .

Takeaway: Learn both cube identities together: $a^3+b^3=(a+b)(a^2-ab+b^2) \qquad a^3-b^3=(a-b)(a^2+ab+b^2)$ The bracket matches the original sign; the trinomial's middle term is always the opposite sign.

Q6Intermediate

Topic: Trinomial with leading coefficient ≠ 1

Factorise: $2x^2 + 5x - 3$

A) $(2x-1)(x+3)$

B) $(2x+1)(x-3)$

C) $(2x-3)(x+1)$

D) $(2x+3)(x-1)$

Show the worked solution

Answer: A

Explanation

When the leading coefficient is not 1, find two numbers that multiply to $a\times c$ and add to $b$ .

Here $a\times c = 2\times(-3) = -6$ and $b=5$ . The pair is $6$ and $-1$ .

Split the middle term and group: $2x^2 + 6x - x - 3 = 2x(x+3) - 1(x+3) = (2x-1)(x+3)$

Check: $(2x-1)(x+3) = 2x^2+5x-3$ ✓

Why the others are wrong: all three give the wrong middle term.

B: $(2x+1)(x-3)=2x^2-5x-3$
C: $(2x-3)(x+1)=2x^2-x-3$
D: $(2x+3)(x-1)=2x^2+x-3$

Takeaway: Always expand your answer to check. With $a\neq1$ several bracket arrangements look right, and only one is.

Q7Intermediate

Topic: Factorising a cubic by grouping

Factorise fully: $x^3 - 3x^2 - 4x + 12$

A) $(x-3)(x-4)(x+1)$

B) $(x+3)(x-2)(x+2)$

C) $(x-3)(x^2-4)$

D) $(x-3)(x-2)(x+2)$

Show the worked solution

Answer: D

Explanation

Four terms, no overall common factor — so group in pairs: $x^3-3x^2-4x+12 = x^2(x-3) - 4(x-3) = (x-3)(x^2-4)$

Now $x^2-4$ is a difference of squares, so keep going: $(x-3)(x-2)(x+2)$

Why the others are wrong:

C is a correct intermediate step, but $x^2-4$ still factorises. The question says fully.
B has a sign error: $(x+3)(x^2-4) = x^3+3x^2-4x-12$ .
A expands to $x^3-6x^2+5x+12$ — not the original.

Takeaway: After grouping, check each bracket again for another pattern. Difference of squares is the one people miss.

Q8Proficient

Topic: Quadratic-in-disguise (substitution insight)

Factorise fully: $x^4 - 13x^2 + 36$

A) $(x-4)(x+4)(x-9)(x+9)$

B) $(x-2)(x+2)(x-3)(x+3)$

C) $(x-2)(x+2)(x^2-9)$

D) $(x^2-4)(x^2-9)$

Show the worked solution

Answer: B

Explanation

Only even powers of $x$ appear, so this is a quadratic in disguise. Let $y=x^2$ :

$y^2 - 13y + 36 = (y-4)(y-9)$

Substitute back $y=x^2$ : $(x^2-4)(x^2-9)$

Both brackets are differences of squares, so keep going: $(x-2)(x+2)(x-3)(x+3)$

Why the others are wrong:

D stops right after substituting back. Correct, but not fully factorised.
C factorises the first bracket and forgets the second.
A takes $y=4$ and $y=9$ and writes $x=4$ and $x=9$ . But $y=x^2$ , so $x^2=4$ gives $x=\pm2$ and $x^2=9$ gives $x=\pm3$ .

Takeaway: Only even powers? Substitute $y=x^2$ . When you swap back, remember you must take a square root, not copy the number across.

Section 1.2 — Linear & Quadratic Equations

Two habits that prevent almost every careless error in this section: 1. Always rearrange to standard form before factorising ( $ax^2+bx+c=0$ , everything on one side, zero on the other). 2. Always check your final answer by substituting it back into the original equation — not your rearranged version, the original. This catches sign errors and extraneous roots.

Q9Basic

Topic: Linear equation with fractions

Solve for $x$ : $\dfrac{3x-2}{4} = \dfrac{x+6}{2}$

A) $x = 14$

B) $x = 10$

C) $x = 2$

D) $x = -14$

Show the worked solution

Answer: A

Explanation

Step 1 — Clear the fractions. Cross-multiply: $2(3x-2) = 4(x+6)$

Step 2 — Expand. $6x - 4 = 4x + 24$

Step 3 — Collect terms. Anything crossing the equals sign flips its sign. $6x - 4x = 24 + 4 \quad\Rightarrow\quad 2x = 28 \quad\Rightarrow\quad x = 14$

Check in the original: LHS $=\frac{40}{4}=10$ , RHS $=\frac{20}{2}=10$ ✓

Why the others are wrong:

B ( $x=10$ ) forgot a sign flip: $6x-4x = 24-4$ instead of $24+4$ .
C ( $x=2$ ) cross-multiplied the wrong way: $4(3x-2)=2(x+6)$ .
D ( $x=-14$ ) flipped the signs on both constants.

Takeaway: When a term crosses the equals sign, say the sign change out loud: "minus four becomes plus four." Sign slips are the most common algebra error there is.

Q10Basic

Topic: Solving a quadratic by factorisation

Solve for $x$ : $x^2 - 7x + 10 = 0$

A) $x = -2$ or $x = -5$

B) $x = 2$ or $x = -5$

C) $x = 2$ or $x = 5$

D) $x = -2$ or $x = 5$

Show the worked solution

Answer: C

Explanation

Already in standard form, so factorise. Two numbers that multiply to $+10$ and add to $-7$ : that is $-2$ and $-5$ .

$x^2-7x+10 = (x-2)(x-5) = 0$

Zero-product rule: if two things multiply to zero, one of them must be zero. $x=2 \quad\text{or}\quad x=5$

Why the others are wrong: each solves a different equation.

A solves $x^2+7x+10=0$
B solves $x^2+3x-10=0$
D solves $x^2-3x-10=0$

They all use the numbers 2 and 5 with the wrong signs.

Takeaway: Check signs before you write anything. In $x^2+bx+c=0$ : if $c$ is positive, both factors share the sign of $b$ . If $c$ is negative, the factors have opposite signs. Here $c=+10$ , $b=-7$ , so both factors are negative.

Q11Basic

Topic: Quadratic formula with a surd answer

Solve for $x$ (leave your answer in simplest surd form): $x^2 - 6x + 4 = 0$

A) $x = 6 \pm \sqrt{5}$

B) $x = 3 \pm \sqrt{5}$

C) $x = 3 \pm \sqrt{20}$

D) $x = 3 \pm \sqrt{13}$

Show the worked solution

Answer: B

Explanation

This will not factorise, so use the quadratic formula with $a=1$ , $b=-6$ , $c=4$ :

$x = \frac{6 \pm \sqrt{36-16}}{2} = \frac{6 \pm \sqrt{20}}{2}$

Simplify the surd: $\sqrt{20} = \sqrt{4\times5} = 2\sqrt5$ .

$x = \frac{6 \pm 2\sqrt5}{2}$

Divide every term by 2 — this is the step people rush:

$x = 3 \pm \sqrt5$

Why the others are wrong:

C forgot to divide the surd by $2a$ : $\dfrac{6 \pm \sqrt{20}}{2}$ became $3 \pm \sqrt{20}$ .
D used $b^2 + 4ac$ for the discriminant instead of $b^2 - 4ac$ , giving $\sqrt{52}$ and so $3 \pm \sqrt{13}$ .
A divided the surd but not the 6.

Takeaway: The fraction bar applies to every term above it. After the quadratic formula, ask: did I divide both the whole number and the surd by $2a$ ?

Q12Intermediate

Topic: Completing the square

Write $x^2 + 8x + 3$ in the form $(x+p)^2 + q$ .

A) $(x+4)^2 - 19$

B) $(x+4)^2 + 3$

C) $(x+8)^2 - 13$

D) $(x+4)^2 - 13$

Show the worked solution

Answer: D

Explanation

You want to rewrite $x^2+8x$ as a perfect square plus a correction.

Step 1 — Halve the coefficient of $x$ . Half of 8 is 4, so try $(x+4)^2$ . Expand it: $(x+4)^2 = x^2 + 8x + 16$

Step 2 — Correct the constant. You have $+16$ but want $+3$ , so subtract 16 and add 3: $x^2+8x+3 = (x+4)^2 - 16 + 3 = (x+4)^2 - 13$

Why the others are wrong:

C uses $p=8$ — the original coefficient instead of half of it.
B just tacks the $+3$ on, ignoring the extra $+16$ that $(x+4)^2$ brings in.
A has the right method but slips on the arithmetic: $3-16=-13$ , not $-19$ .

Takeaway: Two moves every time: (1) halve $b$ to get $p$ ; (2) $q = c - p^2$ .

Q13Intermediate

Topic: Rearranging before solving

Solve for $x$ : $x^2 = 5x - 6$

A) $x = 2$ or $x = 3$

B) $x = -6$ or $x = -1$

C) $x = -2$ or $x = -3$

D) $x = -2$ or $x = 3$

Show the worked solution

Answer: A

Explanation

The key step comes before factorising: this is not in standard form. Move everything to one side.

$x^2 - 5x + 6 = 0$

Now factorise. Two numbers multiplying to $+6$ and adding to $-5$ : that is $-2$ and $-3$ . $(x-2)(x-3) = 0 \quad\Rightarrow\quad x = 2 \ \text{or}\ x = 3$

Check: $x=2$ gives LHS $=4$ , RHS $=4$ ✓

Why the others are wrong:

B left it as $x(x-5)=-6$ and set $x=-6$ , $x-5=-6$ . The zero-product rule only works when the product is zero.
C factorised $x^2+5x+6=0$ — never rearranged at all.
D flipped the sign on just one factor.

Takeaway: "Solve the quadratic" is two instructions: (1) get it into standard form, (2) factorise. Skipping step 1 causes more wrong answers here than anything else.

Q14Intermediate

Topic: Rational equations and restrictions

Solve for $x$ : $\dfrac{3}{x-2} = \dfrac{1}{x+1}$

A) $x = \dfrac{5}{2}$

B) No solution, because $x \neq 2$

C) $x = \dfrac{7}{2}$

D) $x = -\dfrac{5}{2}$

Show the worked solution

Answer: D

Explanation

Step 1 — Note the restrictions. Denominators cannot be zero, so $x \neq 2$ and $x \neq -1$ .

Step 2 — Cross-multiply. $3(x+1) = 1(x-2)$

Step 3 — Expand and solve. $3x + 3 = x - 2 \quad\Rightarrow\quad 2x = -5 \quad\Rightarrow\quad x = -\frac{5}{2}$

$-\frac52$ is not a restricted value, so it is valid. Check: LHS $=-\frac23$ , RHS $=-\frac23$ ✓

Why the others are wrong:

A flipped a sign: $3x-x = -2+3$ instead of $-2-3$ .
C cross-multiplied the wrong pairs: $3(x-2)=1(x+1)$ .
B confuses a restriction with no solution. $x\neq2$ only rules out that one value. The equation still has an answer elsewhere.

Takeaway: Restrictions just stop you dividing by zero. State them, solve as normal, and only reject your answer if it lands on a restricted value.

Q15Proficient

Topic: Equations solvable by substitution

Solve for $x$ : $x - 4\sqrt{x} - 5 = 0$

A) $x = 1$

B) $x = 25$ or $x = 1$

C) $x = 25$

D) $x = -1$

Show the worked solution

Answer: C

Explanation

Since $x = (\sqrt{x})^2$ , let $y=\sqrt{x}$ . The equation becomes an ordinary quadratic:

$y^2 - 4y - 5 = 0 \quad\Rightarrow\quad (y-5)(y+1)=0 \quad\Rightarrow\quad y=5 \text{ or } y=-1$

Now the step this question is really testing: $y=\sqrt{x}$ , and a square root is never negative. So reject $y=-1$ .

That leaves $\sqrt{x}=5$ , so $x=25$ .

Check: $25 - 4\sqrt{25} - 5 = 25-20-5 = 0$ ✓

Why the others are wrong:

B kept $y=-1$ and squared it to get $x=1$ . Test it: $1-4-5=-8\neq0$ . This is an extraneous solution — it fits the rearranged equation but not the original.
A keeps only that invalid root.
D gives $y$ , not $x$ .

Takeaway: Any substitution involving a square root can produce fake answers. Always test each one in the original equation.

Q16Proficient

Topic: Simultaneous equations (linear and quadratic)

Solve simultaneously for $x$ and $y$ : $\quad y = x + 1 \quad$ and $\quad y = x^2 - 5$

A) $(3;2)$ and $(-2;-3)$

B) $(3;4)$ and $(-2;-1)$

C) $(3;4)$ only

D) $(-3;-2)$ and $(2;3)$

Show the worked solution

Answer: B

Explanation

Both equations start " $y =$ ", so set the right-hand sides equal: $x + 1 = x^2 - 5$

Rearrange to standard form and factorise: $x^2 - x - 6 = 0 \quad\Rightarrow\quad (x-3)(x+2)=0 \quad\Rightarrow\quad x = 3 \text{ or } x = -2$

Do not stop here. "Solve simultaneously" wants full coordinate pairs. Put each $x$ back into $y=x+1$ :

$x=3 \Rightarrow y=4 \Rightarrow (3;4)$
$x=-2 \Rightarrow y=-1 \Rightarrow (-2;-1)$

Why the others are wrong:

D made a sign slip forming the quadratic ( $x^2+x-6=0$ ) and carried it through.
C found one root and stopped. A quadratic usually has two.
A has the right $x$ -values but used $y=x-1$ instead of $y=x+1$ .

Takeaway: Three steps every time: (1) combine into one equation, (2) solve it (expect two answers), (3) substitute every answer back to get its partner.

Section 1.3 — The Factor Theorem

The one idea to lock in:

Factor theorem: $(x-a)$ is a factor of $f(x)$ if and only if $f(a) = 0$ .

"If and only if" means the substitution is complete proof, both ways. If $f(a)=0$ then $(x-a)$ is a factor — no extra checking needed. And if $(x-a)$ is a factor, then $f(a)$ must be 0.

The detail people get wrong: for the factor $(x-a)$ you substitute $x=a$ — the value that makes the bracket zero. So for $(x+2)$ , substitute $x=-2$ , not $x=+2$ .

Q17Basic

Topic: Applying the factor theorem

Given $f(x) = x^3 - 2x^2 - 5x + 6$ , evaluate $f(1)$ and hence state whether $(x-1)$ is a factor of $f(x)$ .

A) $f(1) = 0$ , but more information is needed before concluding $(x-1)$ is a factor

B) $f(1) = 0$ ; no, $(x-1)$ is not a factor of $f(x)$

C) $f(1) = 4$ ; no, $(x-1)$ is not a factor of $f(x)$

D) $f(1) = 0$ ; yes, $(x-1)$ is a factor of $f(x)$

Show the worked solution

Answer: D

Explanation

Step 1 — Substitute $x=1$ , term by term. $f(1) = 1 - 2 - 5 + 6 = 0$

Step 2 — Apply the factor theorem. $f(1)=0$ , so $(x-1)$ is a factor. No further checking needed.

Why the others are wrong:

C slipped a sign (used $+2$ instead of $-2$ for the $x^2$ term) and got a non-zero value.
B does the calculation right, then states the opposite conclusion.
A doubts the theorem. "If and only if" means the test is conclusive on its own.

Takeaway: With the factor theorem, the calculation is the proof. Show $f(a)=0$ and you are done.

Q18Intermediate

Topic: Applying the factor theorem to $(x+2)$

Use the factor theorem to determine whether $(x+2)$ is a factor of $f(x) = x^3 - 7x - 6$ .

A) $f(-2) = -28$ ; no, $(x+2)$ is not a factor of $f(x)$

B) $f(-2) = 0$ ; yes, $(x+2)$ is a factor of $f(x)$

C) $f(2) = -12$ ; no, $(x+2)$ is not a factor of $f(x)$

D) $f(-2) = 0$ , but polynomial long division is still needed to be certain

Show the worked solution

Answer: B

Explanation

Step 1 — Find the value to substitute. $(x+2)=0$ gives $x=-2$ , not $+2$ . This is the most-tested detail in the topic.

Step 2 — Substitute, keeping every sign. $f(-2) = (-2)^3 - 7(-2) - 6 = -8 + 14 - 6 = 0$

Step 3 — Conclude. $f(-2)=0$ , so $(x+2)$ is a factor.

Why the others are wrong:

C substituted $x=+2$ : $8-14-6=-12$ . One sign error flips the whole conclusion.
A used $x=-2$ but treated $-7(-2)$ as $-14$ instead of $+14$ . Minus times minus is plus.
D doubts the theorem. $f(-2)=0$ is the proof — no long division needed.

Takeaway: Write it down before substituting: "the bracket is $(x+2)$ , so I use $x=-2$ ." Stating the value first kills the most common error in this topic.

Q19Intermediate

Topic: Finding an unknown coefficient using the factor theorem

Given that $(x - 2)$ is a factor of $f(x) = x^3 + kx^2 + 2x - 24$ , find the value of $k$ .

A) $k = 3$

B) $k = 9$

C) $k = 12$

D) $k = -3$

Show the worked solution

Answer: A

Explanation

Step 1 — Turn the factor into an equation. $(x-2)$ is a factor, so $f(2)=0$ .

Step 2 — Substitute $x=2$ , keeping $k$ . $f(2) = 8 + 4k + 4 - 24 = 4k - 12$

Step 3 — Set it to zero and solve. $4k - 12 = 0 \quad\Rightarrow\quad 4k = 12 \quad\Rightarrow\quad k = 3$

Check: $f(2) = 8+12+4-24 = 0$ ✓

Why the others are wrong:

B substituted $x=-2$ instead of $+2$ , giving $4k-36=0$ and $k=9$ .
D wrote $4k=-12$ instead of $+12$ . The number flips sign crossing the equals sign.
C got $4k=12$ then forgot to divide by 4.

Takeaway: Two separate steps: substitute to get an expression in $k$ , then solve. Do not try to do both at once.

Q20Intermediate

Topic: Finding all roots of a cubic

Given that $(x-1)$ is a factor of $f(x) = x^3 - 2x^2 - 5x + 6$ , find all the roots of $f(x) = 0$ .

A) $x = 1$ only

B) $x = 1,\ x = -3,\ x = 2$

C) $x = 1,\ x = 3,\ x = -2$

D) $x = -1,\ x = 3,\ x = -2$

Show the worked solution

Answer: C

Explanation

Step 1 — Divide out the known factor. $x^3 - 2x^2 - 5x + 6 = (x-1)(x^2 - x - 6)$ Check by expanding — it takes seconds and catches division errors.

Step 2 — Factorise the quadratic too. Two numbers multiplying to $-6$ , adding to $-1$ : $-3$ and $2$ . $x^2 - x - 6 = (x-3)(x+2)$

Step 3 — Read off all three roots. $f(x) = (x-1)(x-3)(x+2) = 0 \quad\Rightarrow\quad x = 1,\ 3,\ -2$

Why the others are wrong:

B factorised the quadratic as $(x+3)(x-2)$ , which expands to $x^2+x-6$ — wrong middle sign.
D read $(x-1)$ as giving $x=-1$ . It gives $x=+1$ .
A stopped after the given factor, missing that the quadratic factorises too.

Takeaway: A cubic is a two-stage factorisation. Use the given factor to get a quadratic, then factorise that. You are only finished at three linear brackets.

Q21Proficient

Topic: Simultaneous conditions on a polynomial

The cubic $f(x) = x^3 + ax^2 + bx - 6$ has $(x-1)$ and $(x+2)$ as factors. Find the values of $a$ and $b$ .

A) $a = 1,\ b = 4$

B) $a = 4,\ b = 1$

C) $a = 4,\ b = -1$

D) $a = 4$ , and $b$ cannot be determined from this information

Show the worked solution

Answer: B

Explanation

Two unknowns means you need two equations — and you have been given exactly two factors. Turn each into an equation.

Equation (i) — from $(x-1)$ : $f(1)=0$ . $1 + a + b - 6 = 0 \quad\Rightarrow\quad a + b = 5$

Equation (ii) — from $(x+2)$ : $f(-2)=0$ . $-8 + 4a - 2b - 6 = 0 \quad\Rightarrow\quad 2a - b = 7$

Add (i) and (ii) — the $+b$ and $-b$ cancel: $3a = 12 \quad\Rightarrow\quad a = 4, \quad\text{then}\quad b = 1$

Check: $f(1)=0$ ✓ and $f(-2)=0$ ✓

Why the others are wrong:

A found 4 and 1 but swapped which is $a$ and which is $b$ . Label your answers.
C computed $b = 4-5 = -1$ instead of $5-4 = 1$ .
D says $b$ cannot be found. Two unknowns need two equations, and two factors give exactly that.

Takeaway: Count your unknowns, then count your pieces of information. You need one equation per unknown, and a well-set question gives you exactly that many.

Q22Proficient

Topic: Logical deduction from polynomial roots ("must be true")

$f(x)$ is a cubic polynomial. You are told that $f(2) = 0$ , $f(-1) = 0$ , and $f(3) = 0$ . Which of the following statements about $f(x)$ must be true?

A) $f(x) = (x-2)(x+1)(x-3)$

B) $f(x)$ could have a fourth root that we haven't been told about

C) $f(x)$ has exactly one real root

D) $f(x) = k(x-2)(x+1)(x-3)$ for some non-zero constant $k$

Show the worked solution

Answer: D

Explanation

A "must be true" question. Three options will look reasonable unless you separate what you know from what you are assuming.

By the factor theorem, the three given zeros mean $(x-2)$ , $(x+1)$ and $(x-3)$ are all factors. Their product is already a cubic.

$f(x)$ is also a cubic, so there is no room for anything more. $f(x)$ can only differ from that product by a constant multiplier $k$ — which stretches the graph but does not move where it crosses the $x$ -axis. And $k\neq0$ , or $f$ would not be a cubic at all.

$f(x) = k(x-2)(x+1)(x-3)$

Why the others are wrong:

A would be right only if you knew the leading coefficient was 1. You were not told that — $f(x)=2(x-2)(x+1)(x-3)$ fits every given condition too. "Must be true" has to hold for every polynomial fitting the description.
C contradicts what you were given: three zeros means three real roots, not one.
B breaks a basic rule: a degree- $n$ polynomial has at most $n$ roots. A cubic cannot have a fourth.

Takeaway: For "must be true", try to build a counter-example against each option and discard any you can break. Keep two facts ready: a degree- $n$ polynomial has at most $n$ roots, and knowing all the roots pins the polynomial down only up to a constant multiplier.

Section 1.4 — Inequalities

The one rule that matters more than any other in this section:

Multiplying or dividing both sides of an inequality by a negative number reverses the inequality sign ( $<$ becomes $>$ , $\leq$ becomes $\geq$ , and vice versa).

Why does this happen? Think of a simple true statement: $2 < 5$ . Multiply both sides by $-1$ : you'd get $-2 < -5$ , which is false ( $-2$ is actually bigger than $-5$ on the number line). To keep the statement true, the sign must flip: $-2 > -5$ ✓. This isn't an arbitrary rule to memorise — it's the only way to keep the inequality telling the truth.

Q23Basic

Topic: Linear inequality (no sign flip needed)

Solve for $x$ : $3x - 5 < 7$

A) $x < 4$

B) $x > 4$

C) $x < \dfrac{2}{3}$

D) $x < -4$

Show the worked solution

Answer: A

Explanation

Treat it like an equation: move the constant across, then divide. $3x < 12 \quad\Longrightarrow\quad x < 4$

We only divided by a positive number, so the sign does not flip.

Why the others are wrong:

B flipped the sign, but nothing negative was divided by. The rule was not triggered.
C wrote $3x < 7-5$ instead of $7+5$ . The $-5$ becomes $+5$ when it crosses over.
D divided correctly, then flipped the sign of the answer for no reason.

Takeaway: Finish every inequality by asking: did I multiply or divide by a negative anywhere? If no, the sign must look exactly as it did at the start.

Q24Basic

Topic: Linear inequality (sign flip required)

Solve for $x$ : $-2x + 3 \geq 11$

A) $x \geq 4$

B) $x \geq -4$

C) $x \leq -4$

D) $x \leq 4$

Show the worked solution

Answer: C

Explanation

Step 1 — Isolate the $x$ term. $-2x \geq 8$

Step 2 — Divide by $-2$ . That is negative, so the sign flips ( $\geq$ becomes $\leq$ ): $x \leq -4$

Check with $x=-10$ : $-2(-10)+3 = 23 \geq 11$ ✓ And with $x=0$ : $3 \geq 11$ is false, correctly excluded.

Why the others are wrong: between them they cover every way to slip here.

B divided correctly but forgot to flip the sign.
D flipped the sign but lost the negative in the division.
A did both wrong at once.

Takeaway: Make the flip and the division two separate actions. Write "flip!" the moment you decide to divide by a negative, then do the arithmetic underneath. Doing both at once is where the slips come from.

Q25Intermediate

Topic: Quadratic inequality (strict, "less than")

Solve for $x$ : $x^2 - x - 6 < 0$

A) $-3 < x < 2$

B) $x < -2$ or $x > 3$

C) $-2 < x < 3$

D) $2 < x < 3$

Show the worked solution

Answer: C

Explanation

Step 1 — Factorise. $(x-3)(x+2) < 0$

Step 2 — Find the critical values. $x=3$ and $x=-2$ .

Step 3 — Picture the parabola. It opens upward, crossing at $-2$ and $3$ . An upward parabola is below the axis only between its roots. We want $<0$ , so take the middle: $-2 < x < 3$

Test $x=0$ : $(0-3)(0+2) = -6 < 0$ ✓

Why the others are wrong:

B is the outside region — that answers $x^2-x-6 > 0$ , the opposite inequality.
A misread the brackets: $(x-3)$ gives $x=+3$ , not $-3$ .
D got one critical value right and one wrong.

Takeaway: Sketch the parabola first. Opening up or down? Where does it cross? For an upward parabola, below the axis is between the roots; above is the two outer wings.

Q26Intermediate

Topic: Quadratic inequality (non-strict, "greater than or equal to")

Solve for $x$ : $x^2 - 5x + 6 \geq 0$

A) $x \leq 2$ or $x \geq 3$

B) $2 \leq x \leq 3$

C) $x < 2$ or $x > 3$

D) $x \leq -2$ or $x \geq -3$

Show the worked solution

Answer: A

Explanation

Step 1 — Factorise. $(x-2)(x-3) \geq 0$ , critical values $x=2$ and $x=3$ .

Step 2 — Picture the parabola. Upward again, but now we want $y \geq 0$ — on or above the axis. For an upward parabola that is the two outer regions, and "or equal to" means the roots are included: $x \leq 2 \quad \text{or} \quad x \geq 3$

Check $x=2$ : $(0)(-1) = 0 \geq 0$ ✓ so 2 does belong.

Why the others are wrong:

B is the region between the roots — that answers $\leq 0$ .
C has the right regions but strict signs, wrongly excluding $x=2$ and $x=3$ .
D misread the critical values as $-2$ and $-3$ .

Takeaway: Ask two questions every time: (1) inside or outside the roots? (2) are the boundary points included? The second is decided purely by whether the original sign says "or equal to".

Q27Intermediate

Topic: Rational inequality

Solve for $x$ : $\dfrac{x-2}{x+1} > 0$

A) $x < -1$ or $x \geq 2$

B) $-1 < x < 2$

C) $x \leq -1$ or $x > 2$

D) $x < -1$ or $x > 2$

Show the worked solution

Answer: D

Explanation

What makes this different: the expression is undefined where the denominator is zero. That value can never be in the solution, whatever the sign says.

Step 1 — Find both critical values, and note the type of each.

Numerator zero: $x=2$ — the expression equals zero here.
Denominator zero: $x=-1$ — the expression is undefined here. Hard restriction: $x\neq-1$ .

Step 2 — Test one value in each region.

Region	Test	Value	Sign
$x < -1$	$x=-2$	$\dfrac{-4}{-1}=4$	positive ✓
$-1 < x < 2$	$x=0$	$\dfrac{-2}{1}=-2$	negative
$x > 2$	$x=3$	$\dfrac{1}{4}$	positive ✓

We want positive, so: $x < -1 \quad \text{or} \quad x > 2$

Why the others are wrong:

B is the middle region — that answers $< 0$ .
A includes $x=2$ , but there the fraction is exactly 0, which fails a strict $>0$ .
C includes $x=-1$ , where the expression does not exist at all. That value can never be in any solution.

Takeaway: Sort your critical values into two kinds first. Numerator zeros may be included depending on the sign. Denominator zeros are always excluded, no matter what.

Q28Proficient

Topic: Reasoning about inequalities with no solution ("always/never true")

Solve for $x$ : $x^2 + 4 < 0$

A) Every real value of $x$ satisfies this inequality

B) No real value of $x$ satisfies this inequality — the solution set is empty

C) $-2 < x < 2$

D) $x < -2$ or $x > 2$

Show the worked solution

Answer: B

Explanation

Do not factorise yet. First ask whether this can be true at all.

For any real $x$ , a square is never negative: $x^2 \geq 0$ . So: $x^2 + 4 \;\geq\; 4$

$x^2+4$ is always at least 4. It can never be less than 0, so no real value of $x$ works. The solution set is empty.

(Algebraically you would need $x^2 < -4$ , and $\sqrt{-4}$ is not a real number — the fingerprint of "no real solutions".)

Why the others are wrong:

C and D run the standard critical-value routine without checking it applies. They quietly work with $x^2-4$ instead of $x^2+4$ , which never crosses the $x$ -axis.
A states the right fact ( $x^2\geq0$ ) but draws the opposite conclusion from it.

Takeaway: Before grinding through any inequality, spend three seconds asking: could this be true for all, some, or no real numbers? Know these on sight: $x^2 \geq 0$ always; $x^2 + (\text{positive})$ is always positive; $\sqrt{x}$ is only real for $x\geq0$ .

Section 1.5 — Exponents & Surds

Two non-negotiable habits before you do anything else: 1. Same base, always. Whenever you see different-looking numbers in an exponent question ( $4, 8, 16$ or $9, 27$ or $6, 2, 3$ ), your first move is to ask "can I rewrite all of these as powers of the same smaller base?" Almost every exponent question on the NBT collapses once you do this. 2. A fractional exponent is two instructions stacked on top of each other. $a^{m/n}$ means "take the $n$ -th root, then raise to the power $m$ " (or the other order — both give the same answer). The denominator is the root; the numerator is the power. Mixing these up is the single most common exponent error students make.

Q29Basic

Topic: Exponent laws (multiplying and dividing powers)

Simplify: $\dfrac{x^5 \cdot x^3}{x^4}$

A) $x^4$

B) $x^{12}$

C) $x^2$

D) $x^{60}$

Show the worked solution

Answer: A

Explanation

The rules: same base multiplied → add the exponents. Same base divided → subtract.

Everything is a power of $x$ , so combine in one move: $\frac{x^5 \cdot x^3}{x^4} = x^{5+3-4} = x^{4}$

Why the others are wrong:

B ( $x^{12}$ ) added all three ( $5+3+4$ ), treating the fraction bar as another multiplication. It means subtract.
C ( $x^2$ ) subtracted $5-3$ as if the dot meant divide, then ignored the $x^4$ entirely.
D ( $x^{60}$ ) multiplied the exponents. That rule is only for a power of a power, like $(x^5)^3$ .

Takeaway: Look at the symbol before you touch the exponents. $\times$ → add. $\div$ or a fraction bar → subtract. Brackets (power of a power) → multiply.

Q30Basic

Topic: Negative exponents

Write $\dfrac{1}{x^{-3}}$ using a positive exponent.

A) $x^{-3}$

B) $-x^3$

C) $x^3$

D) $\dfrac{1}{x^3}$

Show the worked solution

Answer: C

Explanation

The rule: a negative exponent does not make a value negative. It means flip — take the reciprocal. So $a^{-n} = \frac{1}{a^n}$ , and $\frac{1}{a^{-n}} = a^n$ .

$\frac{1}{x^{-3}} = x^3$

In words: "1 over (1 over $x^3$ )" is just $x^3$ . Dividing by a fraction flips it.

Why the others are wrong:

A copied the exponent across without flipping. The instruction says "positive exponent" — if yours is still negative, you have not finished.
B thinks a negative exponent gives a negative value. It does not. $x^{-3}$ and $x^3$ always have the same sign.
D flipped twice, cancelling the first flip. A and D are the same number written two different ways — both equal $\frac{1}{x^3}$ , the reciprocal of the right answer. Making the notation look right does not fix a value that is upside down.

Takeaway: A negative exponent means reciprocate, nothing more. It says nothing about the sign of the answer. Rewrite $a^{-n}$ as $\frac{1}{a^n}$ straight away, before anything else.

Q31Basic

Topic: Fractional (rational) exponents

Evaluate: $8^{2/3}$

A) $\dfrac{16}{3}$

B) $4$

C) $64$

D) $8$

Show the worked solution

Answer: B

Explanation

The rule: in $a^{m/n}$ , the denominator is the root and the numerator is the power.

So for $8^{2/3}$ : cube root first (denominator 3), then square (numerator 2). $8^{2/3} = \left(\sqrt[3]{8}\right)^{2} = 2^{2} = 4$

Root first keeps the numbers small — $\sqrt[3]{8}=2$ is instant, while $8^2=64$ then $\sqrt[3]{64}$ is harder to spot.

Why the others are wrong:

C ( $64$ ) only squared the base and never took the cube root.
A ( $\frac{16}{3}$ ) multiplied $8 \times \frac23$ instead of raising 8 to that power.
D ( $8$ ) took a square root instead of a cube root: $(\sqrt8)^2 = 8$ .

Takeaway: "Denominator = root, numerator = power." Given the choice, always do the root first.

Q32Basic

Topic: Simplifying and combining surds

Simplify: $\sqrt{12} + \sqrt{27}$

A) $\sqrt{12} + \sqrt{27}$ (cannot be simplified further)

B) $\sqrt{39}$

C) $6\sqrt{3}$

D) $5\sqrt{3}$

Show the worked solution

Answer: D

Explanation

The rule: you can only add surds once they have the same number under the root — just like fractions need a common denominator. So simplify each one first by pulling out perfect-square factors.

$\sqrt{12} = \sqrt{4\times3} = 2\sqrt{3} \qquad \sqrt{27} = \sqrt{9\times3} = 3\sqrt{3}$

Now they are like terms, so combine them the way you would $2x+3x$ : $2\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}$

Why the others are wrong:

B ( $\sqrt{39}$ ) assumes $\sqrt{a}+\sqrt{b} = \sqrt{a+b}$ and adds $12+27$ . Never true. Test it: $\sqrt4+\sqrt9 = 5$ , but $\sqrt{13} \approx 3.6$ .
C ( $6\sqrt3$ ) reached $2\sqrt3+3\sqrt3$ then multiplied the coefficients instead of adding. Same slip as writing $2x+3x=6x$ .
A calls it unsimplifiable because 12 and 27 look unrelated. Both hide a factor of $\sqrt3$ .

Takeaway: Surds combine like algebraic terms, but only after each is simplified. Always ask: is there a perfect square (4, 9, 16, 25, …) hiding inside? And remember: $\sqrt{a}+\sqrt{b} \neq \sqrt{a+b}$ , ever.

Q33Intermediate

Topic: Rationalising a binomial denominator

Rationalise the denominator, giving your answer in simplest form: $\dfrac{2}{\sqrt{5}-1}$

A) $\dfrac{2\sqrt{5}+2}{4}$

B) $\dfrac{\sqrt{5}+1}{2}$

C) $\dfrac{\sqrt{5}+1}{3}$

D) $\dfrac{2}{\sqrt{5}+1}$

Show the worked solution

Answer: B

Explanation

The rule: multiply top and bottom by the conjugate — the same expression with the middle sign flipped, $\sqrt5+1$ . This works because $(a-b)(a+b)=a^2-b^2$ kills the surd.

$\frac{2}{\sqrt{5}-1} \times \frac{\sqrt{5}+1}{\sqrt{5}+1} = \frac{2(\sqrt{5}+1)}{5-1} = \frac{2(\sqrt{5}+1)}{4}$

Top and bottom now share a factor of 2. Cancel it: $= \frac{\sqrt{5}+1}{2}$

Why the others are wrong:

A is the correct value one step early — the 2 and the 4 still cancel, and the question asks for simplest form. Rationalising is not finished while a common factor is left over.
C added the squares instead of subtracting: $5+1=6$ . Difference of squares is always a subtraction.
D multiplied only the denominator by the conjugate. You may only multiply a fraction by $\frac{\text{something}}{\text{same something}}=1$ — doing the bottom alone changes its value.

Takeaway: Three checks: (1) conjugate = same terms, flipped middle sign; (2) multiply both top and bottom; (3) cancel any common factor left over.

Q34Intermediate

Topic: Exponents with multiple bases ("same base" insight)

Simplify: $\dfrac{6^{x+1}}{2^{x} \cdot 3^{x-1}}$

A) $18$

B) $12$

C) $9$

D) $36$

Show the worked solution

Answer: A

Explanation

The insight: 6 is not a new base — it is $2\times3$ . Rewrite it and the expression splits into two same-base divisions, and the $x$ 's cancel.

$6^{x+1} = (2\cdot3)^{x+1} = 2^{x+1} \cdot 3^{x+1}$

Now group each base with its partner:

$\frac{2^{x+1} \cdot 3^{x+1}}{2^{x} \cdot 3^{x-1}} = \frac{2^{x+1}}{2^{x}} \times \frac{3^{x+1}}{3^{x-1}} = 2^{1} \times 3^{2} = 18$

Why the others are wrong:

B ( $12$ ) paired each exponent-difference with the wrong base — the 2 with base 2 and the 1 with base 3.
C ( $9$ ) cancelled the 6 against the 2 as if they were plain numbers. Powers with different bases cannot be combined that way.
D ( $36$ ) used the same exponent-difference for both bases. Each base has a different exponent below it, so each gives a different result.

Takeaway: When bases are multiples of each other ( $6=2\times3$ , $12=2^2\times3$ ), break every base into its prime powers first. The expression usually collapses to a plain number and the variable cancels.

Q35Intermediate

Topic: Equations involving surds (and checking for extraneous roots)

Solve for $x$ : $\sqrt{x+3} = x - 3$

A) $x = 1$

B) $x = 1$ or $x = 6$

C) $x = 6$

D) No real solution

Show the worked solution

Answer: C

Explanation

The golden rule of surd equations: squaring can create answers that fit the squared equation but not the original, because squaring hides sign information (both 2 and $-2$ square to 4). So every candidate must be checked in the original.

Step 1 — Square both sides. $x + 3 = x^2 - 6x + 9$

Step 2 — Rearrange and factorise. $x^2 - 7x + 6 = (x-1)(x-6) = 0 \quad\Rightarrow\quad x = 1 \text{ or } 6$

Step 3 — Check both in the original.

$x=1$ : LHS $=\sqrt4=2$ , RHS $=-2$ . $2 \neq -2$ , so reject. This is an extraneous root.
$x=6$ : LHS $=\sqrt9=3$ , RHS $=3$ ✓ accept.

So $x=6$ only.

Why the others are wrong:

B solved correctly but skipped the check — the most-tested surd mistake there is.
A checked but rejected the wrong one.
D expanded $(x-3)^2$ as $x^2-9$ instead of $x^2-6x+9$ . Keep $(a-b)^2$ and $(a-b)(a+b)$ in separate drawers.

Takeaway: Squaring is a "maybe" step, not a reversible one. Every answer it produces is only a candidate until you have watched it satisfy the original equation.

Q36Proficient

Topic: Algebraic insight with exponents (factor before you divide)

Simplify: $\dfrac{2^{n+2} - 2^{n+1}}{2^{n}}$

A) $\dfrac{1}{2}$

B) $1$

C) $2^{n+1}$

D) $2$

Show the worked solution

Answer: D

Explanation

You cannot cancel across a subtraction. But both terms in the numerator share a common factor — this is a factorisation question in an exponent costume.

Step 1 — Factor out the smaller power, $2^{n+1}$ . $2^{n+2} - 2^{n+1} = 2^{n+1}(2 - 1) = 2^{n+1}$

Step 2 — Divide by $2^n$ . $\frac{2^{n+1}}{2^{n}} = 2^{1} = 2$

The whole thing collapses to the constant 2 — the same "looks like it depends on $n$ but does not" pattern as Q34.

Why the others are wrong:

B ( $1$ ) subtracted the leading digits ( $2-1$ ) as if these were like terms. $2^{n+2}$ and $2^{n+1}$ are different powers.
C ( $2^{n+1}$ ) factorised correctly then stopped, forgetting the denominator.
A ( $\frac12$ ) did the final subtraction backwards: $2^{n-(n+1)} = 2^{-1}$ . It is always (top) − (bottom).

Takeaway: Powers of the same base joined by $+$ or $-$ ? Factor out the smaller power. The multiply/divide exponent laws only apply when the powers are joined by $\times$ or $\div$ .

Q37Proficient

Topic: Denesting nested surds (insight / structure recognition)

Simplify: $\sqrt{7 + 4\sqrt{3}}$

A) Cannot be simplified into the form $\sqrt{a} + \sqrt{b}$

B) $1 + \sqrt{6}$

C) $4 + \sqrt{3}$

D) $2 + \sqrt{3}$

Show the worked solution

Answer: D

Explanation

A nested surd like this can sometimes be rewritten as $\sqrt{a}+\sqrt{b}$ . Ask: what would $\sqrt{a}+\sqrt{b}$ have to be so that squaring it gives $7+4\sqrt3$ ?

Step 1 — Square the target form. $\left(\sqrt{a}+\sqrt{b}\right)^2 = a + b + 2\sqrt{ab}$ Match the plain parts and the surd parts separately: $a + b = 7 \qquad 2\sqrt{ab} = 4\sqrt{3}$

Step 2 — Solve for $ab$ . Divide by 2: $\sqrt{ab} = 2\sqrt3$ . Square the whole right side: $ab = (2\sqrt3)^2 = 4 \times 3 = 12$

Step 3 — Two numbers with sum 7 and product 12. That is the trinomial pattern again: $t^2 - 7t + 12 = (t-3)(t-4) \Rightarrow a,b = 3,4$ $\sqrt{7+4\sqrt3} = \sqrt4+\sqrt3 = 2+\sqrt3$

Verify by squaring back: $(2+\sqrt3)^2 = 4+4\sqrt3+3$ ✓

Why the others are wrong:

B forgot to square the coefficient: $ab = 2\times3 = 6$ instead of $4\times3=12$ .
C found $a=4$ , $b=3$ then wrote $a+\sqrt b$ instead of $\sqrt a+\sqrt b$ . Those numbers go under the roots.
A says it cannot be done. True for most nested surds, but not this one — the signal is that $7^2-4^2(3) = 1$ is a perfect square.

Takeaway: $\sqrt{p+q\sqrt r}$ denests when you can find two numbers with sum $p$ and product $\frac{q^2r}{4}$ . Same "sum and product" search as factorising a trinomial.

Section 1.6 — Logarithms

The one conversion that unlocks almost everything: $\log_a N = x \quad \Longleftrightarrow \quad a^{x} = N$ Whenever a log expression confuses you, convert it to this exponential form — the fog usually clears immediately.

The three log laws — and the one habit that protects you from inventing fake ones: $\log_a(xy) = \log_a x + \log_a y \qquad \log_a\!\left(\frac{x}{y}\right) = \log_a x - \log_a y \qquad \log_a(x^n) = n\log_a x$ Notice the pattern: a product inside the log becomes a sum outside it; a quotient becomes a difference; a power becomes a multiplying coefficient. The laws only ever convert multiplicative structure (inside) into additive structure (outside) — never the other way round, and there is no law at all for $\log(x+y)$ or $\log(x-y)$ . If you ever find yourself "simplifying" an addition or subtraction inside a logarithm, stop — that's the single most common way students invent rules that don't exist.

Q38Basic

Topic: Evaluating logarithms (logs as exponent questions)

Evaluate: $\log_2 32$

A) $4$

B) $16$

C) $5$

D) $6$

Show the worked solution

Answer: C

Explanation

Read it as the question it is asking: $\log_2 32$ means "what power of 2 gives 32?"

$2^1=2,\quad 2^2=4,\quad 2^3=8,\quad 2^4=16,\quad 2^5=32$

So $\log_2 32 = 5$ .

(Or: how many times can you halve 32 to reach 1? $32 \to 16 \to 8 \to 4 \to 2 \to 1$ — count the arrows, not the numbers. Five.)

Why the others are wrong:

A ( $4$ ) stopped the halving chain at 2 instead of going all the way to 1.
B ( $16$ ) just divided $32 \div 2$ . A log asks for an exponent, it is not an operation on the two numbers you see.
D ( $6$ ) counted the starting number as a step. Count the moves between numbers, not the numbers themselves.

Takeaway: Translate $\log_a N$ into plain English — " $a$ to what power gives $N$ ?" Then either climb the powers of $a$ , or divide $N$ by $a$ until you reach 1. Either way, count the moves, not the numbers.

Q39Basic

Topic: Evaluating logarithms of fractions (negative results)

Evaluate: $\log_5 \dfrac{1}{25}$

A) $-2$

B) $2$

C) $-\dfrac{1}{2}$

D) $\dfrac{1}{2}$

Show the worked solution

Answer: A

Explanation

Translate first: what power of 5 gives $\frac{1}{25}$ ?

A reciprocal means a negative exponent: $\frac{1}{25} = \frac{1}{5^2} = 5^{-2}$ . So: $\log_5\frac{1}{25} = -2$

Worth remembering: if the number inside a log is a fraction below 1, the answer is negative.

Why the others are wrong:

B ( $2$ ) answered $\log_5 25$ and ignored the fraction. A positive power of 5 can never give a number below 1.
D ( $\frac12$ ) assumes a reciprocal inside gives a reciprocal answer. No such rule exists.
C ( $-\frac12$ ) is that same error with the correct minus sign added — half right, still wrong.

Takeaway: Log of a fraction below 1 is always negative. The link between the number and its log is "base and exponent", never "reciprocal and reciprocal".

Q40Intermediate

Topic: Applying the quotient law

Simplify: $\log_2 40 - \log_2 5$

A) $\log_2 200$

B) $3$

C) $8$

D) $\log_2 35$

Show the worked solution

Answer: B

Explanation

Two logs, same base, joined by subtraction — that is the quotient law: $\log_2 40 - \log_2 5 = \log_2\left(\frac{40}{5}\right) = \log_2 8$

Now evaluate it: what power of 2 gives 8? $2^3=8$ , so the answer is 3.

Why the others are wrong:

D ( $\log_2 35$ ) subtracted the numbers inside ( $40-5$ ). There is no law for adding or subtracting inside a log.
C ( $8$ ) merged correctly to $\log_2 8$ then stopped, giving the number inside instead of the log's value.
A ( $\log_2 200$ ) merged with the wrong operation. Subtraction of logs means division inside; addition means multiplication.

Takeaway: Same base joined by $+$ or $-$ ? Merge them — plus makes a product, minus makes a quotient — then always evaluate the single log you are left with.

Q41Intermediate

Topic: Solving logarithmic equations

Solve for $x$ : $\log_2(x+3) = 4$

A) $x = 1$

B) $x = 5$

C) $x = 13$

D) $x = 19$

Show the worked solution

Answer: C

Explanation

Step 1 — Convert to exponential form. $\log_a(N)=k$ means $a^k = N$ . $\log_2(x+3) = 4 \quad\Longleftrightarrow\quad x + 3 = 2^{4} = 16$

Step 2 — Isolate $x$ . $x = 16 - 3 = 13$

Check: $\log_2 16 = 4$ ✓

Why the others are wrong:

B ( $5$ ) wrote $x+3 = 2\times4$ . The base is raised to the power, not multiplied by it.
A ( $1$ ) deleted the log entirely and solved $x+3=4$ . The base is not decoration.
D ( $19$ ) reached $x+3=16$ then wrote $x = 16+3$ . The 3 flips sign when it crosses over.

Takeaway: Every log equation is two steps: (1) rewrite $\log_a(\ldots)=k$ as $(\ldots)=a^k$ ; (2) solve it with ordinary algebra. The log notation only matters for step 1.

Q42Intermediate

Topic: Logarithms with related bases (the "same base" insight again)

Evaluate: $\log_4 8$

A) $2$

B) $\dfrac{3}{2}$

C) $\dfrac{2}{3}$

D) $\log_4 8$ cannot be written as a simple fraction — it is irrational

Show the worked solution

Answer: B

Explanation

4 and 8 are both powers of 2 ( $4=2^2$ , $8=2^3$ ) — the same common-base trick as Q34, now inside a log.

Let $\log_4 8 = y$ and convert to exponential form: $4^{y} = 8 \quad\Rightarrow\quad 2^{2y} = 2^{3}$

Same base, so the exponents match: $2y = 3 \quad\Rightarrow\quad y = \frac{3}{2}$

Sanity check: $4^1 = 4 < 8 < 16 = 4^2$ , so the answer must sit between 1 and 2 ✓

Why the others are wrong:

A ( $2$ ) guessed from "4 goes into 8 twice" — that is multiplying the base, not raising it to a power.
C ( $\frac23$ ) set up $2y=3$ correctly then flipped the fraction.
D assumes no nice answer exists because 8 is not a whole-number power of 4. Rewriting both as powers of 2 reveals one.

Takeaway: Odd-looking bases are usually powers of the same small prime. Before giving up on a log, ask: can I write both numbers as powers of 2 (or 3, or 5)?

Q43Proficient

Topic: Expressing logarithms in terms of given values (log-law fluency)

If $\log_a 2 = p$ and $\log_a 3 = q$ , express $\log_a 12$ in terms of $p$ and $q$ .

A) $2pq$

B) $p^2 + q$

C) $4p + q$

D) $2p + q$

Show the worked solution

Answer: D

Explanation

Break 12 into the pieces you were given. $12 = 2^2 \times 3$ . Then apply the product law, then the power law:

$\log_a 12 = \log_a(2^{2} \times 3) = \log_a(2^{2}) + \log_a 3 = 2\log_a 2 + \log_a 3$

Substitute $p$ and $q$ : $= 2p + q$

Why the others are wrong:

B ( $p^2+q$ ) squared the log value. The power law brings the exponent out as a coefficient, giving $2p$ , not $p^2$ .
A ( $2pq$ ) multiplied the logs together. A product inside a log becomes a sum outside — logs are never multiplied.
C ( $4p+q$ ) used $2^2=4$ as the coefficient. It is the exponent (2) that comes out front, not the value of the power.

Takeaway: Say each law aloud as you use it: a product inside becomes a sum outside; a power inside becomes a coefficient outside.

Q44Proficient

Topic: Logical reasoning about logarithm laws ("always true")

For all $x > 0$ and $y > 0$ , which of the following statements is always true?

A) $\log(xy) = \log x + \log y$

B) $(\log x)(\log y) = \log(x + y)$

C) $\log(x-y) = \log x - \log y$

D) $\log(x+y) = \log x + \log y$

Show the worked solution

Answer: A

Explanation

The fastest way to test an "always true" claim: pick easy numbers and check both sides. One counterexample kills a statement forever.

Test each with $x=10$ , $y=10$ (since $\log 10 = 1$ exactly):

A: LHS $=\log 100 = 2$ . RHS $=1+1=2$ ✓ survives — this is the genuine product law.
B: LHS $=(1)(1)=1$ . RHS $=\log 20 \approx 1.30$ ✗
C: LHS $=\log(0)$ , which is undefined ✗
D: LHS $=\log 20 \approx 1.30$ . RHS $=2$ ✗

Only A survives.

Why the others are wrong:

C and D are the fake laws: they assume $+$ or $-$ can pass through a log. Logs only turn multiplication into addition, never the other way.
B multiplies two logs together. No log law does that.

Takeaway: For "always true" questions, do not try to prove it under time pressure — test it with easy numbers ( $\log 10 = 1$ , $\log 100 = 2$ ). One counterexample is enough to eliminate an option.

Section 1.7 — Sequences & Series

The one test that tells them apart:

Arithmetic (AP): consecutive terms differ by a constant amount — $T_{n+1} - T_n = d$ (the common difference) is the same every time. Growth is additive: $5, 8, 11, 14,\dots$
Geometric (GP): consecutive terms are related by a constant multiplier — $\dfrac{T_{n+1}}{T_n} = r$ (the common ratio) is the same every time. Growth is multiplicative: $5, 10, 20, 40, \dots$

The four formulas worth knowing cold: $\text{AP: } T_n = a + (n-1)d \qquad S_n = \frac{n}{2}\big[2a + (n-1)d\big]$ $\text{GP: } T_n = a \cdot r^{\,n-1} \qquad S_n = \frac{a(r^{\,n}-1)}{r-1} \quad (r\neq1)$ And sigma notation $\displaystyle\sum_{k=1}^{n} (\cdots)$ is nothing more than shorthand for "add up the expression in the brackets, substituting $k = 1$ , then $k=2$ , ..., all the way up to $k = n$ ." It looks formal, but it's just a compact instruction to write out a sum term by term.

Q45Basic

Topic: $n$ th term of an arithmetic sequence

Find the 10th term of the arithmetic sequence $5, 8, 11, 14, \ldots$

A) $27$

B) $35$

C) $30$

D) $32$

Show the worked solution

Answer: D

Explanation

Step 1 — Identify $a$ and $d$ . First term $a=5$ ; each term is 3 more, so $d=3$ .

Step 2 — Use $T_n = a + (n-1)d$ . Read it as a sentence: start at $a$ , then take $(n-1)$ steps of size $d$ . There is one fewer step than there are terms, because the first term needs no steps.

$T_{10} = 5 + (9)(3) = 32$

Why the others are wrong:

B ( $35$ ) used $a + nd$ — ten steps instead of nine. The most common AP slip.
C ( $30$ ) computed $10\times3$ and forgot the first term entirely.
A ( $27$ ) found the amount added but stopped, never adding $a$ back on.

Takeaway: Two checkpoints: is the multiplier $(n-1)$ and not $n$ ? And did you add $a$ back at the end?

Q46Basic

Topic: $n$ th term of a geometric sequence

Find the 6th term of the geometric sequence $3, 6, 12, 24, \ldots$

A) $96$

B) $192$

C) $48$

D) $36$

Show the worked solution

Answer: A

Explanation

Step 1 — Identify $a$ and $r$ . First term $a=3$ ; each term doubles, so $r=2$ .

Step 2 — Use $T_n = a\cdot r^{\,n-1}$ — the GP twin of the AP formula, still counting $(n-1)$ steps.

$T_6 = 3 \times 2^{5} = 96$

Or walk it: $3\to6\to12\to24\to48\to96$ — five multiplications ✓

Why the others are wrong:

B ( $192$ ) used $r^n$ instead of $r^{n-1}$ — one power too many. The same off-by-one as the AP.
C ( $48$ ) used $r^{n-2}$ — one power too few.
D ( $36$ ) computed $a\times r\times n$ , treating a GP as if it grew by adding. A GP multiplies repeatedly, which is what the exponent captures.

Takeaway: Ask first: does this sequence grow by adding the same amount, or multiplying by the same amount? Getting that wrong means using an entirely wrong kind of formula.

Q47Basic

Topic: Identifying the common ratio of a geometric sequence

What is the common ratio of the geometric sequence $81, 27, 9, 3, \ldots$ ?

A) $-54$

B) $3$

C) $\dfrac{1}{3}$

D) $\dfrac{1}{27}$

Show the worked solution

Answer: C

Explanation

Always the same method: divide a term by the one directly before it, $r = \frac{T_{n+1}}{T_n}$ .

$r = \frac{27}{81} = \frac{1}{3}$

Check a second pair: $\frac{9}{27} = \frac13$ ✓ and $\frac{3}{9} = \frac13$ ✓

It makes sense that $r<1$ — the sequence is shrinking.

Why the others are wrong:

B ( $3$ ) divided the wrong way round, $\frac{81}{27}$ . Always later ÷ earlier, or the ratio comes out upside down.
A ( $-54$ ) computed $27-81$ , a common difference. The word "ratio" tells you to divide.
D ( $\frac{1}{27}$ ) compared non-neighbouring terms, $\frac{3}{81}$ . That gives $r^3$ , not $r$ .

Takeaway: Divide a term by its immediate predecessor, then confirm the same value on a second pair. "Common" means it must be constant.

Q48Intermediate

Topic: Sum of an arithmetic series

Find the sum of the first 20 terms of the arithmetic sequence $4, 7, 10, \ldots$

A) $610$

B) $650$

C) $680$

D) $1300$

Show the worked solution

Answer: B

Explanation

$a=4$ , $d=3$ , $n=20$ . Into $S_n = \frac{n}{2}[2a + (n-1)d]$ :

$S_{20} = \frac{20}{2}\big[8 + 57\big] = 10(65) = 650$

Why the others are wrong:

D ( $1300$ ) is exactly double — the $\frac{n}{2}$ was dropped. A wrong answer that is a clean multiple of the right one usually means a lost factor.
C ( $680$ ) found the last term as $4 + 20(3) = 64$ (the off-by-one from Q45), then used $\frac{20}{2}(4+64)$ . One early slip poisons everything after it.
A ( $610$ ) used a single $a$ instead of $2a$ in the bracket.

Takeaway: Know why each piece is there. $S_n = \frac{n}{2}[2a+(n-1)d]$ is just "(number of terms) × (average of first and last term)". The $2a$ is what builds "first + last" before you know the last term — understanding that makes it hard to drop.

Q49Intermediate

Topic: Sum of a finite geometric series

Find the sum of the first 5 terms of the geometric sequence $2, 6, 18, \ldots$

A) $242$

B) $243$

C) $14$

D) $50$

Show the worked solution

Answer: A

Explanation

Confirm it is geometric: $a=2$ and each term is 3 times the last, so $r=3$ . Into $S_n = \frac{a(r^n-1)}{r-1}$ :

$S_5 = \frac{2(243-1)}{2} = 242$

Why the others are wrong:

B ( $243$ ) dropped the " $-1$ " in the numerator. It misses by exactly one, so glancing at the size of your answer will never catch it — you must check the working.
C ( $14$ ) computed $r^n$ as $3\times5=15$ instead of $3^5$ . Same multiply-versus-power confusion as Q46.
D ( $50$ ) treated it as arithmetic with $d=4$ , built $2,6,10,14,18$ and added. Checking a second ratio kills that story: $\frac{6}{2}=3$ but $\frac{10}{6}\neq3$ .

Takeaway: Three checks on every GP sum: confirm the ratio on two pairs; compute $r^n$ as a power, never $r\times n$ ; and check the working, not just whether the answer "looks about right".

Q50Intermediate

Topic: Sigma (summation) notation

Evaluate: $\displaystyle\sum_{k=1}^{5}(2k+1)$

A) $30$

B) $24$

C) $31$

D) $35$

Show the worked solution

Answer: D

Explanation

Sigma is an instruction, not a mystery symbol: substitute $k=1,2,3,4,5$ into $2k+1$ and add the results.

$3 + 5 + 7 + 9 + 11 = 35$

Why the others are wrong:

A ( $30$ ) summed only the $2k$ part and dropped the " $+1$ " from every term.
C ( $31$ ) added the " $+1$ " once at the end: $2(15)+1$ . It belongs to each term, so it must be added five times. Split properly: $2\sum k + \sum 1 = 30 + 5 = 35$ .
B ( $24$ ) used only four terms, subtracting the limits ( $5-1$ ). Both limits are included: the count is $(\text{top}-\text{bottom})+1 = 5$ .

Takeaway: Write the terms out — do not do it in your head, that is where errors hide. Apply every constant per term, and count terms with $(\text{top}-\text{bottom})+1$ .

Q51Intermediate

Topic: Solving for $n$ in a series sum (forming and solving a quadratic)

How many terms of the arithmetic sequence $3, 7, 11, \ldots$ must be added together for the sum to equal $210$ ?

A) $n = 11$

B) $n = 10$

C) $n = 10$ or $n = -10.5$

D) No whole number of terms gives a sum of $210$

Show the worked solution

Answer: B

Explanation

$a=3$ , $d=4$ . Set the sum equal to 210:

$S_n = \frac{n}{2}\big[6 + 4(n-1)\big] = n(2n+1) = 2n^2+n$ $2n^2+n-210 = 0 \quad\Rightarrow\quad (2n+21)(n-10)=0$

So $n=10$ or $n=-10.5$ .

The step that decides the question: $n$ is a number of terms, so it must be a positive whole number. A sequence cannot have $-10.5$ terms. Reject it.

$n = 10$

Check: $S_{10} = 5(42) = 210$ ✓

Why the others are wrong:

C did the algebra perfectly and reported both roots, never asking whether each could be a real number of terms.
A ( $11$ ) factorised with a sign slip, got $10.5$ , then rounded. A decimal answer for "number of terms" means go back and find the lost sign, not round.
D dropped the minus on $c$ in the discriminant: $-4(2)(210)$ instead of $-4(2)(-210)$ , turning $+1681$ into $-1679$ and making real answers seem to vanish.

Takeaway: When a variable must be a positive whole number — terms, people, years — solving is only half the job. Then ask of each answer: could this actually happen? Discarding the impossible root is often the entire point.

Q52Proficient

Topic: Sum to infinity of a geometric series

Find the sum to infinity of the geometric series $8 + 4 + 2 + 1 + \ldots$

A) $\dfrac{1}{16}$

B) No finite sum exists — the series goes on forever

C) $16$

D) $4$

Show the worked solution

Answer: C

Explanation

The idea: infinitely many terms can still add to a finite total, as long as the terms shrink fast enough. Here $a=8$ and $r=\frac12$ . Since $|r|<1$ , the series converges:

$S_\infty = \frac{a}{1-r} = \frac{8}{\frac{1}{2}} = 16$

Watch the running total: $8, 12, 14, 15, 15.5, 15.75, \ldots$ — climbing by ever smaller amounts, closing in on 16 without passing it.

Why the others are wrong:

B assumes infinitely many terms must total infinity. False when $|r|<1$ . "Goes on forever" describes the number of terms, not the size of the total.
D ( $4$ ) computed $a(1-r)$ instead of $\frac{a}{1-r}$ — the formula assembled upside down.
A ( $\frac{1}{16}$ ) had the right pieces but inverted the fraction: $\frac{1-r}{a}$ instead of $\frac{a}{1-r}$ .

Takeaway: Check $|r|<1$ before using the formula — otherwise no finite sum exists. And remember: "infinitely many things" and "an infinite total" are different statements.

Q53Proficient

Topic: Forming and solving simultaneous equations from sequence terms (insight)

The 3rd term of an arithmetic sequence is $11$ , and the 7th term is $23$ . Find the first term $a$ and the common difference $d$ .

A) $a = 3,\ d = 4$

B) $a = 11,\ d = 3$

C) $a = 5,\ d = 3$

D) $a = 17,\ d = 3$

Show the worked solution

Answer: C

Explanation

Two facts about the sequence are exactly enough to find both unknowns. Translate each into an equation using $T_n = a+(n-1)d$ .

Step 1 — Write both equations. $T_3 = 11: \quad a + 2d = 11 \qquad T_7 = 23: \quad a + 6d = 23$

Step 2 — Subtract to eliminate $a$ . $4d = 12 \quad\Rightarrow\quad d = 3$

Step 3 — Substitute back. $a + 6 = 11 \quad\Rightarrow\quad a = 5$

Check: $T_3 = 11$ ✓ and $T_7 = 23$ ✓

Why the others are wrong:

B treated the given 11 as the first term. "The 3rd term is 11" is a statement about $T_3$ ; it does not hand you $a$ .
A counted the terms between $T_3$ and $T_7$ (three) instead of the steps (four: $T_3\to T_4\to T_5\to T_6\to T_7$ ), giving $3d=12$ .
D found $d=3$ correctly then wrote $a = 11 + 6$ instead of $11 - 6$ .

Takeaway: "Find $a$ and $d$ from two facts" is a simultaneous-equations problem in disguise. Translate each fact with $T_n = a+(n-1)d$ , subtract to eliminate one unknown, solve. The sequence wording changes nothing about the algebra.

Section 1.8 — Financial Mathematics

The crucial distinction — and a secret connection to Section 1.7: $\text{Simple interest (linear growth): } A = P(1 + in)$ $\text{Compound interest (multiplicative growth): } A = P(1+i)^{n}$ Look closely at the compound formula — $P, P(1+i), P(1+i)^2, P(1+i)^3, \ldots$ — and you'll recognise it instantly: it's a geometric sequence, with first term $P$ and common ratio $(1+i)$ . Everything you learned about GPs in Section 1.7 — including the "off-by-one in the exponent" trap — applies directly here. Simple interest, by contrast, adds the same fixed amount every year — it's an arithmetic sequence in disguise.

Depreciation is just "negative growth": $\text{Straight-line (simple) depreciation: } A = P(1 - in)$ $\text{Reducing-balance (compound) depreciation: } A = P(1-i)^{n}$ Same two formula families, same GP/AP structure underneath — only the sign in the bracket flips, because the value is shrinking rather than growing. The single biggest skill this section tests is matching the right formula family to the right scenario description: "compounded annually" / "reducing balance" → exponential; "simple interest" / "straight-line" → linear.

Q54Basic

Topic: Setting up a simple interest expression

Tumi invests R4 000 at a simple interest rate of $6\%$ per annum. Which expression correctly represents the value of her investment after 5 years?

A) $4\,000(1.06)^{5}$

B) $4\,000(1 + 0.06 \times 5)$

C) $4\,000 \times 0.06 \times 5$

D) $4\,000(1+0.06)\times 5$

Show the worked solution

Answer: B

Explanation

"Simple interest" means the linear formula $A = P(1+in)$ . Here $P=4\,000$ , $i=0.06$ , $n=5$ :

$A = 4\,000\big(1 + 0.06 \times 5\big)$

The same fixed interest, $4\,000 \times 0.06 = 240$ , is added every year. That is what "simple" growth means.

Why the others are wrong:

A uses the compound formula on a simple-interest scenario. The word "simple" tells you growth is additive, not exponential.
C gives only the interest earned, not the value of the investment. The question asks for the total, principal included.
D applies one year's growth factor then multiplies by 5. That is neither model — simple interest adds a fixed amount; compound multiplies repeatedly.

Takeaway: First decision in any money question: is the growth linear (simple, straight-line) or multiplicative (compound, reducing balance)? Those words are in the question to tell you. Get the word right and the rest is substitution.

Q55Basic

Topic: Setting up a compound interest expression (the GP connection)

Sipho invests R8 000 at a compound interest rate of $9\%$ per annum, compounded annually. Which expression represents the value of his investment after 4 years?

A) $8\,000(0.09)^{4}$

B) $8\,000(1 + 0.09 \times 4)$

C) $8\,000 \times 1.09 \times 4$

D) $8\,000(1.09)^{4}$

Show the worked solution

Answer: D

Explanation

"Compounded annually" means exponential growth. Each year the whole current balance grows by 9%, so it becomes 109% of itself — multiply by the growth factor $1.09$ . Four years means four multiplications:

$A = 8\,000(1.09)^{4}$

This is just a GP with $r = 1.09$ .

Why the others are wrong:

B uses the simple formula on a compound scenario — the mirror of Q54's trap.
A uses the bare rate $0.09$ as the base instead of $1.09$ . Multiplying by $0.09$ repeatedly would shrink the money to almost nothing. The factor must mean "all you had, plus 9%" = $1.09$ .
C multiplied by 4 instead of raising to the power 4. Compound growth multiplies four times in a row; it does not add up four single years.

Takeaway: Compound interest is a GP. Build the factor as $1+i$ for growth (or $1-i$ for shrinking) — never use the bare rate on its own.

Q56Intermediate

Topic: Setting up a reducing-balance depreciation expression (the mirror of Q55)

A delivery van bought for R240 000 depreciates at a rate of $15\%$ per annum on the reducing-balance method. Which expression gives its value after 3 years?

A) $240\,000(1-0.15)^{3}$

B) $240\,000(1 - 0.15 \times 3)$

C) $240\,000(0.15)^{3}$

D) $240\,000(1+0.15)^{3}$

Show the worked solution

Answer: A

Explanation

"Reducing balance" is the depreciation twin of "compounded annually." The van loses 15% of its current value each year, so it keeps 85% — a factor of $1-0.15 = 0.85$ , applied three times:

$A = 240\,000(1-0.15)^{3} = 240\,000(0.85)^3$

Same shape as Q55, with the sign flipped because the value is shrinking.

Why the others are wrong:

B uses the straight-line formula $P(1-in)$ , which takes a fixed slice of the original value. "Reducing balance" takes a slice of the current value.
C uses the bare rate: $(0.15)^3$ would mean keeping only 15% of the value each year — a far more severe collapse than 15% depreciation.
D uses a growth factor on a shrinking quantity. "Depreciates" means the value falls, so any expression bigger than R240 000 must be wrong.

Takeaway: Start from 100% of the current value, then add the rate if it grows or subtract it if it shrinks. "Compounded/appreciates" → $+$ . "Depreciates/reducing balance/decays" → $-$ . Then sanity-check the direction.

Q57Intermediate

Topic: Working backwards to find the interest rate

An investment of R10 000 grows to R12 100 after 2 years under annual compound interest. Find the annual interest rate.

A) $11\%$

B) $21\%$

C) $10.5\%$

D) $10\%$

Show the worked solution

Answer: D

Explanation

Set up the equation and undo it one layer at a time.

$10\,000(1+i)^2 = 12\,100 \quad\Rightarrow\quad (1+i)^2 = 1.21$

Take the square root. $1.21 = \left(\frac{11}{10}\right)^2$ , so:

$1 + i = 1.1 \quad\Rightarrow\quad i = 0.1 = 10\%$

Check: $10\,000(1.1)^2 = 12\,100$ ✓

Why the others are wrong:

B ( $21\%$ ) is the total growth over both years, reported as if it were annual. Compounding means 10% a year gives 21% over two years, not 20% — the second year grows an already-grown balance.
C ( $10.5\%$ ) took that 21% and halved it, treating the growth as additive instead of multiplicative.
A ( $11\%$ ) found the growth factor $1.1$ and read it off as the rate. The factor is $1+i$ , so you must still subtract 1.

Takeaway: Two things to hold onto: total growth is not (annual rate) × (years) for compound interest; and the growth factor $(1+i)$ and the rate $(i)$ are one subtraction apart.

Q58Intermediate

Topic: Rearranging the compound interest formula to find the principal

An investment grows from a principal $P$ to an amount of R16 000 after 4 years, at a compound interest rate of $i$ per annum, compounded annually. Which expression correctly represents $P$ in terms of $i$ ?

A) $P = \dfrac{16\,000}{1 + 4i}$

B) $P = 16\,000(1+i)^{4}$

C) $P = \dfrac{16\,000}{(1+i)^{4}}$

D) $P = 16\,000 - 4i$

Show the worked solution

Answer: C

Explanation

Start from $A = P(1+i)^n$ with $A = 16\,000$ and $n=4$ :

$16\,000 = P(1+i)^{4}$

Divide both sides by $(1+i)^4$ to isolate $P$ :

$P = \frac{16\,000}{(1+i)^{4}}$

Sense-check: $P$ is the starting amount, so it must be smaller than the R16 000 it grew into. Dividing shrinks it — correct.

Why the others are wrong:

A rearranged the simple interest formula. The question says compounded annually.
B multiplies instead of dividing, making $P$ bigger than $A$ . Impossible — $P$ grew into $A$ .
D subtracts a small flat amount, throwing away the exponential structure entirely.

Takeaway: Undo the operations in reverse order, then sense-check the size: should this quantity be bigger or smaller than the others in the story?

Q59Proficient

Topic: Reasoning about depreciation and inflation together (insight)

A machine valued at $V$ depreciates on the reducing-balance method at $r\%$ per annum (where $0 < r < 100$ ). At the same time, the cost of a brand-new equivalent machine increases (inflates) at the same rate, $r\%$ per annum. After $n$ years, the old machine's depreciated value is $V(1-r)^n$ and the new machine's cost is $V(1+r)^n$ (writing $r$ here as a decimal). Which statement about the ratio $\dfrac{\text{old machine's value}}{\text{new machine's cost}}$ must be true?

A) The ratio equals $\left(\dfrac{1-r}{1+r}\right)^{n}$ , and it becomes smaller with each passing year

B) The two effects cancel exactly: the ratio always equals $1$ , no matter what $n$ is

C) The ratio equals $(1-r^2)^{n}$ , formed by multiplying the two growth factors together

D) The ratio equals $\left(\dfrac{1+r}{1-r}\right)^{n}$ , and it becomes larger with each passing year

Show the worked solution

Answer: A

Explanation

It sounds like losing $r\%$ and gaining $r\%$ should cancel. They do not. Divide the two expressions out and look at what you get.

Step 1 — Form the ratio. $\frac{V(1-r)^{n}}{V(1+r)^{n}} = \left(\frac{1-r}{1+r}\right)^{n}$ The $V$ 's cancel, and a quotient of like powers becomes one power of the quotient.

Step 2 — Look at the base. Since $0<r<1$ , we have $1-r < 1+r$ , so the base $\frac{1-r}{1+r}$ is less than 1. A number below 1 raised to higher powers gets smaller. So the ratio shrinks every year.

Why the others are wrong:

B claims they cancel to 1. Test $r=0.5$ : the ratio is $\frac{0.5}{1.5} = \frac13$ , nowhere near 1.
C multiplied the two factors (giving the difference of squares $1-r^2$ ). A ratio comes from dividing.
D has the right shape but upside down — and its claim that the ratio grows would mean an old machine gains value relative to a new one. That contradicts the story.

Takeaway: Never guess whether two exponential effects cancel — divide them and look at the single base you get. Base below 1 → shrinks toward zero. Base above 1 → grows without bound.

Section 1.9 — Algebraic Insight, Vieta's Formulas & Manipulation Tricks

This section has no new procedures. It rewards you for seeing the structure instead of grinding through the algebra.

The master tool: Vieta's formulas. For a quadratic $ax^2+bx+c=0$ with roots $\alpha$ and $\beta$ : $\alpha + \beta = -\frac{b}{a} \qquad\qquad \alpha\beta = \frac{c}{a}$ These say something remarkable: you can know the sum and product of a quadratic's roots without ever finding the roots themselves — just by reading off its coefficients. Combined with the identity you met in the sample question at the start of this chapter, $a^2+b^2 = (a+b)^2 - 2ab,$ Vieta's formulas let you compute all sorts of expressions built from the roots — $\alpha^2+ \beta^2$ , $\dfrac{1}{\alpha}+\dfrac{1}{\beta}$ , $(\alpha+1)(\beta+1)$ , and more — directly from the coefficients, without the slow, error-prone detour of solving the equation first.

The recurring false belief worth naming, one last time: several sections in this chapter have each, in their own way, warned you against the same imaginary rule — that an operation "distributes" across addition just because it feels like it should: $\sqrt{a}+\sqrt{b} \neq \sqrt{a+b}$ (Q32), $\log(x+y)\neq\log x + \log y$ (Q44), and — as you're about to see — $\dfrac1 {\alpha}+\dfrac1\beta \neq \dfrac{1}{\alpha+\beta}$ too. Recognising this one recurring pattern protects you across an enormous range of NBT questions.

Q60Intermediate

Topic: Vieta's formulas — sum of roots from coefficients

If $\alpha$ and $\beta$ are the roots of $2x^2 - 7x + 3 = 0$ , what is the value of $\alpha + \beta$ ?

A) $7$

B) $\dfrac{7}{2}$

C) $\dfrac{3}{2}$

D) $-\dfrac{7}{2}$

Show the worked solution

Answer: B

Explanation

Vieta's sum formula: for $ax^2+bx+c=0$ , $\ \alpha+\beta = -\frac{b}{a}$ .

Read off $a=2$ , $b=-7$ , and substitute — keeping the minus sign that is already in the formula:

$\alpha + \beta = -\frac{(-7)}{2} = \frac{7}{2}$

Check without finding the roots separately: $2x^2-7x+3 = (2x-1)(x-3)$ , so the roots are $\frac12$ and $3$ , summing to $\frac72$ ✓ — but Vieta got there in one line.

Why the others are wrong:

D ( $-\frac72$ ) dropped the minus already built into the formula, computing $\frac{b}{a}$ . With $b$ also negative, the two minuses make a positive.
C ( $\frac32$ ) used the other Vieta formula, $\frac{c}{a}$ — the product instead of the sum.
A ( $7$ ) forgot to divide by $a$ . That slip is invisible when $a=1$ , which is why it pays to practise on equations where $a \neq 1$ .

Takeaway: Learn the pair together: sum $= -\frac{b}{a}$ , product $= \frac{c}{a}$ — and both divide by $a$ . Saying that sentence as you use them catches all three slips above.

Q61Intermediate

Topic: Building new expressions from a given identity (squaring trick)

If $x - \dfrac{1}{x} = 3$ , what is the value of $x^2 + \dfrac{1}{x^2}$ ?

A) $9$

B) $11$

C) $7$

D) $5$

Show the worked solution

Answer: B

Explanation

You are given something in $x$ and $\frac1x$ , and asked for $x^2$ and $\frac1{x^2}$ . Squaring is what turns one into the other.

Step 1 — Square both sides, expanding fully. $\left(x - \frac{1}{x}\right)^{2} = x^{2} - 2\left(x \cdot \frac{1}{x}\right) + \frac{1}{x^{2}} = x^{2} - 2 + \frac{1}{x^{2}}$

The cross term is beautifully simple: $x \cdot \frac1x = 1$ . That is exactly why squaring works here.

Step 2 — Substitute and solve. $x^{2} - 2 + \frac{1}{x^{2}} = 9 \quad\Rightarrow\quad x^{2}+\frac{1}{x^{2}} = 11$

Why the others are wrong:

A ( $9$ ) dropped the cross term. $(a-b)^2$ is never $a^2+b^2$ — it is $a^2-2ab+b^2$ .
C ( $7$ ) used the wrong sign on the cross term, expanding as $(a+b)^2$ .
D ( $5$ ) forgot to square the right-hand side. Squaring an equation means squaring both sides.

Takeaway: Given a paired expression ( $x$ and $\frac1x$ , or $\sqrt x$ ) and asked for the squares? Square the given equation. The cross term almost always collapses to something simple that links the two.

Q62Proficient

Topic: Constructing a new equation from transformed roots

If $\alpha$ and $\beta$ are the roots of $x^{2} - 5x + 6 = 0$ , form a quadratic equation (with integer coefficients) whose roots are $\alpha + 1$ and $\beta + 1$ .

A) $x^{2} - 7x + 12 = 0$

B) $x^{2} - 5x + 6 = 0$

C) $x^{2} - 3x + 2 = 0$

D) $x^{2} - 7x + 7 = 0$

Show the worked solution

Answer: A

Explanation

You do not need $\alpha$ and $\beta$ separately. Build the new equation from the new sum and product.

Step 1 — Old sum and product (Vieta). $\alpha+\beta = 5 \qquad \alpha\beta = 6$

Step 2 — New sum and product, expanding carefully. $\text{Sum: } (\alpha+1)+(\beta+1) = 5+2 = 7$ $\text{Product: } (\alpha+1)(\beta+1) = \alpha\beta + \alpha + \beta + 1 = 6+5+1 = 12$

Step 3 — Rebuild: $x^2 - (\text{sum})x + (\text{product}) = 0$ $x^{2} - 7x + 12 = 0$

Check the slow way: the old roots are 2 and 3, so the new ones are 3 and 4, and $(x-3)(x-4) = x^2-7x+12$ ✓

Why the others are wrong:

B copies the original equation. Moving the roots moves where the parabola crosses the axis, so the equation must change.
C shifted the roots the wrong way, giving $\alpha-1$ and $\beta-1$ .
D got the sum right but expanded the product as $\alpha\beta+1$ , dropping the cross terms. All four products count when you expand two brackets — the same slip as Q61's option A.

Takeaway: "New roots shifted by a constant" is a Vieta question in disguise. Find the new sum and product, then rebuild with $x^2-(\text{sum})x+(\text{product})=0$ .

Q63Proficient

Topic: Symmetric identities with three variables (extending a known trick)

If $a + b + c = 6$ , $ab+bc+ca = 11$ , and $abc = 6$ , find the value of $a^{2}+b^{2}+c^{2}$ .

A) $58$

B) $36$

C) $25$

D) $14$

Show the worked solution

Answer: D

Explanation

The two-variable identity has a three-variable twin. Recall $a^2+b^2 = (a+b)^2-2ab$ . Expanding $(a+b+c)^2$ gives every square plus every pair, doubled:

$(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$ $\Rightarrow\quad a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca)$

Substitute. Note $abc=6$ is not in this identity — it is a deliberate distraction.

$a^2+b^2+c^2 = 36 - 2(11) = 14$

Why the others are wrong:

B ( $36$ ) gave $(a+b+c)^2$ alone, forgetting to subtract the pairwise term.
A ( $58$ ) added instead of subtracting. The expansion has a $+$ , but rearranging to isolate the squares needs a $-$ .
C ( $25$ ) dropped the factor of 2. Expanding produces each pair twice ( $ab$ and $ba$ ), so both copies must be removed.

Takeaway: Given symmetric combinations (sums, pairwise products) and asked for something that looks like it needs the individual values — look for an identity that connects them directly. You rarely need to find $a$ , $b$ and $c$ themselves.

Q64Proficient

Topic: Logical deduction about the roots of a quadratic ("must be true")

For the equation $x^{2} + bx + c = 0$ , where $b$ and $c$ are real numbers, which statement must be true?

A) If $b = 0$ , the equation has no real roots

B) If $b^{2} - 4c > 0$ , the equation has two real roots of the same sign

C) If $c < 0$ , the equation has two real roots of opposite sign

D) If $c > 0$ , the equation has two real roots of the same sign

Show the worked solution

Answer: C

Explanation

Every option quotes real theory, but only one draws a conclusion its premise guarantees. The others are true only sometimes.

Why C is true. By Vieta, the product of the roots is $\alpha\beta = c$ .

Suppose the roots were not real. Then they form a conjugate pair $p+qi$ and $p-qi$ , whose product is $(p+qi)(p-qi) = p^2+q^2$ a sum of squares — which can never be negative. So non-real roots always have a positive product.

Turn that around: if the product is negative, the roots must be real. And two real numbers with a negative product must have opposite signs. Since $c<0$ gives $\alpha\beta<0$ , option C is forced.

Why the others are wrong — one counterexample each:

B: $x^2-x-6=0$ has $b^2-4c = 25 > 0$ , but its roots are $3$ and $-2$ — opposite signs. A positive discriminant says the roots are real and distinct, nothing about their signs.
D: $x^2+x+1=0$ has $c=1>0$ , but $b^2-4c=-3<0$ — no real roots at all.
A: $x^2-4=0$ has $b=0$ and two real roots, $x=\pm2$ . Removing the linear term says nothing about whether real roots exist.

Takeaway: "Feels plausible" is not evidence. Either prove the claim from a tool you trust, or find a counterexample with the simplest numbers you can. One counterexample settles it.

Q65Proficient

Topic: Vieta's formulas with reciprocal expressions (chapter capstone)

Without solving the equation $3x^{2} - 8x + 5 = 0$ for $x$ , find the value of $\dfrac{1}{\alpha} + \dfrac{1}{\beta}$ , where $\alpha$ and $\beta$ are its roots.

A) $\dfrac{8}{5}$

B) $\dfrac{5}{8}$

C) $\dfrac{8}{3}$

D) $\dfrac{3}{8}$

Show the worked solution

Answer: A

Explanation

The insight: $\frac{1}{\alpha}+\frac{1}{\beta}$ is secretly built from the sum and product of the roots — the two things Vieta gives you free. Combine over a common denominator:

$\frac{1}{\alpha}+\frac{1}{\beta} = \frac{\beta}{\alpha\beta} + \frac{\alpha}{\alpha\beta} = \frac{\alpha+\beta}{\alpha\beta}$

No lone $\alpha$ or $\beta$ survives — which is why you can answer without solving.

Read off sum and product ( $a=3$ , $b=-8$ , $c=5$ ): $\alpha+\beta = \frac{8}{3} \qquad \alpha\beta = \frac{5}{3}$

Substitute: $\frac{\alpha+\beta}{\alpha\beta} = \frac{8/3}{5/3} = \frac{8}{5}$

Why the others are wrong:

B ( $\frac58$ ) built the fraction upside down. The sum goes on top — that is what falls out of the common-denominator step.
C ( $\frac83$ ) assumed $\frac1\alpha+\frac1\beta$ is just $\alpha+\beta$ .
D ( $\frac38$ ) assumed it is $\frac{1}{\alpha+\beta}$ . Both C and D collapse two different quantities into one.

Takeaway — and the closing lesson of this chapter: operations almost never "pass through" addition the way it feels like they should.

$\sqrt{a+b} \neq \sqrt a + \sqrt b \qquad \log(x+y) \neq \log x + \log y$ $\frac{1}{\alpha+\beta} \neq \frac1\alpha+\frac1\beta \qquad (a+b)^2 \neq a^2+b^2$

Every one of these appeared in this chapter in a different disguise, and every one is beaten the same way: write it out fully and let the real identity — usually with an extra cross term — appear. Feel suspicious every time you want to distribute an operation across a sum.

Exam-Bank Extras — Question Types Confirmed in Recent Papers

The NBT MAT draws on a stable question bank, and test-writers consistently report seeing the same question types — often near-identical questions — year after year. The six questions below are modelled directly on types repeatedly confirmed in recent papers that were not yet represented in this chapter. Every answer has been independently machine-verified.

A note on Q71: the remainder theorem sits slightly outside the current CAPS core (which is why the rest of this book works with the factor theorem only), but it has appeared on real NBT papers — so one authentic example is included here, taught from scratch.

Q66Basic

Topic: Sum of the solutions of a quadratic (the ± trap)

What is the sum of the solutions of $x^{2} = 4$ ?

A) $2$

B) $0$

C) $4$

D) $-2$

Show the worked solution

Answer: B

Explanation

Square-rooting both sides of $x^2 = 4$ gives two solutions:

$x = +2 \quad \text{or} \quad x = -2$

The question asks for their sum: $(+2) + (-2) = 0$ .

Why A is wrong: $2$ is what you get if you forget the negative root entirely — the single most common slip with this question type. $(-2) \times (-2) = 4$ just as surely as $2 \times 2 = 4$ . Why D is wrong: $-2$ is the mirror-image slip — the negative root alone. Why C is wrong: $4$ echoes the right-hand side of the equation — no root was ever taken.

Takeaway: $x^2 = k$ (with $k > 0$ ) always has two solutions, $x = \pm\sqrt{k}$ . Whenever a question says "solutions" (plural) of a squared equation, your radar should ping: both roots are in play. This exact question is famous for defeating careless readers — and at least one AI chatbot.

Why the others are wrong: A and D each take one root and stop — $x^2 = 4$ has two. C finds both but adds $2 + 2$ , ignoring that one of them is negative. Two roots that are negatives of each other always sum to zero.

Q67Intermediate

Topic: Custom-defined operations

For real numbers $a$ and $b$ , the operation $\diamond$ is defined by

$a \diamond b = a^{2} - 3b$

Find the value of $(2 \diamond 1) \diamond 3$ .

A) $10$

B) $-2$

C) $-8$

D) $8$

Show the worked solution

Answer: C

Explanation

Work from the inside out. The rule says: square the first number, subtract three times the second.

Inner bracket first: $2 \diamond 1 = 2^{2} - 3(1) = 4 - 3 = 1$

Then feed that result in as the new first number: $1 \diamond 3 = 1^{2} - 3(3) = 1 - 9 = -8$

Why D is wrong: $8$ comes from dropping the minus sign in the final subtraction. Why A is wrong: $10$ comes from computing $2 \diamond (1 \diamond 3)$ — grouping from the right instead of respecting the brackets. Why B is wrong: $-2$ drops the coefficient in " $3b$ " during the second application: $1^2 - 3 = -2$ instead of $1^2 - 3(3) = -8$ .

Takeaway: A "made-up operation" question is pure substitution: the symbol $\diamond$ is just a costume. Put the first value wherever the rule shows $a$ , the second wherever it shows $b$ — and always evaluate the bracketed operation first. These questions test careful reading, not new mathematics.

Why the others are wrong: A adds $3b$ where the definition subtracts it. B subtracts 3 instead of $3b$ . D loses the minus sign at the last step: $1 - 9$ is $-8$ , not $8$ .

Q68Proficient

Topic: Sum and product of two numbers — building $x^2 + y^2$

Two numbers have a sum of $m$ and a product of $n$ . Which expression gives the sum of the squares of the two numbers?

A) $m^{2} + 2n$

B) $m^{2} - 2n$

C) $m^{2} - n$

D) $m - 2n$

Show the worked solution

Answer: B

Explanation

Call the numbers $x$ and $y$ , so $x + y = m$ and $xy = n$ . We want $x^2 + y^2$ .

Square the sum — this is the move that unlocks the whole question: $(x+y)^2 = x^2 + 2xy + y^2 = m^2$

The expansion contains exactly what we want ( $x^2 + y^2$ ) plus an intruder ( $2xy = 2n$ ). Evict the intruder:

$x^2 + y^2 = (x+y)^2 - 2xy = m^{2} - 2n$

Why A is wrong: a sign slip when moving $2xy$ across. Why C is wrong: forgets that the cross term in the expansion is $2xy$ , not $xy$ . Why D is wrong: $m - 2n$ leaves $m$ unsquared — the identity starts from $(x+y)^2 = m^2$ , and that square must survive into the answer.

Takeaway: $x^2 + y^2 = (x+y)^2 - 2xy$ is one of the most reused identities in the entire exam bank — it converts "sum and product" information into "sum of squares" in one line. It is Vieta's world again (compare Q65): symmetric expressions in two numbers can always be rebuilt from their sum and product.

Why the others are wrong: A adds where the expansion subtracts: $(a+b)^2 = a^2 + 2ab + b^2$ , so the $2ab$ must come OFF. C subtracts but drops the factor of 2. D never squares the sum at all.

Q69Proficient

Topic: Factor theorem with two conditions (simultaneous equations)

Both $(x+1)$ and $(x-2)$ are factors of

$f(x) = 2x^{4} - kx^{2} + bx - 4$

Find the value of $k$ .

A) $4$

B) $-6$

C) $2$

D) $-4$

Show the worked solution

Answer: A

Explanation

The factor theorem, twice. Each factor gives one equation, and two unknowns need exactly two.

From $(x+1)$ — substitute $x=-1$ : $2 - k - b - 4 = 0 \implies k + b = -2 \qquad (1)$

From $(x-2)$ — substitute $x=2$ : $32 - 4k + 2b - 4 = 0 \implies 2k - b = 14 \qquad (2)$

Add them so $b$ cancels: $\ 3k = 12 \implies k = 4$

Why the others are wrong:

B ( $-6$ ) is the value of $b$ , not $k$ .
D evaluated $-k(-1)^2$ as $+k$ — a sign error in equation (1).
C ( $k=2$ ) fails the check: it forces $b=-4$ , and then $2k-b = 8 \neq 14$ .

Takeaway: Each known factor gives one free equation via $f(c)=0$ . Two unknowns need two factors — then add or subtract to eliminate the one you do not want.

Q70Proficient

Topic: Exponential equation with two different prime bases

The integers $x$ and $y$ satisfy

$2^{x+2} + 2^{x} = 5^{y+1} - 5^{y}$

Find the value of $x - y$ .

A) $0$

B) $3$

C) $1$

D) $2$

Show the worked solution

Answer: C

Explanation

Logs will not help here. The trick is to factor out the common power on each side.

Left: $\ 2^{x+2} + 2^{x} = 2^{x}(4+1) = 5 \cdot 2^{x}$

Right: $\ 5^{y+1} - 5^{y} = 5^{y}(5-1) = 4 \cdot 5^{y}$

So $5 \cdot 2^x = 4 \cdot 5^y$ , which rearranges to: $2^{x-2} = 5^{y-1}$

Now the prime bases decide it. A power of 2 can equal a power of 5 only if both equal 1 — so both exponents are zero: $x = 2, \qquad y = 1, \qquad x - y = 1$

Check: $16+4 = 20$ and $25-5 = 20$ ✓

Why the others are wrong:

D ( $2$ ) gives $x$ alone — the subtraction never happened.
B ( $3$ ) gives $x+y$ , the wrong combination.
A ( $0$ ) assumes $x=y$ by symmetry, but the constants 5 and 4 differ, which is exactly what offsets the exponents.

Takeaway: Mixed prime bases? Factor each side into (number) × (prime power). The only way $2^m = 5^n$ is $m = n = 0$ — "the bases are prime" is the key, not decoration.

Q71Proficient

Topic: The remainder theorem with two conditions (simultaneous equations)

When $f(x) = 2x^{4} - kx^{2} + bx$ is divided by $(x+1)$ , the remainder is $-3$ . When $f(x)$ is divided by $(x-1)$ , the remainder is $5$ . Find the value of $k$ .

A) $-1$

B) $4$

C) $2$

D) $1$

Show the worked solution

Answer: D

Explanation

This uses the remainder theorem — the factor theorem's close cousin:

Dividing $f(x)$ by $(x-c)$ leaves remainder $f(c)$ .

The factor theorem is just the case where that remainder is zero. So the two facts become $f(-1) = -3$ and $f(1) = 5$ .

Substitute $x=-1$ : $\ 2 - k - b = -3 \qquad (1)$

Substitute $x=1$ : $\ 2 - k + b = 5 \qquad (2)$

Add them so $b$ cancels: $\ 4 - 2k = 2 \implies k = 1$

Why the others are wrong:

B ( $4$ ) is the value of $b$ , not $k$ .
A evaluated $-k(-1)^2$ as $+k$ . The square makes $(-1)^2 = +1$ , so the term stays $-k$ .
C ( $2$ ) is $2k$ — stopped one division early.

Takeaway: Zero is not special. Substitute the root of the divisor and set the result equal to the stated remainder — each division fact becomes one equation. Same skeleton as Q69, where the remainders happened to be zero.

Mixed Practice — Chapter 1

These 30 questions draw from all nine topics in this chapter. There are no section labels and no difficulty tags — just like the real NBT MAT. Work through them in one sitting, without going back to the chapter. Aim for no more than 3 minutes per question. The answer key follows after all 30 questions.

M1Basic

Factorise fully: $6x^2 - x - 2$

A) $(2x+1)(3x-2)$

B) $(2x-1)(3x+2)$

C) $(6x+1)(x-2)$

D) $(3x+1)(2x-2)$

Show the worked solution

Answer: A

Explanation

Step 1 — Look for two numbers multiplying to $6 \times (-2) = -12$ and adding to $-1$ : they are $3$ and $-4$ .

Step 2 — Split the middle term: $6x^2 + 3x - 4x - 2$ .

Step 3 — Group: $3x(2x+1) - 2(2x+1) = (2x+1)(3x-2)$ .

Why the others are wrong:

B signs swapped — this expands to $6x^2+x-2$ , middle term $+x$ .
C the outer pair only — this expands to $6x^2-11x-2$ .
D not fully factorised: $(2x-2)$ still has a common factor of 2, and it expands to $6x^2-4x-2$ .

Takeaway: With a leading coefficient, split the middle term rather than guessing brackets. Always expand your answer back mentally to check the middle term.

M2Intermediate

Factorise fully: $x^3 - 4x^2 - x + 4$

A) $(x+1)(x-1)(x+4)$

B) $(x^2+4)(x-1)$

C) $(x-2)(x+2)(x-4)$

D) $(x-1)(x+1)(x-4)$

Show the worked solution

Answer: D

Explanation

Step 1 — Group in pairs: $x^2(x-4) - 1(x-4)$ .

Step 2 — Take out the common bracket: $(x-4)(x^2-1)$ .

Step 3 — $x^2-1$ is a difference of squares: $(x-1)(x+1)(x-4)$ .

Why the others are wrong:

A sign error on the last factor — gives $(x+4)$ instead of $(x-4)$ .
B treats $x^2-1$ as $x^2+4$ ; it also leaves a quadratic unfactorised.
C factorises $x^2-4$ rather than $x^2-1$ .

Takeaway: Four terms with no common factor is the signal to group in pairs. Then check whether either bracket factorises further — $x^2-1$ almost always does.

M3Basic

Solve for $x$ : $\;3(2x-1) = 2(x+5) + 3$

A) $x = 2$

B) $x = 3$

C) $x = 4$

D) $x = 5$

Show the worked solution

Answer: C

Explanation

Step 1 — Expand both sides: $6x - 3 = 2x + 10 + 3$ .

Step 2 — Tidy the right: $6x - 3 = 2x + 13$ .

Step 3 — Collect: $4x = 16$ , so $x = 4$ .

Why the others are wrong:

A collected the constants as $16-3$ instead of $16+3$ ... i.e. dropped a sign when moving $-3$ across.
B expanded $2(x+5)$ as $2x+5$ , losing a 5.
D divided 16 by 4 incorrectly, or expanded $3(2x-1)$ as $6x-1$ .

Takeaway: Expand fully before collecting anything. Most errors in a linear equation happen in the brackets, not the algebra that follows.

M4Intermediate

Solve: $2x^2 + 3x - 9 = 0$

A) $x = 3$ or $x = -\dfrac{3}{2}$

B) $x = \dfrac{3}{2}$ or $x = -3$

C) $x = -3$ or $x = \dfrac{2}{3}$

D) $x = 3$ or $x = -\dfrac{9}{2}$

Show the worked solution

Answer: B

Explanation

Step 1 — Factorise: $2x^2+3x-9 = (2x-3)(x+3)$ .

Step 2 — Set each factor to zero: $2x-3=0$ or $x+3=0$ .

Step 3 — So $x = \dfrac{3}{2}$ or $x = -3$ .

Why the others are wrong:

A roots of the right factors but with both signs flipped.
C read the factors as $(x+3)(3x-2)$ — the 2 and 3 swapped in the fractional root.
D divided the constant $-9$ by 2 instead of factorising.

Takeaway: Check a quadratic's roots by adding them: they must total $-\dfrac{b}{a} = -\dfrac{3}{2}$ . Here $\dfrac{3}{2}+(-3) = -\dfrac{3}{2}$ ✓

M5Intermediate

Given $p(x) = 2x^3 - 3x^2 - 11x + 6$ , which of the following is a factor of $p(x)$ ?

A) $(x-2)$

B) $(x+3)$

C) $(x-3)$

D) $(x+4)$

Show the worked solution

Answer: C

Explanation

Step 1 — The factor theorem: $(x-c)$ is a factor exactly when $p(c) = 0$ .

Step 2 — Test $x=3$ : $2(27) - 3(9) - 11(3) + 6 = 54 - 27 - 33 + 6 = 0$ ✓

Step 3 — So $(x-3)$ is a factor.

Why the others are wrong:

A $p(2) = 16-12-22+6 = -12 \neq 0$ .
B $p(-3) = -54-27+33+6 = -42 \neq 0$ .
D $p(-4) = -128-48+44+6 = -126 \neq 0$ .

Takeaway: For $(x-c)$ substitute $x=c$ ; for $(x+c)$ substitute $x=-c$ . Getting that sign backwards is the single most common factor-theorem error.

M6Intermediate

Solve: $x^2 + x - 12 \leq 0$

A) $x < -4$ or $x > 3$

B) $-4 < x < 3$

C) $x \leq -4$ or $x \geq 3$

D) $-4 \leq x \leq 3$

Show the worked solution

Answer: D

Explanation

Step 1 — Factorise: $(x+4)(x-3) \leq 0$ .

Step 2 — The critical values are $x=-4$ and $x=3$ .

Step 3 — A parabola opening upwards is below the axis BETWEEN its roots, so $-4 \leq x \leq 3$ .

Why the others are wrong:

A the roots are right but the inequality is reversed — this is where the parabola is positive.
B signs of both roots flipped.
C strict inequalities, but $\leq$ includes the roots themselves.

Takeaway: Sketch the parabola. $\leq 0$ means on or below the $x$ -axis, which for an upward parabola is the interval between the roots, endpoints included.

M7Basic

Simplify: $\dfrac{3^{n+2} - 3^n}{2 \cdot 3^n}$

A) $1$

B) $4$

C) $8$

D) $\dfrac{1}{4}$

Show the worked solution

Answer: B

Explanation

Step 1 — Take out the common factor $3^n$ on top: $3^n(3^2 - 1) = 3^n \cdot 8$ .

Step 2 — The fraction becomes $\dfrac{8 \cdot 3^n}{2 \cdot 3^n}$ .

Step 3 — $3^n$ cancels: $\dfrac{8}{2} = 4$ .

Why the others are wrong:

A cancelled $3^n$ but forgot to divide the 8 by the 2.
C treated $3^{n+2}$ as $3^n + 3^2$ rather than $3^n \times 3^2$ .
D subtracted the exponents instead of factorising.

Takeaway: $3^{n+2}$ means $3^n \times 9$ , never $3^n + 9$ . Factor out the smallest power and the variable exponent disappears.

M8Intermediate

Solve for $x$ : $\;\sqrt{2x+1} = x - 1$

A) $x = 0$ only

B) $x = 0$ or $x = 4$

C) $x = 4$ only

D) No real solution

Show the worked solution

Answer: C

Explanation

Step 1 — Square both sides: $2x + 1 = (x-1)^2 = x^2 - 2x + 1$ .

Step 2 — Rearrange: $x^2 - 4x = 0$ , so $x(x-4)=0$ and $x = 0$ or $x = 4$ .

Check both. $x=0$ : LHS $=\sqrt{1}=1$ , RHS $=-1$ . It fails. $x=4$ : LHS $=\sqrt{9}=3$ , RHS $=3$ ✓

Step 3 — Only $x = 4$ .

Why the others are wrong:

A the extraneous root — squaring introduced it and the check rejects it.
B keeps both roots; $x=0$ does not satisfy the original equation.
D arithmetic slip when expanding $(x-1)^2$ .

Takeaway: Squaring both sides can create roots the original equation never had. After squaring you have not finished until you have substituted every root back.

M9Basic

Evaluate: $\log_5 125$

A) $2$

B) $5$

C) $25$

D) $3$

Show the worked solution

Answer: D

Explanation

Step 1 — $\log_5 125$ asks: 5 to what power gives 125?

Step 2 — $5^3 = 125$ .

Step 3 — So the answer is 3.

Why the others are wrong:

A read it as $125 \div 5$ divided again — not what a logarithm does.
B $5^2 = 25$ , not 125.
C $5^4 = 625$ , too big.

Takeaway: Read every logarithm as a question about a power: $\log_b a = c$ means $b^c = a$ . Say it out loud that way and most log questions become arithmetic.

M10Intermediate

Solve: $\log_2(x+3) + \log_2(x-1) = 5$

A) $x = 5$ or $x = -7$

B) $x = -7$ only

C) $x = 5$ only

D) $x = 7$

Show the worked solution

Answer: C

Explanation

Step 1 — Combine the logs: $\log_2[(x+3)(x-1)] = 5$ .

Step 2 — Undo the log: $(x+3)(x-1) = 2^5 = 32$ .

Step 3 — Expand: $x^2 + 2x - 35 = 0$ , so $(x+7)(x-5)=0$ and $x = 5$ or $x = -7$ .

Check the domain. $x=-7$ makes $x-1$ negative, and you cannot take the log of a negative number. Only $x = 5$ .

Why the others are wrong:

A the rejected root — it solves the quadratic but not the original equation.
B keeps both roots without testing the domain.
D added the logs' arguments instead of multiplying them.

Takeaway: $\log A + \log B = \log(AB)$ — multiply, never add. And every log equation needs a domain check at the end: each argument must be strictly positive.

M11Basic

The 5th term of an arithmetic sequence is $17$ and the common difference is $3$ . What is the first term?

A) $a = 5$

B) $a = 2$

C) $a = 8$

D) $a = 3$

Show the worked solution

Answer: A

Explanation

Step 1 — The $n$ th term of an arithmetic sequence is $T_n = a + (n-1)d$ .

Step 2 — Here $T_5 = a + 4d = 17$ and $d = 3$ .

Step 3 — So $a + 12 = 17$ , giving $a = 5$ .

Why the others are wrong:

B used $a + 5d$ — that is $T_6$ , not $T_5$ .
C added the common difference instead of subtracting it.
D solved $a = 17 + 12$ .

Takeaway: $T_5$ uses $4d$ , not $5d$ : you take four steps to get from the 1st term to the 5th. Off-by-one here is the most common arithmetic-sequence mistake.

M12Intermediate

The first three terms of a geometric sequence are $x+3$ , $x$ , and $x-2$ . Find the value of $x$ .

A) $x = 4$

B) $x = 6$

C) $x = 8$

D) $x = 3$

Show the worked solution

Answer: B

Explanation

Step 1 — In a geometric sequence consecutive ratios are equal: $\dfrac{x}{x+3} = \dfrac{x-2}{x}$ .

Step 2 — Cross-multiply: $x^2 = (x+3)(x-2) = x^2 + x - 6$ .

Step 3 — The $x^2$ cancels: $0 = x - 6$ , so $x = 6$ .

Why the others are wrong:

A treated the sequence as arithmetic — equal differences rather than equal ratios.
C cross-multiplied but expanded $(x+3)(x-2)$ as $x^2-6$ .
D sign error on the constant term.

Takeaway: Geometric means equal RATIOS. Setting $\dfrac{T_2}{T_1} = \dfrac{T_3}{T_2}$ works for any three consecutive terms and turns the question into algebra.

M13Intermediate

R8 000 is invested at 6% per annum, compounded monthly. Which expression gives the value of the investment after 2 years?

A) $8000(1.06)^{24}$

B) $8000(1.005)^{2}$

C) $8000(1.06)^{2}$

D) $8000(1.005)^{24}$

Show the worked solution

Answer: D

Explanation

Step 1 — Compounded monthly means the rate is divided by 12 and the number of periods is multiplied by 12.

Step 2 — Monthly rate: $\dfrac{0.06}{12} = 0.005$ , so the factor is $1.005$ .

Step 3 — Periods: $2 \times 12 = 24$ .

Step 4 — $A = 8000(1.005)^{24}$ .

Why the others are wrong:

A used the annual rate with the monthly number of periods — compounding 12 times a year at 6% each month.
B divided the rate but forgot to multiply the periods.
C ignored the monthly compounding entirely.

Takeaway: Divide the rate and multiply the periods, always together. Changing one without the other is the classic compound-interest error.

M14Proficient

The roots of $3x^2 - 7x + k = 0$ are $\alpha$ and $\beta$ . If $\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{7}{4}$ , find $k$ .

A) $k = 3$

B) $k = 7$

C) $k = 2$

D) $k = 4$

Show the worked solution

Answer: D

Explanation

Step 1 — For $3x^2-7x+k=0$ : $\alpha+\beta = \dfrac{7}{3}$ and $\alpha\beta = \dfrac{k}{3}$ .

Step 2 — $\dfrac{1}{\alpha}+\dfrac{1}{\beta} = \dfrac{\alpha+\beta}{\alpha\beta} = \dfrac{7/3}{k/3} = \dfrac{7}{k}$ .

Step 3 — Set $\dfrac{7}{k} = \dfrac{7}{4}$ , so $k = 4$ .

Why the others are wrong:

A used $\alpha\beta = k$ , forgetting to divide by the leading coefficient 3.
B read $k$ off the middle coefficient.
C inverted the sum-over-product relationship.

Takeaway: $\dfrac{1}{\alpha}+\dfrac{1}{\beta}$ is always $\dfrac{\alpha+\beta}{\alpha\beta}$ — sum over product. Learn that one rearrangement and this whole family of questions collapses.

M15Intermediate

Given that $(x + 3)$ is a factor of $p(x) = x^3 + 2x^2 - 5x + a$ , find the value of $a$ .

A) $a = -6$

B) $a = 6$

C) $a = -30$

D) $a = 0$

Show the worked solution

Answer: A

Explanation

Step 1 — $(x+3)$ is a factor, so by the factor theorem $p(-3) = 0$ .

Step 2 — $p(-3) = (-27) + 2(9) - 5(-3) + a = -27 + 18 + 15 + a = 6 + a$ .

Step 3 — So $6 + a = 0$ and $a = -6$ .

Why the others are wrong:

B substituted $x = 3$ instead of $x = -3$ .
C sign slip on $-5(-3)$ , treating it as $-15$ .
D solved $a = 6$ without moving it across.

Takeaway: $(x+3)$ a factor means substitute $x = -3$ . Write the substitution down before evaluating — the sign is where this goes wrong.

M16Intermediate

Express $2x^2 - 8x + 3$ in the form $a(x - p)^2 + q$ .

A) $2(x-4)^2 - 29$

B) $2(x-2)^2 - 5$

C) $(x-2)^2 - 5$

D) $2(x-2)^2 + 3$

Show the worked solution

Answer: B

Explanation

Step 1 — Take the 2 out of the $x$ terms: $2(x^2 - 4x) + 3$ .

Step 2 — Complete the square inside: $x^2-4x = (x-2)^2 - 4$ .

Step 3 — So $2[(x-2)^2 - 4] + 3 = 2(x-2)^2 - 8 + 3 = 2(x-2)^2 - 5$ .

Why the others are wrong:

A forgot to multiply the $-4$ by the 2 outside.
C halved $-8$ instead of $-4$ when completing the square.
D left the 2 inside the bracket.

Takeaway: The number you subtract inside the bracket gets multiplied by the coefficient outside. Expand your answer back to check you land on the original.

M17Intermediate

Evaluate: $\displaystyle\sum_{k=1}^{5}(3k - 1)$

A) $35$

B) $45$

C) $40$

D) $30$

Show the worked solution

Answer: C

Explanation

Step 1 — Write the terms: $k=1,\dots,5$ gives $2, 5, 8, 11, 14$ .

Step 2 — Add them: $2+5+8+11+14 = 40$ .

Step 3 — (Or: $3\sum k - \sum 1 = 3(15) - 5 = 40$ .)

Why the others are wrong:

A summed $3k$ and forgot to subtract the five $-1$ s.
B subtracted 1 once rather than once per term.
D used $k=0$ to $5$ , adding an extra term.

Takeaway: $\sum(3k-1) = 3\sum k - \sum 1$ , and $\sum 1$ over five terms is 5, not 1. For five terms, writing them out is faster than any formula.

M18Intermediate

A car is purchased for R120 000 and depreciates at 15% per annum on the reducing-balance method. Which expression gives its value after 4 years?

A) $120\,000(0.85)^4$

B) $120\,000 - 4(0.15)(120\,000)$

C) $120\,000(1.15)^4$

D) $120\,000 \times (0.85) \times 4$

Show the worked solution

Answer: A

Explanation

Step 1 — Reducing balance means each year you keep $100\% - 15\% = 85\%$ of the value.

Step 2 — That is a factor of $0.85$ per year, applied 4 times.

Step 3 — $V = 120\,000(0.85)^4$ .

Why the others are wrong:

B straight-line depreciation — the same rand amount each year, which is not what reducing balance means.
C used a growth factor; the car is losing value, not gaining it.
D applied one year's depreciation and then multiplied by 4.

Takeaway: Reducing balance is a POWER, straight line is a MULTIPLE. Depreciating at $r$ gives a factor of $(1-r)$ raised to the number of years.

M19Intermediate

Factorise: $8x^3 - 27$

A) $(2x-3)(4x^2 - 6x + 9)$

B) $(2x-3)(4x^2 + 6x + 9)$

C) $(2x+3)(4x^2 - 6x + 9)$

D) $(2x-3)(2x^2 + 6x + 9)$

Show the worked solution

Answer: B

Explanation

Step 1 — This is a difference of cubes: $(2x)^3 - 3^3$ .

Step 2 — The identity is $a^3-b^3 = (a-b)(a^2+ab+b^2)$ with $a=2x$ , $b=3$ .

Step 3 — $(2x-3)(4x^2 + 6x + 9)$ .

Why the others are wrong:

A used the SUM-of-cubes signs.
C middle term of the trinomial has the wrong sign — that pattern belongs to $a^3+b^3$ .
D squared the bracket instead of using the identity.

Takeaway: $a^3-b^3=(a-b)(a^2+ab+b^2)$ . The bracket matches the original sign; the trinomial's middle term is always the OPPOSITE sign.

M20Proficient

If $x = \sqrt{5 + 2\sqrt{6}}$ , which of the following is equal to $x$ ?

A) $\sqrt{3} + \sqrt{2}$

B) $\sqrt{5} + \sqrt{6}$

C) $2 + \sqrt{3}$

D) $1 + \sqrt{6}$

Show the worked solution

Answer: A

Explanation

Step 1 — Look for $(\sqrt{a}+\sqrt{b})^2 = a + b + 2\sqrt{ab}$ .

Step 2 — Matching $5 + 2\sqrt{6}$ : need $a+b = 5$ and $ab = 6$ , so $a=3$ , $b=2$ .

Step 3 — So $5+2\sqrt6 = (\sqrt3+\sqrt2)^2$ and the square root is $\sqrt3+\sqrt2$ .

Why the others are wrong:

B squares to $11 + 2\sqrt{30}$ , not $5 + 2\sqrt6$ .
C squares to $7 + 4\sqrt3$ .
D squares to $7 + 2\sqrt6$ — right surd, wrong constant.

Takeaway: To un-nest $\sqrt{p + 2\sqrt{q}}$ , find two numbers adding to $p$ and multiplying to $q$ . Then square your answer to confirm.

M21Intermediate

Solve: $2^{x+1} = 5$ . Give the exact answer.

A) $x = \dfrac{\log 5}{\log 2}$

B) $x = \dfrac{\log 5}{\log 2} + 1$

C) $x = \dfrac{\log 5}{\log 2} - 1$

D) $x = \log 5 - \log 2$

Show the worked solution

Answer: C

Explanation

Step 1 — Take logs of both sides: $(x+1)\log 2 = \log 5$ .

Step 2 — So $x + 1 = \dfrac{\log 5}{\log 2}$ .

Step 3 — Therefore $x = \dfrac{\log 5}{\log 2} - 1$ .

Why the others are wrong:

A solved $2^x = 5$ — forgot the $+1$ in the exponent.
B moved the $+1$ to the wrong side. From $x + 1 = \dfrac{\log 5}{\log 2}$ the 1 must be subtracted, not added.
D $\dfrac{\log 5}{\log 2}$ is a quotient, not a difference; $\log 5 - \log 2 = \log 2.5$ .

Takeaway: $\dfrac{\log a}{\log b}$ and $\log a - \log b$ are completely different. The change-of-base rule uses division.

M22Intermediate

The sum of the first $n$ terms of an arithmetic series is $S_n = 2n^2 + 3n$ . What is the 5th term of the series?

A) $23$

B) $17$

C) $19$

D) $21$

Show the worked solution

Answer: D

Explanation

Step 1 — $T_n = S_n - S_{n-1}$ .

Step 2 — $S_5 = 2(25)+3(5) = 65$ and $S_4 = 2(16)+3(4) = 44$ .

Step 3 — $T_5 = 65 - 44 = 21$ .

Why the others are wrong:

A substituted $n=5$ into $S_n$ — that is the SUM of five terms, not the 5th term.
B used $S_5 - S_3$ .
C differentiated the formula instead of differencing it.

Takeaway: A term is the difference of two consecutive sums: $T_n = S_n - S_{n-1}$ . $S_5$ on its own is the total of the first five terms.

M23Basic

Solve: $2(3 - x) < 4x - 10$

A) $x > \dfrac{8}{3}$

B) $x < \dfrac{8}{3}$

C) $x > -\dfrac{8}{3}$

D) $x < 8$

Show the worked solution

Answer: A

Explanation

Step 1 — Expand: $6 - 2x < 4x - 10$ .

Step 2 — Collect: $6 + 10 < 4x + 2x$ , so $16 < 6x$ .

Step 3 — Divide by 6: $x > \dfrac{8}{3}$ .

Why the others are wrong:

B inequality reversed — dividing by a POSITIVE 6 does not flip it.
C did not simplify $\dfrac{16}{6}$ .
D expanded $2(3-x)$ as $6-x$ .

Takeaway: Only multiplying or dividing by a NEGATIVE flips an inequality. Here you divide by $+6$ , so the sign stays as it is.

M24Intermediate

For the geometric series $5 + 5x + 5x^2 + \cdots$ to have a finite sum to infinity, which condition on $x$ is required?

A) $x < 1$

B) $-1 < x < 1$

C) $|x| > 1$

D) $0 < x < 1$

Show the worked solution

Answer: B

Explanation

Step 1 — A geometric series converges exactly when $|r| < 1$ .

Step 2 — Here $r = x$ .

Step 3 — So the condition is $-1 < x < 1$ .

Why the others are wrong:

A allows $x \leq -1$ , where the series diverges.
C the condition for DIVERGENCE.
D excludes the valid negative ratios and $x=0$ .

Takeaway: $|r|<1$ means $-1 < r < 1$ — negative ratios converge too, they just alternate. Do not quietly assume $r$ is positive.

M25Intermediate

Solve: $4^x - 6 \cdot 2^x + 8 = 0$

A) $x = 1$ or $x = 4$

B) $x = 2$ or $x = 8$

C) $x = 1$ or $x = 2$

D) $x = 2$ or $x = 4$

Show the worked solution

Answer: C

Explanation

Step 1 — Note $4^x = (2^2)^x = (2^x)^2$ . Let $y = 2^x$ .

Step 2 — The equation becomes $y^2 - 6y + 8 = 0$ , so $(y-2)(y-4)=0$ .

Step 3 — $y = 2 \Rightarrow 2^x = 2 \Rightarrow x=1$ ; $y = 4 \Rightarrow 2^x = 4 \Rightarrow x=2$ .

Why the others are wrong:

A stopped at the $y$ values and reported them as $x$ .
B found only one root.
D treated $4^x$ as $4 \cdot x$ .

Takeaway: $4^x = (2^x)^2$ turns this into a quadratic. After solving for $y$ , you must still solve $2^x = y$ — the $y$ values are not the answer.

M26Intermediate

$(x - 2)$ is a factor of $p(x) = x^3 + kx^2 - x - 2$ . Find $k$ .

A) $k = -1$

B) $k = 1$

C) $k = 2$

D) $k = -2$

Show the worked solution

Answer: A

Explanation

Step 1 — $(x-2)$ is a factor, so $p(2) = 0$ .

Step 2 — $p(2) = 8 + 4k - 2 - 2 = 4k + 4$ .

Step 3 — $4k + 4 = 0$ gives $k = -1$ .

Why the others are wrong:

B substituted $x=-2$ .
C dropped the constant $-2$ when evaluating.
D solved $4k = 4$ .

Takeaway: $(x-2)$ a factor means substitute $x = +2$ . Evaluate every term, including the constant, before solving.

M27Intermediate

Simplify: $\log_4 9 \cdot \log_3 8$

A) $6$

B) $2$

C) $\dfrac{3}{2}$

D) $3$

Show the worked solution

Answer: D

Explanation

Step 1 — Change to a common base: $\log_4 9 = \dfrac{\log 9}{\log 4}$ and $\log_3 8 = \dfrac{\log 8}{\log 3}$ .

Step 2 — Write in terms of $\log 2$ and $\log 3$ : $\dfrac{2\log 3}{2\log 2} \cdot \dfrac{3\log 2}{\log 3}$ .

Step 3 — Everything cancels except 3.

Why the others are wrong:

A multiplied the 2 and 3 from the powers without cancelling.
B cancelled one factor too many.
C kept the 2 in the denominator of the first fraction.

Takeaway: Write 9 as $3^2$ , 8 as $2^3$ and 4 as $2^2$ first. Products of logs like this almost always collapse to a whole number.

M28Intermediate

In a quadratic sequence, $T_1 = 3$ , $T_2 = 10$ , and the second difference is $6$ . Find $T_3$ .

A) $21$

B) $23$

C) $19$

D) $25$

Show the worked solution

Answer: B

Explanation

Step 1 — The first difference is $T_2 - T_1 = 10 - 3 = 7$ .

Step 2 — The second difference is constant at 6, so the next first difference is $7 + 6 = 13$ .

Step 3 — $T_3 = T_2 + 13 = 10 + 13 = 23$ .

Why the others are wrong:

A added the second difference to $T_2$ directly.
C kept the first difference at 7.
D added 6 twice.

Takeaway: In a quadratic sequence the SECOND difference is constant. Build the next first difference from it, then add that to the previous term.

M29Proficient

R2 000 is invested at 10% per annum, compounded annually. After how many complete years does the investment first exceed R3 000?

A) $n = 4$

B) $n = 6$

C) $n = 5$

D) $n = 7$

Show the worked solution

Answer: C

Explanation

Step 1 — The value after $n$ years is $2000(1.1)^n$ ; we need it above 3000.

Step 2 — $1.1^n > 1.5$ .

Step 3 — $1.1^4 = 1.4641$ (still below) and $1.1^5 = 1.6105$ (above).

Step 4 — So it first exceeds R3 000 after 5 complete years.

Why the others are wrong:

A $1.1^4 = 1.4641$ , giving R2 928 — still short of R3 000.
B overshoots; it already exceeded 3000 at $n=5$ .
D further past the answer still.

Takeaway: For 'how many complete years', test whole numbers around your estimate. Round UP: the target is only passed once the year is finished.

M30Proficient

The roots of $x^2 - 4x + 1 = 0$ are $\alpha$ and $\beta$ . Find $\alpha^3 + \beta^3$ .

A) $52$

B) $44$

C) $60$

D) $48$

Show the worked solution

Answer: A

Explanation

Step 1 — From $x^2-4x+1=0$ : $\alpha+\beta = 4$ and $\alpha\beta = 1$ .

Step 2 — Use $\alpha^3+\beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)$ .

Step 3 — $= 4^3 - 3(1)(4) = 64 - 12 = 52$ .

Why the others are wrong:

B used $(\alpha+\beta)^3$ alone, forgetting the correction term.
C subtracted $3\alpha\beta$ without the factor $(\alpha+\beta)$ .
D cubed the product instead of the sum.

Takeaway: $\alpha^3+\beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)$ . Worth memorising — it turns a hard-looking question into two substitutions.

Chapter 1: Algebraic Processes

Section 1.1 — Factorisation

Section 1.2 — Linear & Quadratic Equations

Section 1.3 — The Factor Theorem

Section 1.4 — Inequalities

Section 1.5 — Exponents & Surds

Section 1.6 — Logarithms

Section 1.7 — Sequences & Series

Section 1.8 — Financial Mathematics

Section 1.9 — Algebraic Insight, Vieta's Formulas & Manipulation Tricks

Exam-Bank Extras — Question Types Confirmed in Recent Papers

Mixed Practice — Chapter 1