Chapter 3 — Exponents, Polynomials and Equivalent Expressions

Answer: D

Explanation

Multiplying powers of the same base adds the exponents:

$$x^5 \cdot x^3 = x^{5+3} = x^8$$

If you ever doubt it, count. $x^5$ is five $x$'s multiplied together and $x^3$ is three, so together there are eight.

Why each wrong option is wrong:

• B, $x^{15}$ — multiplied the exponents. $5 \times 3 = 15$ is the rule for $(x^5)^3$, a power of a power, not a product.
• C, $2 x^{8}$ — added the exponents correctly and then also added the two $x$ terms as though this were $x^5 + x^3$. Multiplying does not produce a coefficient of 2.
• A, $x^{2}$ — subtracted, which is the rule for division.

Takeaway: Multiply → add exponents. Divide → subtract. Power of a power → multiply. When they blur together, count the $x$'s on a small case.

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Answer: B

Explanation

The outside power reaches everything inside the bracket — the 3 as well as the $x^2$:

$$(3x^2)^3 = 3^3 \cdot (x^2)^3 = 27 \cdot x^{2 \times 3} = 27x^6$$

Two separate steps: cube the coefficient, and multiply the exponents.

Why each wrong option is wrong:

• A, $3 x^{6}$ — handled the $x$ correctly and left the 3 untouched. The coefficient is inside the bracket, so it gets cubed too.
• C, $9 x^{6}$ — squared the 3 rather than cubing it. The outer power is 3.
• D, $27 x^{5}$ — added $2 + 3$ instead of multiplying. A power raised to a power multiplies.

Takeaway: The outside exponent applies to every factor inside, coefficient included. Deal with the number and the letter separately so neither gets missed.

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Answer: D

Explanation

A negative exponent means reciprocal, not negative:

$$4^{-2} = \frac{1}{4^2} = \frac{1}{16}$$

Say it as "one over $4^2$". The minus sign moves the power to the bottom of a fraction; it never changes the sign of the answer.

Why each wrong option is wrong:

• A, $-16$ — computed $4^2$ and made it negative. A positive base raised to any power is positive.
• B, $-8$ — multiplied $4 \times 2$ and negated.
• C, $8$ — multiplied instead of raising to a power.

Takeaway: $x^{-a} = \frac{1}{x^a}$. Negative exponent means small and positive, never negative. If your answer to a negative-exponent question is negative, something has gone wrong.

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Answer: 4

Explanation

In $x^{m/n}$, the bottom number is the root and the top is the power. So $8^{2/3}$ is the cube root of 8, squared:

$$8^{2/3} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4$$

You can do it the other way — square first, then take the cube root:

$$8^{2/3} = \sqrt[3]{8^2} = \sqrt[3]{64} = 4$$

Same answer, bigger numbers. Take the root first and the arithmetic stays small, which matters when a question is written to be done without reaching for the calculator every time.

This is a grid-in. Enter 4.

Takeaway: Bottom is the root, top is the power. Root first, then power.

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Section 2 — Expanding

Expanding is multiplying out until no brackets remain. Every term in the first bracket multiplies every term in the second.

$$(a + b)(c + d) = ac + ad + bc + bd$$

Two patterns come up so often they are worth recognising on sight rather than expanding each time:

$$(x + a)(x - a) = x^2 - a^2 \qquad \textbf{difference of two squares}$$ $$(x + a)^2 = x^2 + 2ax + a^2 \qquad \textbf{perfect square}$$

$(x + a)^2$ is not $x^2 + a^2$. That is the most common error in the whole chapter. There is a middle term, and it is $2ax$. Test it with numbers if you ever doubt it: $(3 + 4)^2 = 49$, while $3^2 + 4^2 = 25$.

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Answer: B

Explanation

Multiply every term by every term:

$$x \cdot x = x^2 \qquad x \cdot (-5) = -5x$$ $$3 \cdot x = 3x \qquad 3 \cdot (-5) = -15$$

Collect the two middle terms: $-5x + 3x = -2x$. So

$$(x + 3)(x - 5) = x^2 - 2x - 15$$

Check with a number. At $x = 1$: the original is $(4)(-4) = -16$, and the answer is $1 - 2 - 15 = -16$ ✓

Why each wrong option is wrong:

• A, $x^{2} - 15$ — multiplied the first terms and the last terms and skipped the middle two. Four products, not two.
• C, $x^{2} + 2 x - 15$ — the middle term has the wrong sign. $-5x + 3x$ is $-2x$: the bigger number is the negative one.
• D, $x^{2} - 8 x - 15$ — multiplied the middle terms ($-5 \times 3$) instead of adding them.

Takeaway: Four products, then collect the middle two. Check by substituting $x = 1$: it takes seconds and catches every sign error.

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Answer: D

Explanation

$(2x - 3)^2$ means $(2x - 3)(2x - 3)$. Multiply it out:

$$2x \cdot 2x = 4x^2 \qquad 2x \cdot (-3) = -6x$$ $$-3 \cdot 2x = -6x \qquad -3 \cdot (-3) = +9$$

The two middle terms are both $-6x$, so together they are $-12x$:

$$(2x - 3)^2 = 4x^2 - 12x + 9$$

Check at $x = 1$: the original is $(-1)^2 = 1$, and the answer is $4 - 12 + 9 = 1$ ✓

Why each wrong option is wrong:

• A, $4 x^{2} + 9$ — squared each term separately. $(a - b)^2 \neq a^2 + b^2$; there is always a middle term.
• B, $4 x^{2} - 9$ — used the difference-of-squares pattern, which is for $(2x-3)(2x+3)$ — brackets with opposite signs. Here the signs match, so the middle terms reinforce rather than cancel.
• C, $4 x^{2} - 6 x + 9$ — found $-6x$ once and forgot it appears twice.

Takeaway: $(a \pm b)^2 = a^2 \pm 2ab + b^2$ — the middle term is doubled. Matching signs give a middle term; opposite signs cancel it.

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Answer: A

Explanation

The brackets are identical apart from the sign, which is the difference-of-two- squares pattern:

$$(a + b)(a - b) = a^2 - b^2$$

Here $a = 3x$ and $b = 4$:

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

Expanding the long way shows why the middle disappears: $-12x + 12x = 0$.

Check at $x = 1$: $(7)(-1) = -7$, and $9 - 16 = -7$ ✓

Why each wrong option is wrong:

• B, $9 x^{2} + 16$ — the pattern subtracts. $-4 \times 4 = -16$.
• C, $9 x^{2} - 24 x - 16$ — kept a middle term. With opposite signs the middle terms are equal and opposite, so they always cancel.
• D, $3 x^{2} - 16$ — squared the $x$ but not the 3. $(3x)^2 = 9x^2$; the whole factor gets squared.

Takeaway: Same two terms, opposite signs → $a^2 - b^2$, with no middle term. Square the whole first term, coefficient included.

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Section 3 — Factoring

Factoring is expanding run backwards: turning a sum into a product. It is how you solve quadratics (Chapter 4) and how you simplify fractions (Section 4), so it earns its place twice.

Always look for a common factor first. $2x^2 + 8x$ is $2x(x + 4)$, and pulling the $2x$ out makes everything after it easier.

To factor $x^2 + bx + c$, find two numbers that multiply to $c$ and add to $b$. For $x^2 + 7x + 12$: which pairs multiply to 12? $1 \times 12$, $2 \times 6$, $3 \times 4$. Which adds to 7? $3 + 4$. So it is $(x+3)(x+4)$.

The signs tell you a lot before you start:

$c$	$b$	The two numbers are
positive	positive	both positive
positive	negative	both negative
negative	either	one of each; the bigger takes $b$'s sign

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Answer: B

Explanation

Find two numbers that multiply to $+12$ and add to $-7$.

The constant is positive and the middle term is negative, so both numbers are negative. Pairs multiplying to 12: $-1$ and $-12$ (adds to $-13$), $-2$ and $-6$ (adds to $-8$), $-3$ and $-4$ (adds to $-7$ ✓).

$$x^2 - 7x + 12 = (x - 3)(x - 4)$$

So $x - 3$ is a factor. Check by expanding: $x^2 - 4x - 3x + 12 = x^2 - 7x + 12$ ✓

Why each wrong option is wrong:

• A, $x + 3$ — right number, wrong sign. $(x+3)$ would need a $+3$ in the pair, and $+3$ and $+4$ give a middle term of $+7$.
• C, $x - 12$ — took the constant 12 as a factor. The numbers in the brackets multiply to give 12; they are not 12 itself.
• D, $x + 4$ — the same sign error as A, on the other factor.

Takeaway: Two numbers multiplying to $c$ and adding to $b$. Read the signs off the table before you hunt: positive $c$ with negative $b$ means both are negative.

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Answer: A

Explanation

Take the common factor of 3 out first:

$$3x^2 - 27 = 3(x^2 - 9)$$

Now $x^2 - 9$ is a difference of two squares, since $9 = 3^2$:

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

So the fully factored form is

$$3(x + 3)(x - 3)$$

Check at $x = 0$: the original is $-27$, and $3(3)(-3) = -27$ ✓

Why each wrong option is wrong:

• B, $3 \left(x - 3\right)^{2}$ — a perfect square needs a middle term. This expands to $3x^2 - 18x + 27$, not $3x^2 - 27$. Test it at $x = 0$: it gives $27$, not $-27$.
• C, $\left(x - 9\right) \left(x + 3\right)$ — never pulled the 3 out, then found numbers multiplying to $-27$ and adding to $-6$. That factors $x^2 - 6x - 27$, a different expression.
• D, $3 x^{2} + 27$ — lost the minus sign. $x^2 + 9$ is a sum of squares, and a sum of squares does not factor at all — which is worth knowing, because the test offers it whenever the difference-of-squares pattern is in play.

Takeaway: Common factor first, then look for a pattern in what is left. $a^2 - b^2$ factors; $a^2 + b^2$ does not. And substituting $x = 0$ eliminates wrong options here in one step.

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Answer: C

Explanation

With a leading coefficient the pair of numbers must multiply to $2 \times (-3) = -6$ and add to $+5$. That pair is $+6$ and $-1$.

Split the middle term using them:

$$2x^2 + 6x - x - 3$$

Now factor in pairs:

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

So the factors are $x + 3$ and $2x - 1$. Of the options, $2x - 1$ is there.

Check by expanding: $(x+3)(2x-1) = 2x^2 - x + 6x - 3 = 2x^2 + 5x - 3$ ✓

There is a faster check when you only need to test an option. A factor $2x - 1$ means $x = \frac12$ makes the expression zero:

$$2\left(\tfrac14\right) + 5\left(\tfrac12\right) - 3 = \tfrac12 + \tfrac52 - 3 = 0 \; ✓$$

With a calculator, testing four options this way is often quicker than factoring at all.

Why each wrong option is wrong:

• A, $x - 3$ — would need $x = 3$ to give zero: $18 + 15 - 3 = 30$, not 0.
• B, $2 x + 1$ — needs $x = -\frac12$: $\frac12 - \frac52 - 3 = -5$, not 0.
• D, $x + 1$ — needs $x = -1$: $2 - 5 - 3 = -6$, not 0.

Takeaway: With a leading coefficient, find numbers multiplying to $a \times c$ and adding to $b$, then split the middle term. Or skip all of it: test each option by finding the $x$ that makes it zero and substituting.

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Section 4 — Rational Expressions

A rational expression is a fraction with algebra in it. Simplifying means factoring the top and the bottom and cancelling anything they share.

You may only cancel factors, never terms. In $\dfrac{x + 3}{x}$ you cannot cancel the $x$'s, because the top is a sum, not a product. Factor first, and then cancelling is always safe.

Adding fractions needs a common denominator, exactly as with numbers:

$$\frac{1}{x} + \frac{1}{y} = \frac{y}{xy} + \frac{x}{xy} = \frac{x + y}{xy}$$

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Answer: A

Explanation

Factor both parts. The top is a difference of two squares:

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

For the bottom, two numbers multiplying to 12 and adding to 7 — that is 3 and 4:

$$x^2 + 7x + 12 = (x + 3)(x + 4)$$

So

$$\frac{(x + 3)(x - 3)}{(x + 3)(x + 4)}$$

The $(x + 3)$ is a factor of both, so it cancels:

$$\frac{x - 3}{x + 4}$$

Check at $x = 0$: the original is $\frac{-9}{12} = -\frac34$, and the answer is $\frac{-3}{4} = -\frac34$ ✓

Why each wrong option is wrong:

• B, $\frac{x + 3}{x + 4}$ — cancelled $(x - 3)$ from the top against $(x + 3)$ on the bottom. Only identical factors cancel.
• C, $\frac{x - 3}{x - 4}$ — factored the bottom as $(x+3)(x-4)$, which expands to $x^2 - x - 12$, not what was given.
• D, $- \frac{3}{4}$ — cancelled the $x^2$ terms. They are terms inside a sum, not factors of the whole expression, so they cannot be cancelled. This is exactly the mistake the section warns about — and note it happens to be the value at $x = 0$, which is why the substitution check has to use more than one number.

Takeaway: Factor top and bottom completely, then cancel whole brackets. Never cancel across a $+$ or $-$ sign.

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Answer: C

Explanation

The denominators share nothing, so the common denominator is their product, $x(x+1)$. Scale each fraction to it:

$$\frac{2}{x} = \frac{2(x+1)}{x(x+1)} \qquad \frac{3}{x+1} = \frac{3x}{x(x+1)}$$

Now add the tops:

$$\frac{2(x+1) + 3x}{x(x+1)} = \frac{2x + 2 + 3x}{x(x+1)} = \frac{5x + 2}{x(x+1)}$$

Check at $x = 1$: the original is $\frac21 + \frac32 = 3.5$, and the answer is $\frac{7}{2} = 3.5$ ✓

Why each wrong option is wrong:

• A, $\frac{5}{2 x + 1}$ — added the tops and the bottoms separately. That is never how fractions add: $\frac12 + \frac12$ is 1, not $\frac24$.
• B, $\frac{2 x + 3}{x \left(x + 1\right)}$ — found the right denominator but forgot to scale the numerators when moving them onto it.
• D, $\frac{5}{x \left(x + 1\right)}$ — the same mistake, having just added 2 and 3.

Takeaway: A common denominator means rescaling the tops too. Whatever you multiplied the bottom by, multiply that fraction's top by as well.

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Section 5 — Checking by Substitution

"Equivalent" means equal for every value of the variable. That gives you a technique that turns algebra questions into arithmetic, and it is one of the most valuable habits on this test.

Pick a number, put it in the original, put it in each option, and keep only the options that match.

Rules for choosing well:

• Avoid 0 and 1. They make too many different expressions agree. $x^2$ and $x^3$ and $x$ are all equal at both.
• Avoid numbers already in the problem, which can create accidental matches.
• 2, 3 and 5 are usually ideal.
• If two options survive, test a second number. One survivor is the answer.

This is not a fallback for when you are stuck. On "which expression is equivalent" questions with messy algebra it is often the fastest correct route, and with a calculator permitted it is very hard to get wrong.

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Answer: B

Explanation

Expand the first product:

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

Now subtract the bracket, flipping both its signs:

$$(x^2 - x - 6) - (x^2 - 6) = x^2 - x - 6 - x^2 + 6 = -x$$

Check by substitution, which is the point of this section. At $x = 2$:

• Original: $(4)(-1) - (4 - 6) = -4 - (-2) = -2$
• Answer $-x$: $-2$ ✓

Test the others at $x = 2$ as well: $2x$ gives 4, $-x - 12$ gives $-14$, $x$ gives 2. Only $-x$ matches, and it took ten seconds without any expanding at all.

Why each wrong option is wrong:

• A, $2 x$ — mishandled the subtraction so the $x^2$ terms added rather than cancelled.
• C, $- x - 12$ — subtracted the $x^2$ but not the $-6$, leaving $-6 - 6$.
• D, $x$ — the right size with the wrong sign.

Takeaway: A minus in front of a bracket flips every sign inside it. And when four options differ only in signs, substituting one number settles it faster than expanding.

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Answer: D

Explanation

Rewrite every negative exponent as a fraction:

$$\frac{\dfrac{1}{a} + \dfrac{1}{b}}{\dfrac{1}{ab}}$$

Add the two fractions on top, over the common denominator $ab$:

$$\frac{1}{a} + \frac{1}{b} = \frac{b + a}{ab}$$

So the whole thing is

$$\frac{\dfrac{a + b}{ab}}{\dfrac{1}{ab}} = \frac{a + b}{ab} \times \frac{ab}{1} = a + b$$

Dividing by $\frac{1}{ab}$ is multiplying by $ab$, and the $ab$ cancels.

Or substitute, and skip all of it. Take $a = 2$ and $b = 3$:

$$\frac{\frac12 + \frac13}{\frac16} = \frac{\frac56}{\frac16} = 5$$

Now test the options at $a = 2, b = 3$: option A gives 1, B gives 6, C gives $\frac56$, D gives 5. Only D matches ✓

On a question like this, where the algebra is fiddly and the options are simple, substitution is not the backup plan — it is the better method.

Why each wrong option is wrong:

• A, $1$ — cancelled the negative exponents against each other as though the top and bottom were the same expression.
• B, $a b$ — multiplied $a$ and $b$ rather than adding.
• C, $\frac{a + b}{a b}$ — stopped after combining the numerator and never divided by $\frac{1}{ab}$.

Takeaway: Turn negative exponents into fractions before doing anything else. And when the options are simple but the expression is not, substitute two small numbers — it is faster and more reliable than the algebra.

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Section 6 — Mixed Practice

Exponents, expanding, factoring and fractions, shuffled.

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Answer: B

Explanation

Handle the numbers and the letters separately. $12 \div 3 = 4$, and dividing powers subtracts exponents: $x^{7-4} = x^3$.

$$\frac{12x^7}{3x^4} = 4x^3$$

A divided the exponents. C subtracted the coefficients instead of dividing them. D added the exponents.

Takeaway: Divide the coefficients, subtract the exponents. The coefficients and the powers follow different rules — do them in separate steps.

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Answer: C

Explanation

The outside power reaches every factor, top and bottom:

$$\left(\frac{2x^3}{y^2}\right)^2 = \frac{2^2 (x^3)^2}{(y^2)^2} = \frac{4x^6}{y^4}$$

A left the 2 alone. B added $3 + 2$ instead of multiplying. D left the denominator unsquared.

Takeaway: Squaring a fraction squares the top and the bottom, including every coefficient. Three separate pieces here — check you did all three.

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Answer: D

Explanation

$3^0 = 1$ — anything non-zero to the power 0 is 1. And $2^{-1} = \frac{1}{2}$.

$$1 + \frac12 = \frac32$$

A and B treated $3^0$ as 0. C treated $2^{-1}$ as 0.

Takeaway: $x^0 = 1$, not 0. It is the answer to "how many times have you multiplied by $x$" — none, so you are left with the 1 you started from.

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Answer: A

Explanation

Do the two brackets first, then multiply through by $2x$:

$$(x + 4)(x - 1) = x^2 - x + 4x - 4 = x^2 + 3x - 4$$

$$2x(x^2 + 3x - 4) = 2x^3 + 6x^2 - 8x$$

Check at $x = 1$: the original is $2(1)(5)(0) = 0$, and $2 + 6 - 8 = 0$ ✓

B has the last sign wrong. C multiplied only the first term by $2x$. D lost the 2.

Takeaway: Multiply two brackets first, then distribute the outside factor across every term of the result.

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Answer: B

Explanation

Four terms with no common factor overall — group them in pairs:

$$(x^3 + 2x^2) + (-9x - 18) = x^2(x + 2) - 9(x + 2)$$

Both pairs now contain $(x + 2)$, so pull it out:

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

So $x + 2$ is a factor. Confirm it directly: a factor $x + 2$ means $x = -2$ gives zero.

$$(-8) + 2(4) - 9(-2) - 18 = -8 + 8 + 18 - 18 = 0 \; ✓$$

A needs $x = 2$: $8 + 8 - 18 - 18 = -20$, not 0. C needs $x = -9$, which is nowhere near. D needs $x = 18$.

Takeaway: Four terms with no common factor → group in pairs and look for a shared bracket. To test a proposed factor, set it to zero and substitute — one line, and it settles the question.

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Answer: 14

Explanation

A perfect square looks like

$$(x + a)^2 = x^2 + 2ax + a^2$$

Match it against $x^2 + kx + 49$. The constant tells you $a^2 = 49$, so $a = 7$ (taking the positive value, since $k$ is positive). Then the middle term is

$$k = 2a = 14$$

Check: $(x + 7)^2 = x^2 + 14x + 49$ ✓

This is a grid-in. Enter 14.

Takeaway: In a perfect square the middle coefficient is twice the square root of the constant. Take the root of the constant, then double it.

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Answer: A

Explanation

Factor both parts:

$$2x^2 + 6x = 2x(x + 3) \qquad x^2 - 9 = (x + 3)(x - 3)$$

$$\frac{2x(x + 3)}{(x + 3)(x - 3)} = \frac{2x}{x - 3}$$

Check at $x = 1$: the original is $\frac{2 + 6}{1 - 9} = \frac{8}{-8} = -1$, and the answer is $\frac{2}{-2} = -1$ ✓

B cancelled $(x+3)$ against $(x-3)$. C did not take the $2x$ out properly. D cancelled terms across a sum.

Takeaway: Factor first, cancel whole brackets only. A common factor of $2x$ on top comes out in one piece.

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Answer: 2

Explanation

The aim is to make the top contain a copy of the bottom. The bottom is $2x + 3$; three of those give $6x + 9$, which is nearly the top:

$$6x + 11 = 3(2x + 3) + 2$$

Now split the fraction:

$$\frac{6x + 11}{2x + 3} = \frac{3(2x + 3)}{2x + 3} + \frac{2}{2x + 3} = 3 + \frac{2}{2x + 3}$$

So $c = 2$.

Check at $x = 0$: the original is $\frac{11}{3}$, and the rewritten form is $3 + \frac23 = \frac{11}{3}$ ✓

There is a quicker way when you only need $c$. The two forms must be equal for every $x$, so pick a convenient one. At $x = 0$:

$$\frac{11}{3} = 3 + \frac{c}{3} \quad\Longrightarrow\quad 11 = 9 + c \quad\Longrightarrow\quad c = 2$$

This is a grid-in. Enter 2.

Takeaway: To split a fraction into "whole part plus remainder", work out how many copies of the bottom fit into the top. Or, since the two forms are equal for every $x$, substitute $x = 0$ and solve for the constant in one line.

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Answer: C

Explanation

Simplify each root by pulling out square factors:

$$\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2} \qquad \sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}$$

Both are now multiples of $\sqrt2$, so they add like terms:

$$3\sqrt2 + 2\sqrt2 = 5\sqrt2$$

Check numerically: $\sqrt{18} \approx 4.243$ and $\sqrt8 \approx 2.828$, summing to about $7.071$. And $5\sqrt2 \approx 7.071$ ✓

A and B added the numbers under the roots. $\sqrt{a} + \sqrt{b}$ is never $\sqrt{a+b}$ — test it: $\sqrt9 + \sqrt{16} = 7$, but $\sqrt{25} = 5$. D multiplied the coefficients instead of adding.

Takeaway: Simplify each radical first, then add the ones with matching roots the way you would add like terms. Never add under the root sign.

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Answer: D

Explanation

Expand the left-hand side, keeping $p$ as a letter:

$$(x + p)(x + 4) = x^2 + 4x + px + 4p = x^2 + (p + 4)x + 4p$$

"For all values of $x$" means the two sides are identical, so match the pieces. The $x$ coefficients:

$$p + 4 = 9 \quad\Longrightarrow\quad p = 5$$

Then the constants:

$$q = 4p = 20$$

Check: $(x + 5)(x + 4) = x^2 + 9x + 20$ ✓

A and B copied numbers straight off the question. C added where a multiplication was needed.

Takeaway: "For all values of $x$" means match coefficients term by term. Find the unknown from the $x$ terms first, then use it in the constant.

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Answer: B

Explanation

The bases differ, so no exponent rule applies yet. Rewrite them to match:

$$4^3 = (2^2)^3 = 2^6$$

Now they share a base and dividing subtracts:

$$\frac{2^{10}}{2^6} = 2^{4} = 16$$

Check numerically: $\frac{1024}{64} = 16$ ✓

A subtracted $10 - 3$ without converting the 4 to a power of 2 — the most common error in this whole topic.

Takeaway: Exponent rules need a common base. Rewrite 4, 8, 9, 27 and so on as powers of a smaller base first, then apply the rule.

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Answer: D

Explanation

The common factor is $-2x$. Taking it out flips the sign of every term inside:

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

Check by expanding: $-2x \cdot x = -2x^2$ ✓ and $-2x \cdot (-5) = +10x$ ✓

Substituting is quicker still. At $x = 1$ the original is $-2 + 10 = 8$, and $-2(1)(1-5) = -2(-4) = 8$ ✓

A kept the factor positive, giving $2x^2 - 10x$ — the whole expression negated. B got the inside sign wrong. C divided the $10x$ by $-2$ rather than by $-2x$, leaving a 10 where a 5 belongs.

One more thing worth knowing: $2x(5 - x)$ is also a correct factorisation of this expression, because $5 - x = -(x - 5)$ and the two minus signs cancel. Both forms are right; $-2x(x-5)$ is the conventional one, with the common factor taken out in front.

Takeaway: Pulling out a negative flips every sign inside the bracket. Expand it back, or test $x = 1$, before moving on — and remember $a - b$ and $b - a$ differ only by a sign, so two factorisations can look different and be equal.

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Answer: A

Explanation

Combine the top over a common denominator of $3x$:

$$\frac1x - \frac13 = \frac{3 - x}{3x}$$

Now the whole expression is

$$\frac{3 - x}{3x} \div (x - 3) = \frac{3 - x}{3x(x - 3)}$$

The key step: $3 - x$ is the negative of $x - 3$, so $3 - x = -(x - 3)$ and the brackets cancel, leaving the minus behind:

$$\frac{-(x - 3)}{3x(x - 3)} = -\frac{1}{3x}$$

Check at $x = 1$: the original is $\frac{1 - \frac13}{1 - 3} = \frac{2/3}{-2} = -\frac13$, and $-\frac{1}{3(1)} = -\frac13$ ✓

B lost that minus. D stopped one step early.

Takeaway: $a - b$ and $b - a$ differ only by a minus sign, so they cancel and leave $-1$. Combine the top before dividing, then substitute a number to check the sign.

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