Algebra

Answer: $-4\sqrt{5}$

Write each root in simplest form first, so they are all multiples of $\sqrt{5}\mathclose{:}$ $4\sqrt{5} - 3\sqrt{5} - 5\sqrt{5} = -4\sqrt{5}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Combined the numbers in front but dropped the root → $-4$

Answer: $2 x^{2} - 2 x - y$

Only LIKE terms combine — the same letters to the same powers: the $x^2$ terms give $2 x^{2}\mathclose{,}$ the $x$ terms $- 2 x$ and the $y$ terms $- y\mathclose{,}$ so the answer is $2 x^{2} - 2 x - y\mathclose{.}$

Common mistakes, and the answer each one gives:

• Combined the $x^2$ terms with the $x$ terms as if they were alike → $- y$

Answer: $16 x^{16}$

Raise EVERY factor inside the bracket to the power $4\mathclose{:}$ $2^{4} = 16\mathclose{,}$ and $\left(x^{4}\right)^{4} = x^{16}$ because a power of a power multiplies the exponents. So the answer is $16 x^{16}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Did not raise the number in front to the power → $2 x^{16}$
• Added the exponents instead of multiplying them → $16 x^{8}$

Answer: $x^{2} - 49$

$(p + q)(p - q) = p^2 - q^2\mathclose{,}$ and the two middle terms cancel: $\left(x + 7\right)\left(x - 7\right) = x^{2} - 7^{2} = x^{2} - 49\mathclose{.}$

Common mistakes, and the answer each one gives:

• Treated it as a square, as if both brackets were the same → $x^{2} + 14 x + 49$
• Added the squares instead of subtracting → $x^{2} + 49$

Answer: $3\left(x - 6\right)\left(x - 4\right)$

Take out the common factor $3$ first: $3 x^{2} - 30 x + 72 = 3\left(x^{2} - 10 x + 24\right)\mathclose{.}$ Then look for two numbers whose PRODUCT is $24$ and whose SUM is $-10\mathclose{:}$ they are $-6$ and $-4\mathclose{.}$ So $x^{2} - 10 x + 24 = \left(x - 6\right)\left(x - 4\right)\mathclose{,}$ and the answer is $3\left(x - 6\right)\left(x - 4\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Found the right two numbers but put the opposite signs in the brackets → $3\left(x + 4\right)\left(x + 6\right)$
• Used two numbers with the right product but the wrong sum → $3\left(x - 8\right)\left(x - 3\right)$
• Left out the common factor $3$ → $\left(x - 6\right)\left(x - 4\right)$

Answer: $\dfrac{73}{300}$

Multiplication comes before addition, so work out $\dfrac{4}{5} \times \dfrac{1}{5} = \dfrac{4}{25}$ first, then add $\dfrac{1}{12}\mathclose{:}$ $\dfrac{73}{300}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Worked strictly left to right, adding before multiplying → $\frac{53}{300}$

Answer: $3\sqrt[3]{2}$

Find the largest CUBE factor: $54 = 27 \times 2\mathclose{,}$ and $\sqrt[3]{27} = 3\mathclose{,}$ so $\sqrt[3]{54} = 3\sqrt[3]{2}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the cube factor out without taking its cube root → $27\sqrt[3]{2}$
• Took the square root instead of the cube root → $3\sqrt{6}$

Answer: $\dfrac{9 x + 32}{\left(x + 4\right)\left(x + 3\right)}$

The common denominator is $(x + 4)(x + 3)\mathclose{,}$ so multiply each numerator by the OTHER bracket: $\dfrac{4(x + 3) + 5(x + 4)}{\left(x + 4\right)\left(x + 3\right)} = \dfrac{9 x + 32}{\left(x + 4\right)\left(x + 3\right)}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added the numerators and the denominators separately → $\dfrac{9}{2 x + 7}$
• Multiplied each numerator by its OWN denominator instead of the other one → $\dfrac{9 x + 31}{\left(x + 4\right)\left(x + 3\right)}$

Answer: $\dfrac{x + 1}{x + 2}$

Dividing by a fraction is multiplying by its reciprocal: $\dfrac{x + 1}{x - 7} \times \dfrac{x - 7}{x + 2}\mathclose{.}$ The $x - 7$ then cancels, leaving $\dfrac{x + 1}{x + 2}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied straight across without turning the second fraction over → $\dfrac{x^{2} + 3 x + 2}{x^{2} - 14 x + 49}$
• Turned the FIRST fraction over instead of the second → $\dfrac{x + 2}{x + 1}$

Answer: $4$

Multiply both sides by $\left(x + 5\right)\left(x - 4\right)\mathclose{:}$ $5 x - 11 = A\left(x - 4\right) + B\left(x + 5\right)\mathclose{.}$ Put $x = -5$ to lose the $B$ term: $-36 = A \cdot (-9)\mathclose{,}$ so $A = 4\mathclose{.}$ Put $x = 4\mathclose{:}$ $9 = B \cdot 9\mathclose{,}$ so $B = 1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave B instead → $1$
• Substituted $x = 5$ instead of $x = -5$ → $14$

Answer: $- 24 a^{5} b^{2}$

Multiply the numbers, then each letter: powers of the SAME letter multiply by ADDING exponents. $4 \cdot (-6) = -24\mathclose{,}$ $a^{3} \cdot a^{2} = a^{5}$ and $b^{1} \cdot b^{1} = b^{2}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied the exponents instead of adding them → $- 24 a^{6} b$
• Added the numbers in front instead of multiplying them → $- 2 a^{5} b^{2}$

Answer: $x^{11}$

Multiplying adds the exponents and dividing subtracts them: $7 + 9 - 5 = 11\mathclose{,}$ so the answer is $x^{11}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied the exponents instead of adding them → $x^{58}$
• Divided the exponents instead of subtracting → $x^{\frac{16}{5}}$

Answer: $2\sqrt{110}$

$\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}\mathclose{,}$ so this is $\sqrt{440} = 2\sqrt{110}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added the numbers under the roots → $\sqrt{42}$

Answer: $14 - 18 x$

Multiply EVERY term inside each bracket by the number in front: $10 - 20 x$ and $2 x + 4\mathclose{,}$ then collect: $14 - 18 x\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied only the first term in each bracket → $- 18 x$

Answer: $10 x^{2} + 33 x - 7$

Multiply every term in the first bracket by every term in the second: $5x \cdot 2x = 10 x^{2}\mathclose{,}$ the two cross terms give $35 x - 2 x = 33 x\mathclose{,}$ and $-1 \cdot 7 = -7\mathclose{.}$ Together: $10 x^{2} + 33 x - 7\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied only the first terms and the last terms → $10 x^{2} - 7$

Answer: $\left(x - 7\right)\left(x + 7\right)$

$x^{2} - 49 = x^{2} - 7^{2}$ is a difference of two squares, $p^2 - q^2 = (p - q)(p + q)\mathclose{,}$ so the answer is $\left(x - 7\right)\left(x + 7\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Wrote it as a perfect square → $\left(x - 7\right)^{2}$

Answer: $\dfrac{x - 3}{x - 2}$

Factorise both: $\dfrac{\left(x - 7\right) \left(x - 3\right)}{\left(x - 7\right) \left(x - 2\right)}\mathclose{.}$ The factor $x - 7$ is common to both and cancels, leaving $\dfrac{x - 3}{x - 2}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Cancelled the wrong bracket, keeping the common one → $\frac{x - 7}{x - 2}$
• Cancelled the $x^2$ terms, which are terms and not factors → $\frac{10 x - 21}{9 x - 14}$

Answer: $\dfrac{2\sqrt{3}}{3}$

Multiply top and bottom by $\sqrt{3}\mathclose{,}$ since $\sqrt{3} \cdot \sqrt{3} = 3\mathclose{:}$ $\dfrac{2\sqrt{3}}{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied only the denominator by the root → $\dfrac{2}{3}$
• Moved the root to the top without dividing by it → $2\sqrt{3}$

Answer: $x^{\frac{15}{2}}$

$\sqrt{x} = x^{\frac{1}{2}}\mathclose{,}$ and dividing by $x^{-4}$ subtracts $-4\mathclose{,}$ which ADDS $4\mathclose{:}$ $3 + \frac{1}{2} + 4 = \frac{15}{2}\mathclose{,}$ so the answer is $x^{\frac{15}{2}}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Treated $\sqrt{x}$ as $x^2$ → $x^{9}$
• Subtracted the negative exponent as if it were positive → $\frac{1}{\sqrt{x}}$

Answer: $35$

Put each value in BRACKETS, so the minus sign is squared too: $5\left(-4\right)^{2} - 2\left(-4\right)\left(-5\right) + \left(-5\right) = 80 - 40 - 5 = 35\mathclose{.}$

Common mistakes, and the answer each one gives:

• Squared the number but not its minus sign, so $a^2$ came out negative → $-125$
• Doubled $a$ instead of squaring it → $-85$

Answer: $\dfrac{525}{242}$

Multiplication comes before addition, so work out $\dfrac{9}{11} \times \dfrac{9}{11} = \dfrac{81}{121}$ first, then add $\dfrac{3}{2}\mathclose{:}$ $\dfrac{525}{242}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Worked strictly left to right, adding before multiplying → $\frac{459}{242}$

Answer: $3\sqrt{5}$

Find the largest square factor: $45 = 9 \times 5\mathclose{,}$ and $\sqrt{ab} = \sqrt{a}\sqrt{b}\mathclose{,}$ so $\sqrt{45} = \sqrt{9} \cdot \sqrt{5} = 3\sqrt{5}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the square factor out without taking its root → $9\sqrt{5}$
• Added the roots of the two factors instead of multiplying them → $3 + \sqrt{5}$

Answer: $\dfrac{-6}{x - 4}$

The denominators are the same, so keep the denominator and subtract the numerators — the WHOLE second one, in brackets: $\dfrac{3 x - 5 - \left(3 x + 1\right)}{x - 4} = \dfrac{-6}{x - 4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Did not take the minus sign through the whole second numerator → $\dfrac{-4}{x - 4}$
• Subtracted the numerators the wrong way round → $\dfrac{6}{x - 4}$

Answer: $\dfrac{x - 3}{x - 5}$

Multiply the numerators and the denominators: the $x - 1$ below the first fraction and above the second is a common FACTOR, so it cancels, leaving $\dfrac{x - 3}{x - 5}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Turned the second fraction over, as if dividing → $\dfrac{x^{2} - 8 x + 15}{x^{2} - 2 x + 1}$

Answer: $-1$

Multiply both sides by $\left(x - 1\right)\left(x - 2\right)\mathclose{:}$ $3 x - 7 = A\left(x - 2\right) + B\left(x - 1\right)\mathclose{.}$ Put $x = 1$ to lose the $B$ term: $-4 = A \cdot (-1)\mathclose{,}$ so $A = 4\mathclose{.}$ Put $x = 2\mathclose{:}$ $-1 = B \cdot 1\mathclose{,}$ so $B = -1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave A instead → $4$
• Substituted $x = -2$ instead of $x = 2$ → $\dfrac{13}{3}$

Answer: $9 x^{2} + 42 x + 49$

$(p + q)^2 = p^2 + 2pq + q^2\mathclose{,}$ so $\left(3 x + 7\right)^{2} = \left(3x\right)^{2} + 2 \cdot 3x \cdot 7 + 7^{2} = 9 x^{2} + 42 x + 49\mathclose{.}$

Common mistakes, and the answer each one gives:

• Squared each term and left out the middle term → $9 x^{2} + 49$
• Gave the middle term the wrong sign → $9 x^{2} - 42 x + 49$
• Forgot to double the middle term → $9 x^{2} + 21 x + 49$

Answer: $\left(x + 3\right)\left(x + 8\right)$

Look for two numbers whose PRODUCT is $24$ and whose SUM is $11\mathclose{:}$ they are $3$ and $8\mathclose{.}$ So $x^{2} + 11 x + 24 = \left(x + 3\right)\left(x + 8\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Found the right two numbers but put the opposite signs in the brackets → $\left(x - 8\right)\left(x - 3\right)$
• Used two numbers with the right product but the wrong sum → $\left(x + 4\right)\left(x + 6\right)$

Answer: $81 x^{16}$

Raise EVERY factor inside the bracket to the power $4\mathclose{:}$ $3^{4} = 81\mathclose{,}$ and $\left(x^{4}\right)^{4} = x^{16}$ because a power of a power multiplies the exponents. So the answer is $81 x^{16}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Did not raise the number in front to the power → $3 x^{16}$
• Added the exponents instead of multiplying them → $81 x^{8}$

Answer: $x^{2} - 2 x - y$

Only LIKE terms combine — the same letters to the same powers: the $x^2$ terms give $x^{2}\mathclose{,}$ the $x$ terms $- 2 x$ and the $y$ terms $- y\mathclose{,}$ so the answer is $x^{2} - 2 x - y\mathclose{.}$

Common mistakes, and the answer each one gives:

• Combined the $x^2$ terms with the $x$ terms as if they were alike → $- x - y$

Answer: $6\sqrt{3}$

Write each root in simplest form first, so they are all multiples of $\sqrt{3}\mathclose{:}$ $3\sqrt{3} + 5\sqrt{3} - 2\sqrt{3} = 6\sqrt{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added and subtracted the numbers under the roots → $3\sqrt{10}$
• Combined the numbers in front but dropped the root → $6$