2. Algebraic fractions

Answer: $\frac{12 \left(x + 5\right)}{\left(x + 3\right) \left(x + 6\right)}$

The common denominator is $(x+6)(x+3)$. $\dfrac{4(x+3) + 8(x+6)}{(x+6)(x+3)} = \frac{12 \left(x + 5\right)}{\left(x + 3\right) \left(x + 6\right)}$.

Common mistakes, and the answer each one gives:

• Added numerators and denominators separately → $\frac{12}{2 x + 9}$
• Used $(x+6+3)$ as the common denominator → $\frac{12}{x + 9}$

Answer: $36 x^{2} - 36$

$(p \pm q)^2 = p^2 \pm 2pq + q^2$ and $(p+q)(p-q) = p^2 - q^2$. Here that gives $36 x^{2} - 36$.

Common mistakes, and the answer each one gives:

• Kept a middle term; in this product the two cancel → $36 x^{2} + 72 x - 36$
• Added the squares instead of subtracting → $36 x^{2} + 36$
• Squared only the first term → $36 x^{2} - 6$

Answer: $\left(x + 3\right) \left(x + 4\right)$

Find two numbers whose PRODUCT is $12$ and whose SUM is $-7$: they are $-3$ and $-4$. So the factorisation is $\left(x + 3\right) \left(x + 4\right)$.

Common mistakes, and the answer each one gives:

• Used the numbers with their signs unchanged in the brackets → $x^{2} - 7 x + 12$
• Matched the sum to the constant and the product to the middle term → $x^{2} + 6 x - 7$

Answer: $x^{9}$

Multiplying adds the exponents and dividing subtracts them: $8 + 6 - 5 = 9$, so the answer is $x^{9}$.

Common mistakes, and the answer each one gives:

• Multiplied the exponents instead of adding them → $x^{43}$
• Divided the exponents instead of subtracting → $x^{\frac{14}{5}}$

Answer: $\frac{x - 5}{x + 4}$

Dividing by a fraction is multiplying by its reciprocal: $\dfrac{x - 5}{x - 1} \times \dfrac{x - 1}{x + 4}$. The $x - 1$ then cancels, leaving $\frac{x - 5}{x + 4}$.

Common mistakes, and the answer each one gives:

• Multiplied straight across without flipping the second fraction → $\frac{x^{2} - x - 20}{x^{2} - 2 x + 1}$
• Flipped the FIRST fraction instead of the second → $\frac{x^{2} - 2 x + 1}{x^{2} - x - 20}$

Answer: $\frac{5 x + 19}{\left(x + 2\right) \left(x + 5\right)}$

The common denominator is $(x+5)(x+2)$. $\dfrac{2(x+2) + 3(x+5)}{(x+5)(x+2)} = \frac{5 x + 19}{\left(x + 2\right) \left(x + 5\right)}$.

Common mistakes, and the answer each one gives:

• Added numerators and denominators separately → $\frac{5}{2 x + 7}$
• Used $(x+5+2)$ as the common denominator → $\frac{5}{x + 7}$

Answer: $16 x^{2} - 48 x + 36$

$(p \pm q)^2 = p^2 \pm 2pq + q^2$ and $(p+q)(p-q) = p^2 - q^2$. Here that gives $16 x^{2} - 48 x + 36$.

Common mistakes, and the answer each one gives:

• Squared each term and left out the middle term → $16 x^{2} + 36$
• Gave the middle term the wrong sign → $16 x^{2} + 48 x + 36$
• Doubled the whole bracket instead of squaring it → $8 x - 12$

Answer: $\left(x - 6\right) \left(x + 2\right)$

Find two numbers whose PRODUCT is $-12$ and whose SUM is $4$: they are $-2$ and $6$. So the factorisation is $\left(x - 6\right) \left(x + 2\right)$.

Common mistakes, and the answer each one gives:

• Used the numbers with their signs unchanged in the brackets → $x^{2} + 4 x - 12$
• Matched the sum to the constant and the product to the middle term → $x^{2} - 5 x + 4$

Answer: $x^{3}$

Multiplying adds the exponents and dividing subtracts them: $4 + 3 - 4 = 3$, so the answer is $x^{3}$.

Common mistakes, and the answer each one gives:

• Multiplied the exponents instead of adding them → $x^{8}$
• Divided the exponents instead of subtracting → $x^{\frac{7}{4}}$

Answer: $\frac{x + 6}{x + 7}$

Dividing by a fraction is multiplying by its reciprocal: $\dfrac{x + 6}{x + 5} \times \dfrac{x + 5}{x + 7}$. The $x + 5$ then cancels, leaving $\frac{x + 6}{x + 7}$.

Common mistakes, and the answer each one gives:

• Multiplied straight across without flipping the second fraction → $\frac{x^{2} + 13 x + 42}{x^{2} + 10 x + 25}$
• Flipped the FIRST fraction instead of the second → $\frac{x^{2} + 10 x + 25}{x^{2} + 13 x + 42}$

Answer: $\frac{7 x + 19}{\left(x + 2\right) \left(x + 3\right)}$

The common denominator is $(x+2)(x+3)$. $\dfrac{5(x+3) + 2(x+2)}{(x+2)(x+3)} = \frac{7 x + 19}{\left(x + 2\right) \left(x + 3\right)}$.

Common mistakes, and the answer each one gives:

• Added numerators and denominators separately → $\frac{7}{2 x + 5}$
• Used $(x+2+3)$ as the common denominator → $\frac{7}{x + 5}$

Answer: $36 x^{2} - 84 x + 49$

$(p \pm q)^2 = p^2 \pm 2pq + q^2$ and $(p+q)(p-q) = p^2 - q^2$. Here that gives $36 x^{2} - 84 x + 49$.

Common mistakes, and the answer each one gives:

• Squared each term and left out the middle term → $36 x^{2} + 49$
• Gave the middle term the wrong sign → $36 x^{2} + 84 x + 49$
• Doubled the whole bracket instead of squaring it → $12 x - 14$

Answer: $\left(x - 5\right) \left(x + 9\right)$

Find two numbers whose PRODUCT is $-45$ and whose SUM is $-4$: they are $-9$ and $5$. So the factorisation is $\left(x - 5\right) \left(x + 9\right)$.

Common mistakes, and the answer each one gives:

• Used the numbers with their signs unchanged in the brackets → $x^{2} - 4 x - 45$
• Matched the sum to the constant and the product to the middle term → $x^{2} + 3 x - 4$

Answer: $x^{7}$

Multiplying adds the exponents and dividing subtracts them: $6 + 5 - 4 = 7$, so the answer is $x^{7}$.

Common mistakes, and the answer each one gives:

• Multiplied the exponents instead of adding them → $x^{26}$
• Divided the exponents instead of subtracting → $x^{\frac{11}{4}}$

Answer: $\frac{x + 3}{x}$

Dividing by a fraction is multiplying by its reciprocal: $\dfrac{x + 3}{x + 7} \times \dfrac{x + 7}{x}$. The $x + 7$ then cancels, leaving $\frac{x + 3}{x}$.

Common mistakes, and the answer each one gives:

• Multiplied straight across without flipping the second fraction → $\frac{x^{2} + 3 x}{x^{2} + 14 x + 49}$
• Flipped the FIRST fraction instead of the second → $\frac{x^{2} + 14 x + 49}{x^{2} + 3 x}$

Answer: $\frac{11 x + 20}{\left(x + 1\right) \left(x + 4\right)}$

The common denominator is $(x+4)(x+1)$. $\dfrac{8(x+1) + 3(x+4)}{(x+4)(x+1)} = \frac{11 x + 20}{\left(x + 1\right) \left(x + 4\right)}$.

Common mistakes, and the answer each one gives:

• Added numerators and denominators separately → $\frac{11}{2 x + 5}$
• Used $(x+4+1)$ as the common denominator → $\frac{11}{x + 5}$

Answer: $x^{2} - 81$

$(p \pm q)^2 = p^2 \pm 2pq + q^2$ and $(p+q)(p-q) = p^2 - q^2$. Here that gives $x^{2} - 81$.

Common mistakes, and the answer each one gives:

• Kept a middle term; in this product the two cancel → $x^{2} + 18 x - 81$
• Added the squares instead of subtracting → $x^{2} + 81$
• Squared only the first term → $x^{2} - 9$

Answer: $\left(x + 4\right) \left(x + 6\right)$

Find two numbers whose PRODUCT is $24$ and whose SUM is $-10$: they are $-4$ and $-6$. So the factorisation is $\left(x + 4\right) \left(x + 6\right)$.

Common mistakes, and the answer each one gives:

• Used the numbers with their signs unchanged in the brackets → $x^{2} - 10 x + 24$
• Matched the sum to the constant and the product to the middle term → $x^{2} + 9 x - 10$

Answer: $x^{5}$

Multiplying adds the exponents and dividing subtracts them: $3 + 4 - 2 = 5$, so the answer is $x^{5}$.

Common mistakes, and the answer each one gives:

• Multiplied the exponents instead of adding them → $x^{10}$
• Divided the exponents instead of subtracting → $x^{\frac{7}{2}}$

Answer: $\frac{x}{x + 3}$

Dividing by a fraction is multiplying by its reciprocal: $\dfrac{x}{x - 7} \times \dfrac{x - 7}{x + 3}$. The $x - 7$ then cancels, leaving $\frac{x}{x + 3}$.

Common mistakes, and the answer each one gives:

• Multiplied straight across without flipping the second fraction → $\frac{x^{2} + 3 x}{x^{2} - 14 x + 49}$
• Flipped the FIRST fraction instead of the second → $\frac{x^{2} - 14 x + 49}{x^{2} + 3 x}$

Answer: $\frac{8 x + 17}{\left(x + 1\right) \left(x + 4\right)}$

The common denominator is $(x+1)(x+4)$. $\dfrac{3(x+4) + 5(x+1)}{(x+1)(x+4)} = \frac{8 x + 17}{\left(x + 1\right) \left(x + 4\right)}$.

Common mistakes, and the answer each one gives:

• Added numerators and denominators separately → $\frac{8}{2 x + 5}$
• Used $(x+1+4)$ as the common denominator → $\frac{8}{x + 5}$

Answer: $4 x^{2} - 4$

$(p \pm q)^2 = p^2 \pm 2pq + q^2$ and $(p+q)(p-q) = p^2 - q^2$. Here that gives $4 x^{2} - 4$.

Common mistakes, and the answer each one gives:

• Kept a middle term; in this product the two cancel → $4 x^{2} + 8 x - 4$
• Added the squares instead of subtracting → $4 x^{2} + 4$
• Squared only the first term → $4 x^{2} - 2$

Answer: $\left(x - 6\right) \left(x + 1\right)$

Find two numbers whose PRODUCT is $-6$ and whose SUM is $5$: they are $-1$ and $6$. So the factorisation is $\left(x - 6\right) \left(x + 1\right)$.

Common mistakes, and the answer each one gives:

• Used the numbers with their signs unchanged in the brackets → $x^{2} + 5 x - 6$
• Matched the sum to the constant and the product to the middle term → $x^{2} - 6 x + 5$

Answer: $x^{7}$

Multiplying adds the exponents and dividing subtracts them: $5 + 8 - 6 = 7$, so the answer is $x^{7}$.

Common mistakes, and the answer each one gives:

• Multiplied the exponents instead of adding them → $x^{34}$
• Divided the exponents instead of subtracting → $x^{\frac{13}{6}}$

Answer: $\frac{x - 2}{x - 7}$

Dividing by a fraction is multiplying by its reciprocal: $\dfrac{x - 2}{x - 4} \times \dfrac{x - 4}{x - 7}$. The $x - 4$ then cancels, leaving $\frac{x - 2}{x - 7}$.

Common mistakes, and the answer each one gives:

• Multiplied straight across without flipping the second fraction → $\frac{x^{2} - 9 x + 14}{x^{2} - 8 x + 16}$
• Flipped the FIRST fraction instead of the second → $\frac{x^{2} - 8 x + 16}{x^{2} - 9 x + 14}$

Answer: $\frac{11 x + 19}{\left(x + 1\right) \left(x + 5\right)}$

The common denominator is $(x+5)(x+1)$. $\dfrac{9(x+1) + 2(x+5)}{(x+5)(x+1)} = \frac{11 x + 19}{\left(x + 1\right) \left(x + 5\right)}$.

Common mistakes, and the answer each one gives:

• Added numerators and denominators separately → $\frac{11}{2 x + 6}$
• Used $(x+5+1)$ as the common denominator → $\frac{11}{x + 6}$

Answer: $16 x^{2} - 16$

$(p \pm q)^2 = p^2 \pm 2pq + q^2$ and $(p+q)(p-q) = p^2 - q^2$. Here that gives $16 x^{2} - 16$.

Common mistakes, and the answer each one gives:

• Kept a middle term; in this product the two cancel → $16 x^{2} + 32 x - 16$
• Added the squares instead of subtracting → $16 x^{2} + 16$
• Squared only the first term → $16 x^{2} - 4$

Answer: $\left(x + 2\right) \left(x + 8\right)$

Find two numbers whose PRODUCT is $16$ and whose SUM is $-10$: they are $-8$ and $-2$. So the factorisation is $\left(x + 2\right) \left(x + 8\right)$.

Common mistakes, and the answer each one gives:

• Used the numbers with their signs unchanged in the brackets → $x^{2} - 10 x + 16$
• Matched the sum to the constant and the product to the middle term → $x^{2} + 9 x - 10$

Answer: $x^{8}$

Multiplying adds the exponents and dividing subtracts them: $7 + 6 - 5 = 8$, so the answer is $x^{8}$.

Common mistakes, and the answer each one gives:

• Multiplied the exponents instead of adding them → $x^{37}$
• Divided the exponents instead of subtracting → $x^{\frac{13}{5}}$

Answer: $\frac{x - 1}{x - 5}$

Dividing by a fraction is multiplying by its reciprocal: $\dfrac{x - 1}{x + 1} \times \dfrac{x + 1}{x - 5}$. The $x + 1$ then cancels, leaving $\frac{x - 1}{x - 5}$.

Common mistakes, and the answer each one gives:

• Multiplied straight across without flipping the second fraction → $\frac{x^{2} - 6 x + 5}{x^{2} + 2 x + 1}$
• Flipped the FIRST fraction instead of the second → $\frac{x^{2} + 2 x + 1}{x^{2} - 6 x + 5}$