2.2.1 Factorisation and Algebraic Fractions, PT3 Practice

2.2.1 Factorisation and Algebraic Fractions, PT3 Practice

Question 1:
(a)(i) Factorise 18a + 3
(a)(ii) Expand –3 (–y + 5)
(b) Express $\frac{5}{6y}-\frac{3x-5}{12y}$ as a single fraction in its simplest form.

Solution:
(a)(i) 18+ 3 = 3(6a + 1)

(a)(ii) –3 (–+ 5) = 3y – 15

(b)
$\begin{array}{l}\frac{5}{6y}-\frac{3x-5}{12y}=\frac{5×2}{6y×2}-\frac{\left(3x-5\right)}{12y}\\ \text{}=\frac{10-3x+5}{12y}\\ \text{}=\frac{15-3x}{12y}\\ \text{}=\frac{\overline{)3}\left(5-x\right)}{4\overline{)12}y}\\ \text{}=\frac{5-x}{4y}\end{array}$

Question 2:
(a) Expand:
(i) 3 (–a + c)
(ii) –5 (c)
(b) Factorise 4+ 2
(c) Simplify: $\frac{3x+6}{{x}^{2}-4}÷\frac{x+2}{x-2}$

Solution:
(a)(i) 3 (–+ c) = –3a + 3c

(a)(ii) –5 (c) = –5a + 5c

(c)
$\begin{array}{l}\frac{3x+6}{{x}^{2}-4}÷\frac{x+2}{x-2}=\frac{3\left(\overline{)x+2}\right)}{\left(x+2\right)\overline{)\left(x-2\right)}}×\frac{\overline{)x-2}}{\overline{)x+2}}\\ \text{}=\frac{3}{x+2}\end{array}$

Question 3:
(a) Factorise:
(i) 5m + 25
(ii) 7x + 9xy
(b) Simplify: $\frac{4x-12}{4y}÷\frac{{x}^{2}-9}{yz}$

Solution:
(a)(i)5m + 25 = 5 (m + 5)

(a)(ii)7x + 9xy = x (7 + 9y)

$\begin{array}{l}\text{(b)}\frac{4x-12}{4y}÷\frac{{x}^{2}-9}{yz}=\frac{\overline{)4}\overline{)\left(x-3\right)}}{\overline{)4}\overline{)y}}×\frac{\overline{)y}z}{\left(x+3\right)\overline{)\left(x-3\right)}}\\ \text{}=\frac{z}{x+3}\end{array}$

Question 4:
(a) Factorise completely:
4 – 100n2
(b) Express $\frac{4}{5x}-\frac{7-10y}{15x}$  as a single fraction in its simplest form.

Solution:
(a)
4 – 100n2= (2 + 10n)(2 – 10n)
(b)
$\begin{array}{l}\frac{4}{5x}-\frac{7-10y}{15x}=\frac{4×3}{5x×3}-\frac{\left(7-10y\right)}{15x}\\ \text{}=\frac{12-7+10y}{15x}\\ \text{}=\frac{5+10y}{15x}\\ \text{}=\frac{\overline{)5}\left(1+2y\right)}{3\overline{)15}x}\\ \text{}=\frac{1+2y}{3x}\end{array}$