# 2.2.3 Factorisation and Algebraic Fractions, PT3 Practice

Question 9:
(a) Factorise 12x2 – 27y2
(b)   Simplify $\frac{3{m}^{2}-10m+3}{{m}^{2}-9}÷\frac{3m-1}{m+3}.$

Solution:
(a)
12x– 27y2 = 3 (4x2 – 9y2)
= 3(2x – 3y) (2x + 3y)

(b)
$\begin{array}{l}\frac{3{m}^{2}-10m+3}{{m}^{2}-9}÷\frac{3m-1}{m+3}=\frac{\overline{)\left(3m-1\right)}\overline{)\left(m-3\right)}}{\overline{)\left(m+3\right)}\overline{)\left(m-3\right)}}×\frac{\overline{)m+3}}{\overline{)3m-1}}\\ \text{}=1\end{array}$

Question 10:

Solution:
$\begin{array}{l}\frac{8m+mn}{3m}÷\frac{{n}^{2}-64}{24}\\ =\frac{8m+mn}{3m}×\frac{24}{{n}^{2}-64}\\ =\frac{m\left(8+n\right)}{3m}×\frac{24}{{n}^{2}-{8}^{2}}\\ =\frac{\overline{)m}\overline{)\left(8+n\right)}}{\overline{)3}\overline{)m}}×\frac{{\overline{)24}}_{8}}{\left(n-8\right)\overline{)\left(n+8\right)}}\\ =\frac{8}{n-8}\end{array}$

Question 11:
(a)
Expand: x (2 + y)

(b)

Solution:
(a)
x (2 + y) = 2x + xy

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