# 2.2.2 Factorisation and Algebraic Fractions, PT3 Practice

Question 5:
(a) Simplify:
(m – 4n)(m + 4n) – m2
(b) Simplify: $\frac{3x-3y}{x+y}×\frac{2x+2y}{6x}$

Solution:
(a)
(m – 4n)(+ 4n) – m2
= m2 + 4mn – 4mn – 4n2m2
= 0

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

  
      
               
Question 6:
   (a) Simplify each of the following:
       
        $\begin{array}{l}\text{(i)}\frac{12mn}{32}\\ \text{(ii)}\frac{{x}^{2}-xy}{x}\end{array}$        
   (b) Express         $\frac{1}{2q}-\frac{2p-7}{6q}$          as a single fraction in its simplest form.
       
       

   Solution:

           $\begin{array}{l}\text{(a)(i)}\frac{12mn}{32}=\frac{3mn}{8}\\ \text{(a)(ii)}\frac{{x}^{2}-xy}{x}=\frac{\overline{)x}\left(x-y\right)}{\overline{)x}}=x-y\end{array}$        
       
       

   (b)

           $\begin{array}{l}\frac{1}{2q}-\frac{2p-7}{6q}=\frac{1×3}{2q×3}-\frac{\left(2p-7\right)}{6q}\\ \text{}=\frac{3-2p+7}{6q}\\ \text{}=\frac{10-2p}{6q}\\ \text{}=\frac{\overline{)2}\left(5-p\right)}{3\overline{)6}q}\\ \text{}=\frac{5-p}{3q}\end{array}$      
 
  
 
 
 
       
       
Question 7:
       
(a) Factorise 2ae + 3af – 6de – 9df
       
(b)   Simplify         $\frac{{a}^{2}-{b}^{2}}{{\left(a+b\right)}^{2}}$        
       
       

   Solution:
       
(a)
       
2ae + 3af – 6de – 9df = (2+ 3f ) – 3d (2e + 3f)
       
= (2+ 3f ) (a – 3d)
       
       

   (b)
       
        $\begin{array}{l}\frac{{a}^{2}-{b}^{2}}{{\left(a+b\right)}^{2}}=\frac{\overline{)\left(a+b\right)}\left(a-b\right)}{\overline{)\left(a+b\right)}\left(a+b\right)}\\ \text{}=\frac{a-b}{a+b}\end{array}$      
 
  
 
 
  
      
       
Question 8:
       
(a) Factorise –8c2 – 12ac.
       
(b)   Simplify         $\frac{ae+ad-2be-2bd}{{a}^{2}-4{b}^{2}}.$        
       
       

   Solution:
       
(a)
       
–8c2– 12ac
       
= –4c (2c + 3a)
       
       

   (b)
       
        $\begin{array}{l}\frac{ae+ad-2be-2bd}{{a}^{2}-4{b}^{2}}=\frac{a\left(e+d\right)-2b\left(e+d\right)}{\left(a+2b\right)\left(a-2b\right)}\\ \text{}=\frac{\left(e+d\right)\overline{)\left(a-2b\right)}}{\left(a+2b\right)\overline{)\left(a-2b\right)}}\\ \text{}=\frac{e+d}{a+2b}\end{array}$