Question
Factorise the following:
$\Big(2\text{x}+\frac{1}{3}\Big)^2-\Big(\text{x}-\frac{1}{2}\Big)^2$
$\Big(2\text{x}+\frac{1}{3}\Big)^2-\Big(\text{x}-\frac{1}{2}\Big)^2$
✓
Answer
$\Big(2\text{x}+\frac{1}{3}\Big)^2-\Big(\text{x}-\frac{1}{2}\Big)^2$
$=\Bigg[\Big(2\text{x}+\frac{1}{3}\Big)-\Big(\text{x}-\frac{1}{2}\Big)\Bigg]\Bigg[\Big(2\text{x}+\frac{1}{3}\Big)+\Big(\text{x}-\frac{1}{2}\Big)\Bigg]$
[Using identity, $\mathrm{a}^2-\mathrm{b}^2=(\mathrm{a}-\mathrm{b})(\mathrm{a}+\mathrm{b})$ ]
$=\Big(2\text{x}-\text{x}+\frac{1}{3}+\frac{1}{2}\Big)\Big(2\text{x}+\text{x}+\frac{1}{3}-\frac{1}{2}\Big)$
$=\Big(\text{x}+\frac{2+3}{6}\Big)\Big(3\text{x}+\frac{2-3}{6}\Big)$
$=\Big(\text{x}+\frac{5}{6}\Big)\Big(3\text{x}-\frac{1}{6}\Big)$
$=\Bigg[\Big(2\text{x}+\frac{1}{3}\Big)-\Big(\text{x}-\frac{1}{2}\Big)\Bigg]\Bigg[\Big(2\text{x}+\frac{1}{3}\Big)+\Big(\text{x}-\frac{1}{2}\Big)\Bigg]$
[Using identity, $\mathrm{a}^2-\mathrm{b}^2=(\mathrm{a}-\mathrm{b})(\mathrm{a}+\mathrm{b})$ ]
$=\Big(2\text{x}-\text{x}+\frac{1}{3}+\frac{1}{2}\Big)\Big(2\text{x}+\text{x}+\frac{1}{3}-\frac{1}{2}\Big)$
$=\Big(\text{x}+\frac{2+3}{6}\Big)\Big(3\text{x}+\frac{2-3}{6}\Big)$
$=\Big(\text{x}+\frac{5}{6}\Big)\Big(3\text{x}-\frac{1}{6}\Big)$
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