Question
Factorise: $\frac{1}{3}\text{x}^2-2\text{x}-9$
✓
Answer
$\frac{1}{3}\text{x}^2-2\text{x}-9$
$=\frac{1}{3}\text{x}^2-3\text{x}+\text{x}-9$
$=\text{x}\Big(\frac{\text{x}}{3}-3\Big)+(\text{x}-9)$
$=\frac{\text{x}}{3}(\text{x}-9)+(\text{x}-9)$
$=(\text{x}-9)\Big(\frac{\text{x}}{3}+1\Big)$
$=(\text{x}-9)\frac{(\text{x}+3)}{3}$
$=\frac{1}{3}(\text{x}-9)(\text{x}+3)$
$=\frac{1}{3}\text{x}^2-3\text{x}+\text{x}-9$
$=\text{x}\Big(\frac{\text{x}}{3}-3\Big)+(\text{x}-9)$
$=\frac{\text{x}}{3}(\text{x}-9)+(\text{x}-9)$
$=(\text{x}-9)\Big(\frac{\text{x}}{3}+1\Big)$
$=(\text{x}-9)\frac{(\text{x}+3)}{3}$
$=\frac{1}{3}(\text{x}-9)(\text{x}+3)$
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