Question
Factorize: $2\text{x}^2-\frac{5}{6}\text{x}+\frac{1}{12}$
✓
Answer
$2\text{x}^2-\frac{5}{6}\text{x}+\frac{1}{12}$
Splitting the middle term, $=2\text{x}^2-\text{x}^2-\text{x}^3+\frac{1}{12}$
$\Big[\therefore-\frac{5}{6}=-\frac{1}{2}-\frac{1}{3} \ \text{also} \ -\frac{1}{2}\times-\frac{1}{3}=2\times\frac{1}{12}\Big]$
$=\text{x}\Big(2\text{x}-\frac{1}{2}\Big)-\frac{1}{6}\Big(2\text{x}-\frac{1}{2}\Big)$
$=\Big(2\text{x}-\frac{1}{2}\Big)\Big(\text{x}-\frac{1}{6}\Big)$
$\therefore2\text{x}^2-56\text{x}+\frac{1}{12}=\Big(2\text{x}-\frac{1}{2}\Big)\Big(\text{x}-\frac{1}{6}\Big)$
Splitting the middle term, $=2\text{x}^2-\text{x}^2-\text{x}^3+\frac{1}{12}$
$\Big[\therefore-\frac{5}{6}=-\frac{1}{2}-\frac{1}{3} \ \text{also} \ -\frac{1}{2}\times-\frac{1}{3}=2\times\frac{1}{12}\Big]$
$=\text{x}\Big(2\text{x}-\frac{1}{2}\Big)-\frac{1}{6}\Big(2\text{x}-\frac{1}{2}\Big)$
$=\Big(2\text{x}-\frac{1}{2}\Big)\Big(\text{x}-\frac{1}{6}\Big)$
$\therefore2\text{x}^2-56\text{x}+\frac{1}{12}=\Big(2\text{x}-\frac{1}{2}\Big)\Big(\text{x}-\frac{1}{6}\Big)$
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