Question
Factorize: $\text{x}^2-\sqrt{3}\text{x}-6$
✓
Answer
$\text{x}^2-\sqrt{3}\text{x}-6$ Splitting the middle term, $=\text{x}^2-2\sqrt{3}\text{x}+\sqrt{3}\text{x}-6$
$\big[\therefore-\sqrt{3}=-2\sqrt{3}+\sqrt{3} \ \text{also} \ -2\sqrt{3}\times\sqrt{3}=-6\big]$
$=\text{x}\big(\text{x}-2\sqrt{3}\big)+\sqrt{3}\big(\text{x}-2\sqrt{3}\big)$
$=\big(\text{x}-2\sqrt{3}\big)\big(\text{x}+\sqrt{3}\big)$
$\therefore\text{x}^2-\sqrt{3}\text{x}-6$
$=\big(\text{x}-2\sqrt{3}\big)\big(\text{x}+\sqrt{3}\big)$
$\big[\therefore-\sqrt{3}=-2\sqrt{3}+\sqrt{3} \ \text{also} \ -2\sqrt{3}\times\sqrt{3}=-6\big]$
$=\text{x}\big(\text{x}-2\sqrt{3}\big)+\sqrt{3}\big(\text{x}-2\sqrt{3}\big)$
$=\big(\text{x}-2\sqrt{3}\big)\big(\text{x}+\sqrt{3}\big)$
$\therefore\text{x}^2-\sqrt{3}\text{x}-6$
$=\big(\text{x}-2\sqrt{3}\big)\big(\text{x}+\sqrt{3}\big)$
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