Question
Factorize:
$\text{x}^2+2\sqrt{3}\text{x}-24$
$\text{x}^2+2\sqrt{3}\text{x}-24$
✓
Answer
$\text{x}^2+2\sqrt{3}\text{x}-24$
Splitting the middle term,
$=\text{x}^2+4\sqrt{3}\text{x}-2\sqrt{3}\text{x}-24$
$\big[\therefore2\sqrt{3}=4\sqrt{3}-2\sqrt{3} \ \text{also} \ 4\sqrt{3}\big(-2\sqrt{3}\big)=-24\big]$
$=\text{x}\big(\text{x}+4\sqrt{3}\big)-2\sqrt{3}\big(\text{x}+4\sqrt{3}\big)$
$=\big(\text{x}+4\sqrt{3}\big)\big(\text{x}-2\sqrt{3}\big)$
$\therefore\text{x}^2+2\sqrt{3}\text{x}-24$
$=\big(\text{x}+4\sqrt{3}\big)\big(\text{x}-2\sqrt{3}\big)$
Splitting the middle term,
$=\text{x}^2+4\sqrt{3}\text{x}-2\sqrt{3}\text{x}-24$
$\big[\therefore2\sqrt{3}=4\sqrt{3}-2\sqrt{3} \ \text{also} \ 4\sqrt{3}\big(-2\sqrt{3}\big)=-24\big]$
$=\text{x}\big(\text{x}+4\sqrt{3}\big)-2\sqrt{3}\big(\text{x}+4\sqrt{3}\big)$
$=\big(\text{x}+4\sqrt{3}\big)\big(\text{x}-2\sqrt{3}\big)$
$\therefore\text{x}^2+2\sqrt{3}\text{x}-24$
$=\big(\text{x}+4\sqrt{3}\big)\big(\text{x}-2\sqrt{3}\big)$
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