Question
Factorise: $\sqrt{5}\text{x}^2+2\text{x}-3\sqrt{5}$
✓
Answer
We have: We have to split $2$ into numbers such that their sum is $2$ and product is $(-15)$,
i.e., $\sqrt{5}\times\big(-3\sqrt{5}\big)$ Clearly, $5 + (-3) = 2$ and $5 \times (-3) = -15$.
$\therefore\ \sqrt{5}\text{x}^2+2\text{x}-3\sqrt{5}$
$=\sqrt{5}\text{x}^2+5\text{x}-3\text{x}-3\sqrt{5}$
$=\sqrt{5}\text{x}\big(\text{x}+\sqrt{5}\big)-3\big(\text{x}+\sqrt{5}\big)$
$=\big(\text{x}+\sqrt{5}\big)\big(\sqrt{5}\text{x}-3\big)$
i.e., $\sqrt{5}\times\big(-3\sqrt{5}\big)$ Clearly, $5 + (-3) = 2$ and $5 \times (-3) = -15$.
$\therefore\ \sqrt{5}\text{x}^2+2\text{x}-3\sqrt{5}$
$=\sqrt{5}\text{x}^2+5\text{x}-3\text{x}-3\sqrt{5}$
$=\sqrt{5}\text{x}\big(\text{x}+\sqrt{5}\big)-3\big(\text{x}+\sqrt{5}\big)$
$=\big(\text{x}+\sqrt{5}\big)\big(\sqrt{5}\text{x}-3\big)$
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