Question

Factorise $\left(i\right)x^2+3\sqrt{3}x+6$
$\left(ii\right)6x^2+\sqrt{5}x-10$

Answer 1

(i)$x^2+3\sqrt{3}x+6$

This can be written as

$\Rightarrow x^2+\sqrt{3}x+2\sqrt{3}x+6$

$\Rightarrow x(x+\sqrt{3})+2\sqrt{3}(x+\sqrt{3})$

$\Rightarrow (x+\sqrt{3})(x+2\sqrt{3})$

Therefore, $\Rightarrow x^2+3\sqrt{3}x+6=(x+\sqrt{3})(x+2\sqrt{3})$

(ii)$6x^2+\sqrt{5}x-10$

This can be written as

$\Rightarrow 6(x^2+\frac{\sqrt{5}}{6}x-\frac{5}{3})$

$\Rightarrow 6(x^2-\frac{\sqrt{5}}{2}x+\frac{2\sqrt{5}}{3}x-\frac{5}{3})$

$\Rightarrow 6\left[x\right(x-\frac{\sqrt{5}}{2})+\frac{2\sqrt{5}}{3}(x-\frac{\sqrt{5}}{2}\left)\right]$

$\Rightarrow 6(x+\frac{2\sqrt{5}}{3})(x-\frac{\sqrt{5}}{2})$

Therefore, $\Rightarrow 6x^2+\sqrt{5}x-10=6(x+\frac{2\sqrt{5}}{3})(x-\frac{\sqrt{5}}{2})$