Question

Factorise $\sqrt{3}{x}^2+11x+6\sqrt{3}$.

Answer 1

This is of the form $a{x}^2+bx+c$, where $a=\sqrt{3};b=11$ and $c=6\sqrt{3}$. We may now observe that
$ac=\sqrt{3}\times 6\sqrt{3}=3\times 6=18$
We can choose $p=9,q=2$, so that $pq=ac$ and $p+q=b$. We thus obtain
$\sqrt{3}{x}^2+11x+6\sqrt{3}=\sqrt{3}{x}^2+2x+9x+6\sqrt{3}$
$=\sqrt{3}{x}^2+2x+3\sqrt{3}\sqrt{3}x+6\sqrt{3}=x(\sqrt{3}x+2)+3\sqrt{3}(\sqrt{3}x+2)=(\sqrt{3}x+2)(x+3\sqrt{3})$