Question

Factorise the polynomial $x^2+3\sqrt{3}x+6$ by splitting the middle term

• A. $(x+\sqrt{3})(x-2\sqrt{3})$
• B. $(x-\sqrt{3})(x+2\sqrt{3})$
• C. $(x+\sqrt{3})(x+2\sqrt{3})$
• D. $(x-\sqrt{3})(x-2\sqrt{3})$

Answer 1

The correct option is C $(x+\sqrt{3})(x+2\sqrt{3})$

We have, $x^2+3\sqrt{3}x+6=x^2+(\sqrt{3}+2\sqrt{3})x+6$ ($\because$ $\sqrt{3}+2\sqrt{3}=3\sqrt{3}$ and $\left(\sqrt{3}\right)\left(2\sqrt{3}\right)=6$ )
$=x^2+\sqrt{3}x+2\sqrt{3}x+\left(\sqrt{3}\right)\left(2\sqrt{3}\right)$
$=\{x^2+\sqrt{3}x\}+\{2\sqrt{3}x+(\sqrt{3}\left)\right(2\sqrt{3}\left)\right\}$
$=x(x+\sqrt{3})+2\sqrt{3}(x+\sqrt{3})$
$=(x+\sqrt{3})(x+2\sqrt{3})$
$\therefore x^2+3\sqrt{3}x+6=(x+\sqrt{3})(x+2\sqrt{3})$