Question

Factorize:
$\frac{1}{3}{x}^2-2x-9$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ \frac{1}{3}{x}^2-2x-9 \\ =\frac{x^2-6x-27}{3} \\ =\frac{1}{3}\left(x^2-6x-27\right) \end{gathered}$$
Now, we have to split ($-$6) into two numbers such that their sum is ($-$6) and their product is ($-$27).
Clearly, $\left(-9\right)+3=-6\mathrm{and}\left(-9\right)\times 3=-27$.

$$\begin{gathered} \therefore \frac{1}{3}{x}^2-2x-9=\frac{1}{3}\left[(x^2+-6x-27)\right] \\ =\frac{1}{3}\left[(x^2+3x-9x-27)\right] \\ =\frac{1}{3}\left[x(x+3)-9(x+3)\right] \\ =\frac{1}{3}(x+3)(x-9) \end{gathered}$$