Question

Factorise:
$\frac{2}{3}{x}^2-\frac{17}{3}x-28$

Answer 1

$$\begin{gathered} \frac{2}{3}{x}^2-\frac{17}{3}x-28=\frac{2}{3}{x}^2-8x+\frac{7}{3}x-28 \\ =2x\left(\frac{1}{3}x-4\right)+7\left(\frac{1}{3}x-4\right) \\ =\left(\frac{1}{3}x-4\right)\left(2x+7\right) \end{gathered}$$

Hence, factorisation of $\frac{2}{3}{x}^2-\frac{17}{3}x-28$ is $\left(\frac{1}{3}x-4\right)\left(2x+7\right)$.