Question

Factorise:
$6{\left(2x-\frac{3}{x}\right)}^2+7\left(2x-\frac{3}{x}\right)-20$

Answer 1

$$\begin{gathered} 6{\left(2x-\frac{3}{x}\right)}^2+7\left(2x-\frac{3}{x}\right)-20=6{\left(2x-\frac{3}{x}\right)}^2+15\left(2x-\frac{3}{x}\right)-8\left(2x-\frac{3}{x}\right)-20 \\ =\left[3\left(2x-\frac{3}{x}\right)\right]\left[2\left(2x-\frac{3}{x}\right)+5\right]-4\left[2\left(2x-\frac{3}{x}\right)+5\right] \\ =\left[2\left(2x-\frac{3}{x}\right)+5\right]\left[3\left(2x-\frac{3}{x}\right)-4\right] \\ =\left(4x-\frac{6}{x}+5\right)\left(6x-\frac{9}{x}-4\right) \end{gathered}$$

Hence, factorisation of $6{\left(2x-\frac{3}{x}\right)}^2+7\left(2x-\frac{3}{x}\right)-20$ is $\left(4x-\frac{6}{x}+5\right)\left(6x-\frac{9}{x}-4\right)$.