Question

Factorise:
16p³ − 4p

Answer 1

We have:
$$\begin{gathered} 16p^3-4p=4p\left(4p^2-1\right) \\ =4p\left\{{\left(2p\right)}^2-{\left(1\right)}^2\right\} \\ =4p\left(2p+1\right)\left(2p-1\right) \end{gathered}$$

∴ $16p^3-4p=4p\left(2p+1\right)\left(2p-1\right)$