Question

Factorise: $x^2+\frac{x}{4}-\frac{1}{8}$

Answer 1

We need to factorise $x^2+\frac{x}{4}-\frac{1}{8}$:

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

Hence. the factors of the given polynomial $x^2+\frac{x}{4}-\frac{1}{8}$ are $\left(x+\frac{1}{2}\right)$ and $\left(x-\frac{1}{4}\right)$