# How do you factor 20x^2 - 5?

Aug 29, 2016

$20 {x}^{2} - 5 = 5 \left(4 {x}^{2} - 1\right)$ $=$ $5 \left(2 x - 1\right) \left(2 x + 1\right)$

#### Explanation:

So we have $20 {x}^{2} - 5 = 5 \left(4 {x}^{2} - 1\right)$; dividing thru by $5$.

$5 \left(4 {x}^{2} - 1\right)$

And $4 {x}^{2} - 1$ is the difference between 2 squares, $2 x$ and $1$.

$\left(4 {x}^{2} - 1\right)$ $=$ $\left(2 x - 1\right) \left(2 x + 1\right)$

Putting it all together:

$20 {x}^{2} - 5 = 5 \left(2 x - 1\right) \left(2 x + 1\right)$

So there are roots at $\pm \frac{1}{2}$.