# How do you factor 1/49 - x^2?

Mar 5, 2018

$\left(\frac{1}{49} - {x}^{2}\right)$ =
$\left(\frac{1}{7} + x\right) \cdot \left(\frac{1}{7} - x\right)$

#### Explanation:

$\frac{1}{49}$ is the same as (${1}^{2} / {7}^{2}$). $\frac{1}{49}$ and ${x}^{2}$ are both perfect squares so we use the difference of squares (factoring).
$\left({a}^{2} - {b}^{2}\right) = \left(a + b\right) \cdot \left(a - b\right)$
If you 'foil' the answer, you get $\left({a}^{2} - a b + a b - {b}^{2}\right)$. $\left(- a b\right)$ and $\left(+ a b\right)$ cancel, resulting in the original $\left({a}^{2} - {b}^{2}\right)$.
For the problem, the same process applies:
1. $\left(\frac{1}{49} - {x}^{2}\right)$
2. $\left(\frac{1}{7} + x\right) \cdot \left(\frac{1}{7} - x\right)$
To check, $\left[\left(\frac{1}{7} \cdot \frac{1}{7}\right) - \left(\frac{1}{7}\right) x + \left(\frac{1}{7}\right) x - \left(x \cdot x\right)\right] = \left(\frac{1}{49} - {x}^{2}\right)$