# How do you factor  16-1/4k^2?

Sep 23, 2016

$16 - \frac{1}{4} {k}^{2} = \left(4 - \frac{1}{2} k\right) \left(4 + \frac{1}{2} k\right)$

#### Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Note that $16 = {4}^{2}$ and $\frac{1}{4} {k}^{2} = {\left(\frac{1}{2} k\right)}^{2}$ are both perfect squares.

So we find:

$16 - \frac{1}{4} {k}^{2} = {4}^{2} - {\left(\frac{1}{2} k\right)}^{2} = \left(4 - \frac{1}{2} k\right) \left(4 + \frac{1}{2} k\right)$