# Factor this unless it is fully factored?

## $18 r {s}^{2} - 2 r$

Apr 30, 2018

$2 r \left(9 {s}^{2} - 1\right)$

#### Explanation:

$18 r {s}^{2} - 2 r = 2 r \left(9 {s}^{2} - 1\right)$

Apr 30, 2018

$18 r {s}^{2} - 2 r = 2 r \left(3 s - 1\right) \left(3 s + 1\right)$

#### Explanation:

Given:

$18 r {s}^{2} - 2 r$

Note that both terms are divisible by $2 r$, so we can separate that out as a factor...

$18 r {s}^{2} - 2 r = 2 r \left(9 {s}^{2} - 1\right)$

Note that both $9 {s}^{2} = {\left(3 s\right)}^{2}$ and $1 = {1}^{2}$ are perfect squares.

So we can use the difference of squares identity:

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

with $A = 3 s$ and $B = 1$ to find:

$9 {s}^{2} - 1 = {\left(3 s\right)}^{2} - {1}^{2} = \left(3 s - 1\right) \left(3 s + 1\right)$

Putting it all together:

$18 r {s}^{2} - 2 r = 2 r \left(3 s - 1\right) \left(3 s + 1\right)$