Factor this unless it is fully factored?

Question

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

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Answer 1 · 2018-04-30T06:10:36

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

$18 r s^{2} - 2 r = 2 r (9 s^{2} - 1)$ $18rs^2-2r=2r(9s^2-1)$

Answer 2 · 2018-04-30T06:11:21

$18 r s^{2} - 2 r = 2 r (3 s - 1) (3 s + 1)$ $18rs^2-2r = 2r(3s-1)(3s+1)$

Given:

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

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

$18 r s^{2} - 2 r = 2 r (9 s^{2} - 1)$ $18rs^2-2r = 2r(9s^2-1)$

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

So we can use the difference of squares identity:

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

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

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

Putting it all together:

$18 r s^{2} - 2 r = 2 r (3 s - 1) (3 s + 1)$ $18rs^2-2r = 2r(3s-1)(3s+1)$