How do you factor 36n²-9?

Mar 3, 2016

$36 {n}^{2} - 9 = 9 \cdot \left(2 n - 1\right) \cdot \left(2 n + 1\right)$

Explanation:

$36 {n}^{2} - 9 = 9 \cdot \left(4 {n}^{2} - 1\right) = 9 \cdot \left({\left(2 n\right)}^{2} - {1}^{2}\right) =$
$= 9 \cdot \left(2 n - 1\right) \cdot \left(2 n + 1\right)$

First we can see that both expressions are divisible by $9$.

Next step is to see that in the parenethesis we have the substraction of 2 squares which can be written as:

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

Mar 3, 2016

9(2n - 1 )(2n + 1 )

Explanation:

First step is to take out common factor of 9.

$36 {n}^{2} - 9 = 9 \left(4 {n}^{2} - 1\right)$

now $4 {n}^{2} - 1 \textcolor{b l u e}{\text{ is a difference of squares }}$

Consider how this factors: ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

hence: $4 {n}^{2} - 1 = \left(2 n - 1\right) \left(2 n + 1\right)$

$\Rightarrow 36 {n}^{2} - 9 = 9 \left(2 n - 1\right) \left(2 n + 1\right)$