How do you factor 36a ^ { 2} - 36a b - 27b ^ { 2}?

May 19, 2017

9(2a - 3b)(2a + b)

Explanation:

Given, $36 {a}^{2} - 36 a b - 27 {b}^{2} = 9 \left(4 {a}^{2} - 4 a b - 3 {b}^{2}\right)$

$\Rightarrow 9 \left(4 {a}^{2} - 6 a b + 2 a b - 3 {b}^{2}\right)$

$\Rightarrow 9 \left[2 a \left(2 a - 3 b\right) + b \left(2 a - 3 b\right)\right]$

$\Rightarrow 9 \left(2 a - 3 b\right) \left(2 a + b\right)$

May 19, 2017

$9 \left(2 a + b\right) \left(2 a - 3 b\right)$

Explanation:

$36 {a}^{2} - 36 a b - 27 {b}^{2} = 9 \left(4 {a}^{2} - 4 a b - 3 {b}^{2}\right)$
->$9$ is a highest common factor for $36 \mathmr{and} 27$

we try to find the product of factor for (4 * 3 = 12), then we get
$12 = 6 \cdot 2 , - 6 + 2 = - 4$
thereofore we replace $- 4 a b \to - 6 a b + 2 a b$

$9 \left(4 {a}^{2} - 4 a b - 3 {b}^{2}\right) = 9 \left(4 {a}^{2} - 6 a b + 2 a b - 3 {b}^{2}\right)$
We separate into 2 portions and factor it.
1. $4 {a}^{2} - 6 a b = 2 a \left(2 a - 3 b\right)$
2. $2 a b - 3 {b}^{2} = b \left(2 a - 3 b\right)$
it is correct when both of them have a same of $\left(2 a - 3 b\right)$,
therefore,
$\left(4 {a}^{2} - 6 a b + 2 a b - 3 {b}^{2}\right) = 2 a \left(2 a - 3 b\right) + b \left(2 a - 3 b\right) = \left(2 a + b\right) \left(2 a - 3 b\right)$

therefore,
$9 \left(4 {a}^{2} - 6 a b + 2 a b - 3 {b}^{2}\right) = 9 \left(2 a + b\right) \left(2 a - 3 b\right)$