# How do you factor completely a^2 - 4b^2 - 4a + 4?

Sep 2, 2016

${a}^{2} - 4 {b}^{2} - 4 a + 4 = \left(a - 2 b - 2\right) \left(a + 2 b - 2\right)$

#### Explanation:

The difference of squares identity can be written:

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

We use this with $A = \left(a - 2\right)$ and $B = 2 b$ as follows:

${a}^{2} - 4 {b}^{2} - 4 a + 4 = \left({a}^{2} - 4 a + 4\right) - 4 {b}^{2}$

$\textcolor{w h i t e}{{a}^{2} - 4 {b}^{2} - 4 a + 4} = {\left(a - 2\right)}^{2} - {\left(2 b\right)}^{2}$

$\textcolor{w h i t e}{{a}^{2} - 4 {b}^{2} - 4 a + 4} = \left(\left(a - 2\right) - 2 b\right) \left(\left(a - 2\right) + 2 b\right)$

$\textcolor{w h i t e}{{a}^{2} - 4 {b}^{2} - 4 a + 4} = \left(a - 2 b - 2\right) \left(a + 2 b - 2\right)$