# How do you factor completely a^2 - 1 - 4b^2 - 4b?

Dec 18, 2016

${a}^{2} - 1 - 4 {b}^{2} - 4 b = \left(a - 2 b - 1\right) \left(a + 2 b + 1\right)$

#### Explanation:

The difference of squares identity can be written:

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

Use this with $A = a$ and $B = 2 b + 1$ as follows:

${a}^{2} - 1 - 4 {b}^{2} - 4 b = {a}^{2} - \left(4 {b}^{2} + 4 b + 1\right)$

$\textcolor{w h i t e}{{a}^{2} - 1 - 4 {b}^{2} - 4 b} = {a}^{2} - {\left(2 b + 1\right)}^{2}$

$\textcolor{w h i t e}{{a}^{2} - 1 - 4 {b}^{2} - 4 b} = \left(a - \left(2 b + 1\right)\right) \left(a + \left(2 b + 1\right)\right)$

$\textcolor{w h i t e}{{a}^{2} - 1 - 4 {b}^{2} - 4 b} = \left(a - 2 b - 1\right) \left(a + 2 b + 1\right)$