# How do you factor completely a^2 - 121b^2?

Jun 13, 2018

${a}^{2} - 121 {b}^{2} = \left(a - 11 b\right) \left(a + 11 b\right)$

#### Explanation:

The difference of sqares identity can be written:

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

Given:

${a}^{2} - 121 {b}^{2}$

Note that both ${a}^{2}$ and $121 {b}^{2} = {\left(11 b\right)}^{2}$ are perfect sqares.

So we can use the difference of squares identity with $A = a$ and $B = 11 b$ to find:

${a}^{2} - 121 {b}^{2} = {a}^{2} - {\left(11 b\right)}^{2} = \left(a - 11 b\right) \left(a + 11 b\right)$

Jun 13, 2018

Shown below...

#### Explanation:

Use difference of two squares:

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

$\implies \left({\left(a\right)}^{2} - {\left(11 b\right)}^{2}\right)$

$\implies \left(a + 11 b\right) \left(a - 11 b\right)$