# How do you factor completely 25a^4-49b^2?

Apr 19, 2017

$25 {a}^{4} - 49 {b}^{2} = \left(5 {a}^{2} - 7 b\right) \left(5 {a}^{2} + 7 b\right)$

#### Explanation:

Note that both $25 {a}^{4} = {\left(5 {a}^{2}\right)}^{2}$ and $49 {b}^{2} = {\left(7 b\right)}^{2}$ are perfect squares.

The difference of squares identity can be written:

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

Using this with $A = 5 {a}^{2}$ and $B = 7 b$ we find:

$25 {a}^{4} - 49 {b}^{2} = {\left(5 {a}^{2}\right)}^{2} - {\left(7 b\right)}^{2}$

$\textcolor{w h i t e}{25 {a}^{4} - 49 {b}^{2}} = \left(5 {a}^{2} - 7 b\right) \left(5 {a}^{2} + 7 b\right)$

There are no simpler factors.