# How do you factor 9b^4-16c^4?

Dec 24, 2015

Recognize that this is a difference of squares, which factor as follows:

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

$9 {b}^{4} - 16 {c}^{4}$

$\implies {\left(3 {b}^{2}\right)}^{2} - {\left(4 {c}^{2}\right)}^{2} = \left(3 {b}^{2} + 4 {c}^{2}\right) \left(3 {b}^{2} - 4 {c}^{2}\right)$

Note that this is a fine final answer, but that $3 {b}^{2} - 4 {c}^{2}$ can also be treated as a difference of squares.

$3 {b}^{2} - 4 {c}^{2}$

$\implies {\left(b \sqrt{3}\right)}^{2} - {\left(2 c\right)}^{2} = \left(b \sqrt{3} + 2 c\right) \left(b \sqrt{3} - 2 c\right)$

So,

$9 {b}^{4} - 16 {c}^{4} = \left(3 {b}^{2} + 4 {c}^{2}\right) \left(b \sqrt{3} + 2 c\right) \left(b \sqrt{3} - 2 c\right)$

or

$9 {b}^{4} - 16 {c}^{4} = \left(3 {b}^{2} + 4 {c}^{2}\right) \left(3 {b}^{2} - 4 {c}^{2}\right)$