How do you factor completely: 49t^6 - 4k^8?

Jul 13, 2015

This is a difference of squares:

$49 {t}^{6} - 4 {k}^{8} = {\left(7 {t}^{3}\right)}^{2} - {\left(2 {k}^{4}\right)}^{2} = \left(7 {t}^{3} - 2 {k}^{4}\right) \left(7 {t}^{3} + 2 {k}^{4}\right)$

Explanation:

The difference of squares identity is:

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

Use this with $a = 7 {t}^{3}$ and $b = 2 {k}^{4}$ as follows:

$49 {t}^{6} - 4 {k}^{8}$

$= {\left(7 {t}^{3}\right)}^{2} - {\left(2 {k}^{4}\right)}^{2}$

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

$= \left(a - b\right) \left(a + b\right)$

$= \left(7 {t}^{3} - 2 {k}^{4}\right) \left(7 {t}^{3} + 2 {k}^{4}\right)$