# How do you factor 7h^4-7p^4?

May 27, 2017

$7 \left({h}^{2} + {p}^{2}\right) \left(h + p\right) \left(h - p\right)$

#### Explanation:

When factorising :

1) Firstly, look for common factors

2) Secondly look for difference of squares

In this case

$7 {h}^{4} - 7 {p}^{4}$

by inspection we see $7$ is a common factor, so:

$= 7 \left({h}^{4} - {p}^{4}\right)$

now look for difference of squares

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

if both powers are even then we have DoS

$7 \left({h}^{4} - {p}^{4}\right) = 7 \left({h}^{2} + {p}^{2}\right) \left({h}^{2} - {p}^{2}\right)$

now look and see if we can factorise anything further.

the first bracket can not be factorised using real numbers, but the second bracket is DoS again.

so we have

$7 \left({h}^{2} + {p}^{2}\right) \left(h + p\right) \left(h - p\right)$

and that is now fully factorised.

***If you allow complex numbers then the first bracket can be factorised using DoS

$= 7 \left(h + i p\right) \left(h - i p\right) \left(h + p\right) \left(h - p\right)$