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

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Answer 1 · 2017-05-27T10:03:25

#7(h^2+p^2)(h+p)(h-p)#

When factorising :

1) Firstly, look for common factors

2) Secondly look for difference of squares

In this case

#7h^4-7p^4#

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

#=7(h^4-p^4)#

now look for difference of squares

#ie" "a^2-b^2=(a+b)(a-b)#

if both powers are even then we have DoS

#7(h^4-p^4)=7(h^2+p^2)(h^2-p^2)#

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(h^2+p^2)(h+p)(h-p)#

and that is now fully factorised.

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

#=7(h+ip)(h-ip)(h+p)(h-p)#