# Factoring Completely...

• April 18th 2008, 08:52 PM
eraser851
Factoring Completely...
Alright, I've got one more question. Could someone explain how to factor this completely?

$16z^4-x^4z^4$
• April 18th 2008, 09:52 PM
Aryth
Well, we have:

$16z^4 - x^4z^4$

1. Factor out a $z^4$

$z^4(16 - x^4)$

2. Factor the difference of squares:

$z^4(4 + x^2)(4 - x^2)$

3. Factor the other difference of squares:

$z^4(4 + x^2)((2 + x)(2 - x))$

And there you go.
• April 18th 2008, 09:53 PM
eraser851
Ah, thank you so much!
That helps!