1. ## Factoring with exponents

81y^4-w^4y^4

2. $\displaystyle 81y^4-w^4y^4$

$\displaystyle y^4(81-w^4)$

$\displaystyle y^4(9+w^2)(9-w^2)$

$\displaystyle y^4(9+w^2)(3+w)(3-w)$

$\displaystyle -y^4(w^2+9)(w+3)(w-3)$

3. Stroodle, many thanks for your quick response and help!

4. Originally Posted by Stroodle
$\displaystyle 81y^4-w^4y^4$

$\displaystyle y^4(81-w^4)$

$\displaystyle y^4(9+w^2)(9-w^2)$

$\displaystyle y^4(9+w^2)(3+w)(3-w)$

$\displaystyle -y^4(w^2+9)(w+3)(w-3)$
Stroodle,

In trying to understand how this is factored, how did you get to the second step, please?

$\displaystyle y^4(81-w^4)$

Could you break that down and explain it a little further for me, please?

5. Anyone else is welcome to jump in to and explain as well.

6. $\displaystyle (81-w^4)$

this is a difference of squares

$\displaystyle (9-w^2)(9+w^2)$
if you FOIL this the middle terms will cancel out

$\displaystyle 81 + w^2 -w^2-w^4$
leaving

$\displaystyle 81-w^4$

7. Originally Posted by bigwave
$\displaystyle (81-w^4)$

this is a difference of squares

$\displaystyle (9-w^2)(9+w^2)$
if you FOIL this the middle terms will cancel out

$\displaystyle 81 + w^2 -w^2-w^4$
leaving

$\displaystyle 81-w^4$
Thank you, Bigwave.