# Factoring with exponents

• Dec 16th 2009, 07:18 AM
dkpeppard
Factoring with exponents

81y^4-w^4y^4

• Dec 16th 2009, 07:37 AM
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)\$
• Dec 16th 2009, 07:38 AM
dkpeppard
Stroodle, many thanks for your quick response and help!
• Dec 17th 2009, 06:30 AM
dkpeppard
Quote:

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?

• Dec 17th 2009, 09:45 AM
dkpeppard
Anyone else is welcome to jump in to and explain as well.

• Dec 17th 2009, 10:49 AM
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
\$
• Dec 17th 2009, 10:53 AM
dkpeppard
Quote:

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.