Please help me factor the following: 81y^4-w^4y^4 Thanks in advance for your help.
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$\displaystyle 81y^4-w^4y^4$ $\displaystyle y^4(81-w^4)$ $\displaystyle y^4(9+w^2)(9-w^2)$ $\displaystyle y^4(9+w^2)(3+w)(3-w)$ $\displaystyle -y^4(w^2+9)(w+3)(w-3)$
Stroodle, many thanks for your quick response and help!
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? Thank you in advance for your help.
Last edited by dkpeppard; Dec 17th 2009 at 06:30 AM. Reason: updated equation referenced
Anyone else is welcome to jump in to and explain as well. Thanks in advance.
$\displaystyle (81-w^4)$ 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 $
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.
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