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Mathematics 20 Online
OpenStudy (anonymous):

This can be factored. but how ? x^4 + 2 x^2 y^2 + 9y ^4

OpenStudy (anonymous):

@Mertsj

OpenStudy (mertsj):

What makes you think it can be factored?

OpenStudy (mertsj):

There are no factors of 9 whose sum is 2

OpenStudy (anonymous):

my profesor said there is a technique for this, but not quad. formula

OpenStudy (mertsj):

So you want the factors based on the roots?

OpenStudy (anonymous):

he said diff of squares

OpenStudy (anonymous):

we thought its prime. but he insisted that it can.

OpenStudy (mertsj):

(x^2-2xy+3y^2)(x^2+2xy+3y^2) according to wolframalpha.com

OpenStudy (anonymous):

@amistre64 @jim_thompson5910

OpenStudy (anonymous):

how about x ^4 + 4y ^4 ? can you show process ?

OpenStudy (anonymous):

@Luis_Rivera

OpenStudy (mertsj):

ok. Try this trick: \[x^4+2x^2y^2+9y^4=x^4+6x^2y^2+9y^4-4x^2y^2=(x^2+3y^2)^2-(2xy)^2\]

OpenStudy (anonymous):

so i would expand the ( x^2 +3y^2)^2 right ?

OpenStudy (mertsj):

No. You would replace (x^2+3y^2 ) with a. Replace 2xy with b. Then you would factor that and after you factored it, resubstitute.

OpenStudy (mertsj):

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