Factoring Polynomials: Start Here

[paulmdrdo1]

how would i start factoring this

$m^8-n^8-2m^6n^2+2n^6m^2$

 

[MarkFL]

Re: factoring polynomial

You really need to post your attempts at these, this way we can see where you are at. :D

I would look at factoring the first two terms as the difference of squares, and the last two terms have a common factor as well (I would factor out $-2m^2n^2$)...what do you get?

 

[paulmdrdo1]

Re: factoring polynomial

this is where i can get to

$(m^4+n^4)(m^4-n^4)-2n^2m^2(m^4-n^4)$

$(m^2-n^2)(m^2+n^2)(m^4+n^4)-2n^2m^2(m^2-n^2)(m^2+n^2)$

$(m-n)(m+n)(m^2+n^2)(m^4+n^4)-2n^2m^2(m-n)(m+n)(m^2+n^2)$

 

[MarkFL]

Re: factoring polynomial

I would take this path:

\(\displaystyle \left(m^4+n^4 \right)\left(m^4-n^4 \right)-2m^2n^2\left(m^4-n^4 \right)\)

\(\displaystyle \left(m^4-n^4 \right)\left(m^4-2m^2n^2+n^4 \right)\)

Do you recognize that the second factor is a square of a binomial?

 

[paulmdrdo1]

Re: factoring polynomial

yes this is answer

$(m-n)^{3}(m+n)^{3}(m^2+n^2)$

 

[MarkFL]

Re: factoring polynomial

yes this is answer

$(m-n)^{3}(m+n)^{3}(m^2+n^2)$

Yes, good job! (Sun)