# Dividing polynomials with a remainder

• Sep 25th 2012, 12:18 PM
koolaid123
Dividing polynomials with a remainder
• Sep 25th 2012, 12:22 PM
TheEmptySet
Re: Dividing polynomials with a remainder
Factorize the numerator!

By grouping we get

$\displaystyle x^4+x^3y-xy^3-y^4 = x^3(x+y)-y^3(x+y) = (x+y)(x^3-y^3)$

This should get you started. Note the 2nd factor is the difference of cubes and can be factored as well.

after you have factored both the numerator and denominator the entire denominator should reduce out.

Try to work this out and post back with your work if you get stuck.
• Sep 25th 2012, 12:55 PM
johnsomeone
Re: Dividing polynomials with a remainder
You can almost do this "by inspection".

I see $\displaystyle (x^4 - y^4)$, so I see an $\displaystyle (x^2 - y^2)$ in there.

The rest of the numerator is $\displaystyle x^3y-xy^3$, which factors into $\displaystyle xy(x^2 - y^2)$.
• Sep 25th 2012, 02:51 PM
Soroban
Re: Dividing polynomials with a remainder
Hello, koolaid123!

Quote:

$\displaystyle \text{Divide: }\:(x^4+x^3y-xy^3 - y^4) \div (x^2-y^2)$

I don't suppose you tried long division . . .

$\displaystyle \begin{array}{ccccccccccccc} &&&&&& x^2 &+& xy &+&y^2 \\ && --&--&--&--&--&--&--&--&-- \\x^2-y^2 & | & x^4 &+& x^3y && &-& xy^3 & - & y^4 \\ && x^4 && &-& x^2y^2 \\&& --&--&--&--&-- \\ &&&& x^3y &+& x^2y^2 &-& xy^3 \\ &&&& x^3y && &-& xy^3 \\ &&&& --&--&--&--&-- \\ &&&&&& x^2y^2 && &-& y^4 \\ &&&&&& x^2y^2 && &-& y^4 \\ &&&&&& --&--&--&--&-- \end{array}$