# Thread: Dividing polynomials with a remainder

Divide: by

pllzzzz help

2. ## Re: Dividing polynomials with a remainder

Factorize the numerator!

By grouping we get

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

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.

3. ## Re: Dividing polynomials with a remainder

You can almost do this "by inspection".

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

The rest of the numerator is $x^3y-xy^3$, which factors into $xy(x^2 - y^2)$.

4. ## Re: Dividing polynomials with a remainder

Hello, koolaid123!

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

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

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