# Math Help - Dividing polynomials

1. ## Dividing polynomials

Hi all,

Got another problem I'd love to get some feedback on. I wasn't sure how to do long division code on here so I decided to use paint :P.

I'm unsure how to proceed because I'm pretty sure I can't subtract x^2y - xy as there not like terms but there has to be away so hopefully someone can point that out . Its prob something simple I'm missing as per usual lol.

2. Originally Posted by Substince
Hi all,

Got another problem I'd love to get some feedback on. I wasn't sure how to do long division code on here so I decided to use paint :P.

I'm unsure how to proceed because I'm pretty sure I can't subtract x^2y - xy as there not like terms but there has to be away so hopefully someone can point that out . Its prob something simple I'm missing as per usual lol.

Are you sure that the first term is $x^2$ ? If it is actually $x^3$ then the problem becomes very much simpler, because $x^3 + x^2y -9xy^2-9y^3 = (x+y)(x^2-9y^2).$

If you really want the first term to be $x^2$ then the problem is definitely not "something simple". It takes you into some quite advanced mathematics involving Gröbner bases.

3. Hello, Substince!

I quite certain that you have a typo.

As written, the division is possible, but very awkward.

. . - - - . . . . . . . . . .
I bet that it says: . $(x^3 + x^2y - 9x^2y - 9y^3) \div (x + y)$

. . $\begin{array}{ccccccccc}
&&&& x^2 &&& - & 9y^3 \\
&& -- & -- & -- & -- & -- & -- & -- \\
x+y & \bigg) & x^3 &+& x^2y &-& 9xy^2 &-& 9y^3 \\
&& x^3 &+& x^2y \\
&& -- & -- & -- \\
&&&&& -& 9xy^2 &-& 9y^3 \\
&&&&& -&9xy^2 &-& 9y^3 \\
&&&&& -- & -- & -- & -- \end{array}$

edit: Too slow again.
.

4. It can be divided as written.
But is there anyone this planet who would want to? .**

Note that: . $x^2 + x^2y - 9xy^2 - 9y^3 \;=\;(y+1)x^2 - 9xy^2 - 9y^3$

. . $\begin{array}{cccccccc}
&&&& (y+1)x &-& (10y+1)y \\
&& ---- & -- & ----- & -- & ----- \\
x+y & \bigg) & (y+1)x^2 &-& 9xy^2 &-& 9y^2 \\
&& (y+1)x^2 &+& (y+1)xy \\
&& ---- & -- & ----- \\
&&& - & (10y+1)xy & - & 9y^2 \\
&&& - & (10y+1)xy &-& (10y+1)y^2 \\
&&& -- & ----- & -- & ----- \\
&&&&&& 10y^3 - 8y^2

\end{array}$

** .Okay, it's me . . .

5. Thanks for the comments guys! You have confirmed my suspicions as I also thought it was x^3 but the book does state x^2 and considering this in an intro to dividing polynomials it must be a typo. My teacher did warn me that there was errors in the book when she gave it to me (lol)... and this is one hell of an error! I've been trying to figure this thing out for days :P.

Thanks to you all I'm just gonna switch x^2 with x^3 and call it a day.