Dividing polynomial by binomial How can I divide this using long division?
$$\frac{ax^3-a^2x^2-bx^2+b^2}{ax-b}$$
Edit
Sorry guys I wrote it wrong... Fixed it now.
 A: You may have more luck computing the reciprocal, $\dfrac{ax^3 - a^2x^2 - bx^2 + b^2}{ax-b}$, and then inverting it again at the end.
(When you've done that, try artificially 'setting' $ax = b$ in the expression $ax^3 - a^2x^2 - bx^2 + b^2$ and see what you get. What happens, and why?)
A: Hint: See if $(ax-b)$ is a factor of $(ax^3-a^2x^2-bx^2+b^2)$ by writing:
$$(ax^3-a^2x^2-bx^2+b^2)=(ax-b)(\square x^2+\square x+\square)$$
and see if you can fill in the $\square$s.
A: $$
\begin{array}{rccccccccccccc}
& &  x^2 & - & ax & - & b \\[12pt]
ax-b & ) & ax^3 & - & (a^2+b)x^2 & + & b^2 \\
& & ax^3 & - & bx^2 \\[12pt]
& & & & -a^2x^2 & + & b^2 \\
& & & & -a^2x^2 & + & abx \\[12pt]
& & & & & & -abx & + & b^2 \\
& & & & & & -abx & + & b^2 \\[12pt]
& & & & & & & & 0
\end{array}
$$
A: Here's one way to approach it. You can try grouping the elements in the numerator. Note that, you should try grouping in the sense that $ax - b$ must appear. So that they can calcel themselves out. So here it goes:
$$\dfrac{\color{red}{ax^3} \color{blue}{- a^2 x^2} \color{red}{- bx^2} \color{blue}{+ b^2}}{ax - b} = \dfrac{(ax^3 -bx^2) + (b^2 - a^2 x^2)}{ax - b}$$
I think you should be able to go from here. If not, I'll give you a hint:


*

*Try factoring.

*Try to apply the difference of squares fomula.

