How do I solve $(x^3-4x^2+5x-6)/(x^2-x-6)=4$ algebraically? How do I solve $\frac {(x^3-4x^2+5x-6)}{(x^2-x-6)=4}$ algebraically?
I tried:
$4(x^2-x-6)=x^3-4x^2+5x-6$
$4x^2-4x-24=x^3-4x^2+5x-6$
$x^3-8x^2+9x+18=0$
I don't know how to solve this algebraically.
 A: $$x^3-4x^2+5x-6 = (x-3)(x^2-x+2)$$
$$x^2-x-6 = (x-3)(x+2)$$
so the equation becomes: $$\frac{x^2-x+2}{x+2} =4 $$
$$x^2-x+2=4x+8$$
$$x^2-5x-6=0$$
$$(x-6)(x+1)=0$$
$x=6$ or $-1$
A: Use the fact that:
$$\frac{ (x^3-4x^2+5x-6)}{(x^2-x-6)}=(x-3)+\frac{8}{(x+2)} $$
Therefore your solving:
$$ (x-3)+\frac{8}{x+2}=4 \\ x^2-5x-6+8=4x+8\\ x^2-5x-6=0$$
Can you solve this?
A: $\dfrac{x^3-4x^2+5x-6}{x^2-x-6}=\dfrac{x^3-4x^2+5x-6}{(x-3)(x+2)}$
Now we are going to use Factor Theorem, to determine one of the factors of the numerator, which I call call $h(x)$.
(Note: Just in case you're not familiar with the Factor Theorem, it says that a polynomial $p(x)$ has a factor $(x-c) \iff p(c)=0$ )
We note that $h(3)=0$ and hence $x-3$ is a factor. Then we apply polynomial long division to obtain the other factors. You should get $x^2-x+2$ as your quotient.
$\dfrac{x^3-4x^2+5x-6}{(x-3)(x+2)}=\dfrac{(x-3)(x^2-x+2)}{(x-3)(x+2)}$
.
Then the $(x-3)$'s  will cancel leaving 
$\dfrac{(x^2-x+2)}{(x+2)}$
Going back to the original question,
$\dfrac{(x^2-x+2)}{(x+2)}=4 \\
\implies x^2-x+2=4x+8\\
\implies x^2-5x-6=0\\
\implies (x-6)(x+1)=0\\
\implies x=6\text{ or } x=-1$
