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.

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?
