<s>[INST] Finding the intervals of increase and decrease of $\frac{x^4 - x^3 -8}{x^2 - x - 6}$ How can I find the intervals of increase and decrease of $\frac{x^4 - x^3 -8}{x^2 - x - 6}$?
I tried to find the derivative by the quotient rule to obtain the critical points but the formula was getting complicated, I know that $D_f = \mathbb{R} \setminus \{-2,3\}$  but then what?
Could anyone help me please?
 [/INST] One can simplify calculation of derivative by partitioning fraction:
$$\frac{x^4 - x^3 -8}{x^2 - x - 6}=Ax^2+Bx+C+\frac{D}{x-3}+\frac{E}{x+2}$$
$$\frac{x^4 - x^3 -8}{x^2 - x - 6}=x^2+6+\frac{46}{5(x-3)}-\frac{16}{5(x+2)}$$
$$\left(\frac{x^4 - x^3 -8}{x^2 - x - 6}\right)'=2x-\frac{46}{5(x-3)^2}+\frac{16}{5(x+2)^2}$$
Unfortunately, for this problem one needs to have common denominator, which makes almost no sense in proposed way of calculation. Common denominator is $5(x-3)^2(x+2)^2$ and numerator is $$2x(x-3)^2(x+2)^2-46(x+2)^2+16(x-3)^2=2(x^5-2x^4-11x^3+9x^2+8x-4)$$ Then getting answer requires solving fifth degree algebraic equation, which I suppose cannot be done in closed-form.
If there is typo error in some one of signs before $x^3$ or $x$ in problem, then this way of solution leads to cubic equation.
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