Factor: $x^3-4x^2-11x-6$ Factor: $$x^3-4x^2-11x-6$$
The answer is $(x-6)(x+1)^2$ but how do I get to it?
I tried many different methods but did not solve it correctly. Pleade help?
 A: It's enough to find three roots, cause $w(a) = 0$ means that exists some polynomial $q(x)$, that $w(x) = (x-a)q(x)$. And your polynomial degree is 3.
You can check $0, 1, -1$ first. It's the easiest to do and in tasks often works. After that you needs something stronger:
Hint: Rational root theorem and not necessarily in this case Horner's method to receive polynomial with degree two.

A: By the rational root theorem, we guess the factors include at least one of the following:
$$x-6$$
$$x+6$$
$$x-2$$
$$x+2$$
$$x-3$$
$$x+3$$
$$x-1$$
$$x+1$$
Where did I get those?


*

*Take the absolute value of the last term (in descending order), $6$, and the absolute value of the last term, $1$.

*Take the factors of each:
$$6: \color{red}{1,2,3}$$
$$1: \color{green}{1}$$


*Consider $x-a$ where $a$ may be any of the ff:


$$+\frac{\color{red}{6}}{\color{green}{1}}$$
$$-\frac{\color{red}{6}}{\color{green}{1}}$$
$$+\frac{\color{red}{3}}{\color{green}{1}}$$
$$-\frac{\color{red}{3}}{\color{green}{1}}$$
$$+\frac{\color{red}{2}}{\color{green}{1}}$$
$$-\frac{\color{red}{2}}{\color{green}{1}}$$
$$+\frac{\color{red}{1}}{\color{green}{1}}$$
$$-\frac{\color{red}{1}}{\color{green}{1}}$$
