Factor: $x^3-4x^2-11x-6$

Question

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?

Answer 1

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.

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Answer 2

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?

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

2. Take the factors of each:

$$6: \color{red}{1,2,3}$$

$$1: \color{green}{1}$$

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

$$+\frac{\color{red}{6}}{\color{green}{1}}$$

$$-\frac{\color{red}{6}}{\color{green}{1}}$$

$$+\frac{\color{red}{3}}{\color{green}{1}}$$

$$-\frac{\color{red}{3}}{\color{green}{1}}$$

$$+\frac{\color{red}{2}}{\color{green}{1}}$$

$$-\frac{\color{red}{2}}{\color{green}{1}}$$

$$+\frac{\color{red}{1}}{\color{green}{1}}$$

$$-\frac{\color{red}{1}}{\color{green}{1}}$$