# How do you factor x^3 - 4x^2 - 11x + 2 = 0?

Aug 18, 2016

${x}^{3} - 4 {x}^{2} - 11 x + 2 = \left(x + 2\right) \left(x - 3 - 2 \sqrt{2}\right) \left(x - 3 + 2 \sqrt{2}\right)$

#### Explanation:

$f \left(x\right) = {x}^{3} - 4 {x}^{2} - 11 x + 2$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ must be expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $2$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2$

Trying each in turn, we find:

$f \left(- 2\right) = - 8 - 4 \left(4\right) + 11 \left(2\right) + 2 = - 8 - 16 + 22 + 2 = 0$

So $x = - 2$ is a zero and $\left(x + 2\right)$ a factor:

${x}^{3} - 4 {x}^{2} - 11 x + 2 = \left(x + 2\right) \left({x}^{2} - 6 x + 1\right)$

Factor the remaining quadratic by completing the square and using the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

as follows:

${x}^{2} - 6 x + 1$

$= {x}^{2} - 6 x + 9 - 8$

$= {\left(x - 3\right)}^{2} - {\left(2 \sqrt{2}\right)}^{2}$

$= \left(\left(x - 3\right) - 2 \sqrt{2}\right) \left(\left(x - 3\right) + 2 \sqrt{2}\right)$

$= \left(x - 3 - 2 \sqrt{2}\right) \left(x - 3 + 2 \sqrt{2}\right)$

Putting it all together:

${x}^{3} - 4 {x}^{2} - 11 x + 2 = \left(x + 2\right) \left(x - 3 - 2 \sqrt{2}\right) \left(x - 3 + 2 \sqrt{2}\right)$