# How do you factor P(x)= x^4 - 6x^3 - 8x^2?

Apr 30, 2018

See explanation.

#### Explanation:

First you can notice that ${x}^{2}$ is a common factor, so we can write that:

$P \left(x\right) = {x}^{2} \cdot \left({x}^{2} - 6 x - 8\right)$

To check if the second expression can be factorized we have to calculate its discriminant:

The expression is: ${x}^{2} - 6 x - 8$, so:

$a = 1$, $b = - 6$, $c = - 8$,

the discriminant is:

$\Delta = {\left(- 6\right)}^{2} - 4 \cdot 1 \cdot \left(- 8\right) = 36 + 32 = 68$

The discriminant is greater than zero, so the equation has 2 distinct real roots:

${x}_{1} = \frac{6 - \sqrt{68}}{2} = 3 - 2 \sqrt{17}$

and

${x}_{2} = \frac{6 + \sqrt{68}}{2} = 3 + 2 \sqrt{17}$

So the complete factorization is:

$P \left(x\right) = {x}^{2} \cdot \left(x - 3 + 2 \sqrt{17}\right) \cdot \left(x - 3 - 2 \sqrt{17}\right)$