# How do you factor x^3 - 7x - 6?

Feb 25, 2017

${x}^{3} - 7 x - 6 = \left(x - 1\right) \left(x - 3\right) \left(x + 2\right)$

#### Explanation:

Given:

${x}^{3} - 7 x - 6$

Notice that if you reverse the signs of the coefficients on the terms of odd degree, then their sum is $0$. That is:

$- 1 + 7 - 6 = 0$

Hence $x = - 1$ is a zero and $\left(x + 1\right)$ a factor:

${x}^{3} - 7 x - 6 = \left(x + 1\right) \left({x}^{2} - x - 6\right)$

To factor the remaining quadratic, find a pair of factors of $6$ which differ by $1$. The pair $3 , 2$ works and hence we find:

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

Putting it all together:

${x}^{3} - 7 x - 6 = \left(x - 1\right) \left(x - 3\right) \left(x + 2\right)$