# How do you factor x^3 - 4x^2 -7x = -10?

Oct 10, 2015

$\left(x - 1\right) \left(x - 5\right) \left(x + 2\right) = 0$

#### Explanation:

Re-writing ${x}^{3} - 4 {x}^{2} - 7 x = - 10$ as
${x}^{3} - 4 {x}^{2} - 7 x + 10 = 0$

We notice that the sum of the coefficients is zero
$\Rightarrow x = 1$ is a solution to this equation
$\Rightarrow \left(x - 1\right)$ is a factor of ${x}^{3} - 4 x - 7 x + 10$

Dividing ${x}^{3} - 4 x - 7 x + 10$ by $\left(x - 1\right)$ by polynomial long division or synthetic division gives:
$\textcolor{w h i t e}{\text{XXX}} \left(x - 1\right) \left({x}^{2} - 3 x - 10\right)$

If there is a factoring of $\left({x}^{2} - 3 - 10\right)$ with integer coefficients
then we need to find a pair of factors of $\left(- 10\right)$ which add up to $\left(- 3\right)$
With a bit of thought we can come up with $\left(- 5 , 2\right)$
which gives us the complete factoring:
color(white)("XXX")(x-1)((x-5)(x+2)