# How do you factor (x^3 - 7x^2 + 14x - 8)?

Apr 8, 2015

There doesn't seem to be a grouping that will work.

The rational zero theorem tells us that tlhe possible rational zeros are $\pm 1 , \pm 2 , \pm 4 , \pm 8$.

We can see that ${\left(1\right)}^{3} - 7 {\left(1\right)}^{2} + 14 \left(1\right) - 8 = 1 - 7 + 14 - 8 = 0$, so 1 is a zero.

The factor theorem then tells us that $x - 1$ is a factor.
Do the division:

$\frac{{x}^{3} - 7 {x}^{2} + 14 x - 8}{x - 1} = {x}^{2} - 6 x + 8$

So
$\left({x}^{3} - 7 {x}^{2} + 14 x - 8\right) = \left(x - 1\right) \left({x}^{2} - 6 x + 8\right) = \left(x - 1\right) \left(x - 2\right) \left(x - 4\right)$