# How do you factor x^3 - 9x^2 + 24x - 20?

Aug 18, 2016

${x}^{3} - 9 {x}^{2} + 24 x - 20 = \left(x - 2\right) \left(x - 2\right) \left(x - 5\right)$

#### Explanation:

$f \left(x\right) = {x}^{3} - 9 {x}^{2} + 24 x - 20$

By the rational root theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $20$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 4 , \pm 5 , \pm 10 , \pm 20$

Trying each in turn, we find:

$f \left(2\right) = 8 - 9 \left(4\right) + 24 \left(2\right) - 20 = 8 - 36 + 48 - 20 = 0$

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

${x}^{3} - 9 {x}^{2} + 24 x - 20 = \left(x - 2\right) \left({x}^{2} - 7 x + 10\right)$

Note that $7 = 2 + 5$ and $10 = 2 \cdot 5$

So we find:

${x}^{2} - 7 x + 10 = \left(x - 2\right) \left(x - 5\right)$

Putting it all together:

${x}^{3} - 9 {x}^{2} + 24 x - 20 = \left(x - 2\right) \left(x - 2\right) \left(x - 5\right)$