How do you factor y= x^3-5x^2-2x+24 ?

Jan 2, 2016

Use the rational root theorem to get started, then factor the remaining quadratic to find:

${x}^{3} - 5 {x}^{2} - 2 x + 24 = \left(x + 2\right) \left(x - 4\right) \left(x - 3\right)$

Explanation:

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

By the rational root theorem, any rational zeros of $f \left(x\right)$ must be expressible in the for $\frac{p}{q}$ for integers $p$, $q$ with $p$ a divisor of the constant term $24$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are the factors of $24$, namely:

$\pm 1 , \pm 2 , \pm 3 , \pm 4 , \pm 6 , \pm 12 , \pm 24$

Try each in turn:

$f \left(1\right) = 1 - 5 - 2 + 24 = 18$

$f \left(- 1\right) = - 1 - 5 + 2 + 24 = 20$

$f \left(2\right) = 8 - 20 - 4 + 24 = 8$

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

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

${x}^{3} - 5 {x}^{2} - 2 x + 24 = \left(x + 2\right) \left({x}^{2} - 7 x + 12\right)$

We can factor ${x}^{2} - 7 x + 12$ by noting that $4 \times 3 = 12$ and $4 + 3 = 7$, so:

${x}^{2} - 7 x + 12 = \left(x - 4\right) \left(x - 3\right)$

Putting it all together:

${x}^{3} - 5 {x}^{2} - 2 x + 24 = \left(x + 2\right) \left(x - 4\right) \left(x - 3\right)$