# How do you find all roots for x^3 - x^2 - 14x + 24 = 0?

Jun 27, 2016

$x = 2$, $x = - 4$ and $x = 3$

#### Explanation:

$f \left(x\right) = {x}^{3} - {x}^{2} - 14 x + 24$

By the rational root theorem, any rational zeros of $f \left(x\right)$ must be expressible in the form $\frac{p}{q}$ 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:

$\pm 1 , \pm 2 , \pm 3 , \pm 4 , \pm 6 , \pm 8 , \pm 12 , \pm 24$

Trying each in turn, we find:

$f \left(1\right) = 1 - 1 - 14 + 24 = 10$

$f \left(- 1\right) = - 1 - 1 + 14 + 24 = 36$

$f \left(2\right) = 8 - 4 - 28 + 24 = 0$

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

${x}^{3} - {x}^{2} - 14 x + 24 = \left(x - 2\right) \left({x}^{2} + x - 12\right)$

We can factor the remaining quadratic by finding a pair of factors of the absolute value $12$ of the constant term which differ by the coefficient $1$ of the middle term. The pair $4 , 3$ works.

Hence we find:

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

Putting it together:

${x}^{3} - {x}^{2} - 14 x + 24 = \left(x - 2\right) \left(x + 4\right) \left(x - 3\right)$

which has zeros $x = 2$, $x = - 4$ and $x = 3$