# Is it possible to factor y=x^4 - 2x^3 - 13x^2 + 14x + 24 ? If so, what are the factors?

##### 1 Answer
Jan 19, 2016

Use the rational root theorem to help find the first two factors, then divide and factor the remaining quadratic to find:

$y = {x}^{4} - 2 {x}^{3} - 13 {x}^{2} + 14 x + 24$

$= \left(x + 1\right) \left(x - 2\right) \left(x - 4\right) \left(x + 3\right)$

#### Explanation:

Let $f \left(x\right) = {x}^{4} - 2 {x}^{3} - 13 {x}^{2} + 14 x + 24$

By the rational root theorem, any rational roots of $f \left(x\right) = 0$ must be of the form $\frac{p}{q}$ for some integers $p$ and $q$ with $p$ as factor of the constant term $24$ and $q$ a factor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

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

Try the first few in turn:

$f \left(1\right) = 1 - 2 - 13 + 14 + 24 = 24$

$f \left(- 1\right) = 1 + 2 - 13 - 14 + 24 = 0$

$f \left(2\right) = 16 - 16 - 52 + 28 + 24 = 0$

So $x = - 1$ and $x = 2$ are zeros and $\left(x + 1\right)$ and $\left(x - 2\right)$ are factors:

${x}^{4} - 2 {x}^{3} - 13 {x}^{2} + 14 x + 24$

$= \left(x + 1\right) \left({x}^{3} - 3 {x}^{2} - 10 x + 24\right)$

$= \left(x + 1\right) \left(x - 2\right) \left({x}^{2} - x - 12\right)$

To factor the remaining quadratic, find a pair of factors of $12$ that differ by $1$. The pair $4 , 3$ works, hence we find:

$= \left(x + 1\right) \left(x - 2\right) \left(x - 4\right) \left(x + 3\right)$