# How do you factor x^4-2x^3-12x^2+18x+27?

Jul 31, 2015

Use the rational roots theorem to find:

${x}^{4} - 2 {x}^{3} - 12 {x}^{2} + 18 x + 27 = \left(x + 1\right) \left(x + 3\right) \left(x - 3\right) \left(x - 3\right)$

#### Explanation:

Let $f \left(x\right) = {x}^{4} - 2 {x}^{3} - 12 {x}^{2} + 18 x + 27$

By the rational roots theorem, any rational roots of $f \left(x\right) = 0$ must be of the form $\frac{p}{q}$ where $p , q$ are integers, $q \ne 0$, $p$ a divisor of $27$ and $q$ a divisor of $1$.

So the only possible rational roots are:

$\pm 1$, $\pm 3$, $\pm 9$, $\pm 27$

$f \left(1\right) = 1 - 2 - 12 + 18 + 27 = 32$
$f \left(- 1\right) = 1 + 2 - 12 - 18 + 27 = 0$
$f \left(3\right) = 81 - 54 - 108 + 54 + 27 = 0$
$f \left(- 3\right) = 81 + 54 - 108 - 54 + 27 = 0$

So $x = - 1$, $x = 3$ and $x = - 3$ are roots of $f \left(x\right) = 0$ and $\left(x + 1\right)$, $\left(x - 3\right)$ and $\left(x + 3\right)$ are factors of $f \left(x\right)$.

The remaining factor must be $\left(x - 3\right)$ in order that when multiplied by the other factors the coefficient of the ${x}^{4}$ term is $1$ and the constant term $27$.

graph{x^4-2x^3-12x^2+18x+27 [-10, 10, -5, 5]}