# How do you factor y= 3x^5 -6x^2 -27x-18 ?

Mar 11, 2016

$3 {x}^{5} - 6 {x}^{2} - 27 x - 18$

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

$= 3 \left(x + 1\right) \left(x + 1\right) \left(x - 2\right) \left(x - \sqrt{3} i\right) \left(x + \sqrt{3} i\right)$

#### Explanation:

First separate out the common scalar factor $3$ to find:

$y = 3 {x}^{5} - 6 {x}^{2} - 27 x - 18 = 3 \left({x}^{5} - 2 {x}^{2} - 9 x - 6\right)$

Next let $f \left(x\right) = {x}^{5} - 2 {x}^{2} - 9 x - 6$

Notice that if the signs of the coefficients are reversed on the terms of odd degree then their sum is zero. That is:

$- 1 - 2 + 9 - 6 = 0$

Hence $f \left(- 1\right) = 0$, so $x = - 1$ is a zero and $\left(x + 1\right)$ a factor:

${x}^{5} - 2 {x}^{2} - 9 x - 6 = \left(x + 1\right) \left({x}^{4} - {x}^{3} + {x}^{2} - 3 x - 6\right)$

Notice that if we reverse the signs of the coefficients of the terms of odd degree in the remaining quartic factor then the sum is zero. That is:

$1 + 1 + 1 + 3 - 6 = 0$

Hence we have another factor $\left(x + 1\right)$:

${x}^{4} - {x}^{3} + {x}^{2} - 3 x - 6 = \left(x + 1\right) \left({x}^{3} - 2 {x}^{2} + 3 x - 6\right)$

Let $g \left(x\right) = {x}^{3} - 2 {x}^{2} + 3 x - 6$

By the rational root theorem, any rational zeros of $g \left(x\right)$ must be expressible in the form $\frac{p}{q}$ for integers $p$, $q$ with $p$ a divisor of the constant term $- 6$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that its only possible rational zeros are:

$\pm 1$, $\pm 2$, $\pm 3$, $\pm 6$

By examining the sum of the coefficients we can rule out $\pm 1$, but we find:

$g \left(2\right) = 8 - 8 + 6 - 6 = 0$

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

${x}^{3} - 2 {x}^{2} + 3 x - 6 = \left(x - 2\right) \left({x}^{2} + 3\right)$

The remaining quadratic factor has no linear factors with Real coefficients since ${x}^{2} + 3 \ge 3 > 0$ for all $x \in \mathbb{R}$. If we allow Complex coefficients then it factors as $\left(x - \sqrt{3} i\right) \left(x + \sqrt{3} i\right)$

Putting it all together:

$3 {x}^{5} - 6 {x}^{2} - 27 x - 18$

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

$= 3 \left(x + 1\right) \left(x + 1\right) \left(x - 2\right) \left(x - \sqrt{3} i\right) \left(x + \sqrt{3} i\right)$