# How do you find the roots of 12x^3+31x^2-17x-6=0?

Jan 8, 2017

Use the rational roots theorem, then factor to find:

$x = \frac{2}{3} \text{ }$ or $\text{ "x = -1/4" }$ or $\text{ } x = - 3$

#### Explanation:

$f \left(x\right) = 12 {x}^{3} + 31 {x}^{2} - 17 x - 6$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are 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 $12$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{12} , \pm \frac{1}{6} , \pm \frac{1}{4} , \pm \frac{1}{3} , \pm \frac{1}{2} , \pm \frac{2}{3} , \pm \frac{3}{4} , \pm 1 , \pm \frac{3}{2} , \pm 2 , \pm 3 , \pm 6$

Notice that the signs of the coefficients of $f \left(x\right)$ are in the pattern $+ + - -$. By Descartes' Rule of Signs, since there is one change of sign, this cubic has exactly one positive Real zero.

So let's look among the positive possibilities:

$f \left(1\right) = 12 + 31 - 17 - 6 = 20$

$f \left(\frac{1}{2}\right) = 12 \left(\frac{1}{8}\right) + 31 \left(\frac{1}{4}\right) - 17 \left(\frac{1}{2}\right) - 6 = \frac{6 + 31 - 34 - 24}{4} = - \frac{21}{4}$

$f \left(\frac{2}{3}\right) = 12 \left(\frac{8}{27}\right) + 31 \left(\frac{4}{9}\right) - 17 \left(\frac{2}{3}\right) - 6 = \frac{32 + 124 - 102 - 54}{9} = 0$

So $x = \frac{2}{3}$ is a zero and $\left(3 x - 2\right)$ a factor:

$12 {x}^{3} + 31 {x}^{2} - 17 x - 6 = \left(3 x - 2\right) \left(4 {x}^{2} + 13 x + 3\right)$

To factor the remaining quadratic, use an AC method:

Look for a pair of factors of $A C = 4 \cdot 3 = 12$ with sum $B = 13$.

The pair $12 , 1$ works.

Use this pair to split the middle term and factor by grouping:

$4 {x}^{2} + 13 x + 3 = \left(4 {x}^{2} + 12 x\right) + \left(x + 3\right)$

$\textcolor{w h i t e}{4 {x}^{2} + 13 x + 3} = 4 x \left(x + 3\right) + 1 \left(x + 3\right)$

$\textcolor{w h i t e}{4 {x}^{2} + 13 x + 3} = \left(4 x + 1\right) \left(x + 3\right)$

So the other two zeros are:

$x = - \frac{1}{4} \text{ }$ and $\text{ } x = - 3$