# How do you find all the zeros of f(x)=12x^3+31x^2-17x-6?

Aug 3, 2016

$f \left(x\right)$ has zeros: $\frac{2}{3}$, $- \frac{1}{4}$, $- 3$

#### Explanation:

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

By the rational root 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.

So the only possible rational zeros are:

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

That's rather a lot of possibilities to check, so let's search by approximately binary chop:

$f \left(0\right) = - 6$

$f \left(1\right) = 12 + 31 - 17 - 7 = 19$

$f \left(\frac{1}{2}\right) = \frac{3}{2} + \frac{31}{4} - \frac{17}{2} - 6 = \frac{6 + 31 - 34 - 24}{4} = - \frac{21}{4}$

$f \left(\frac{3}{4}\right) = 12 \left(\frac{27}{64}\right) + 31 \left(\frac{9}{16}\right) - 17 \left(\frac{3}{4}\right) - 6 = \frac{81 + 279 - 204 - 96}{16} = \frac{15}{4}$

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

So $x = \frac{2}{3}$ is a zero and $\left(3 x - 2\right)$ is 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 $4 {x}^{2} + 13 x + 2$ we can 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:

$4 {x}^{2} + 13 x + 3 = 4 {x}^{2} + 12 x + x + 3 = 4 x \left(x + 3\right) + 1 \left(x + 3\right) = \left(4 x + 1\right) \left(x + 3\right)$

So the remaining zeros are $x = - \frac{1}{4}$ and $x = - 3$