# How do you use the rational roots theorem to find all possible zeros of F(x) = 6x^3 - 20x^2 + 11x + 7?

Jul 30, 2016

Zeros: $\frac{7}{3}$, $\frac{1}{2} \pm \frac{\sqrt{3}}{2}$

#### Explanation:

$F \left(x\right) = 6 {x}^{3} - 20 {x}^{2} + 11 x + 7$

By the rational roots theorem, any rational zeros of $F \left(x\right)$ must be expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $7$ and $q$ a divisor of the coefficient $6$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{6} , \pm \frac{1}{3} , \pm \frac{1}{2} , \pm 1 , \pm \frac{7}{6} , \pm \frac{7}{3} , \pm \frac{7}{2} , \pm 7$

Trying each in turn, we (eventually) find:

$F \left(\frac{7}{3}\right) = 6 \left(\frac{343}{27}\right) - 20 \left(\frac{49}{9}\right) + 11 \left(\frac{7}{3}\right) + 7$

$= \frac{686 - 980 + 231 + 63}{9} = 0$

So $x = \frac{7}{3}$ is a zero and $\left(3 x - 7\right)$ a factor:

$6 {x}^{3} - 20 {x}^{2} + 11 x + 7$

$= \left(3 x - 7\right) \left(2 {x}^{2} - 2 x - 1\right)$

Use the quadratic formula to find the zeros of the remaining quadratic:

$x = \frac{2 \pm \sqrt{{\left(- 2\right)}^{2} - 4 \left(2\right) \left(- 1\right)}}{2 \cdot 2}$

$= \frac{2 \pm \sqrt{4 + 8}}{4}$

$= \frac{2 \pm 2 \sqrt{3}}{4}$

$= \frac{1}{2} \pm \frac{\sqrt{3}}{2}$