# How do you use the Rational Zeros theorem to make a list of all possible rational zeros, and use the Descarte's rule of signs to list the possible positive/negative zeros of f(x)=3x^3+3x^2-11x-10?

Jan 2, 2018

$x = - 2 , x = \frac{3 \pm \sqrt{69}}{6}$

#### Explanation:

The pattern of signs of $f \left(x\right)$ is $+ + - -$, meaning there is one sign change, and Descartes' rule tells us that there must be one positive root.

The pattern of signs of $f \left(- x\right)$ is $- + + -$. This has two changes, which according to Descartes' rule means that there is either two or no negative roots.

The rational roots theorem says that we can find all the possible rational roots of a polynomial by dividing the factors of the last term with all the factors of the first term (note that you can use both positive and negative). The factors of $10$ are $1 , 2 , 5 , 10$ and the factors of $3$ are $1 , 3$. This means we have the following possible rational roots:
$\pm 1 , \pm 2 , \pm 5 , \pm 10 , \pm \frac{1}{3} , \pm \frac{2}{3} , \pm \frac{5}{3} \pm \frac{10}{3}$

We'll start by trying the positive values, since Descartes' rule told us the polynomial has exactly one positive root. Unfortunately, none of these values are a root, which means that the positive root we know exists is irrational.

Now we'll have to hope that there are two negative roots (we worked out there could either be two or none), and try the negative values. Trying these, we find that $f \left(- 2\right)$ is a root.

This means we can use polynomial long division to factor like so:
$\left(x + 2\right) \left(3 {x}^{2} - 3 x - 5\right)$

To find the remaining negative zero we know exists, we could try all the other negative rational values, but none of them are roots, so the other negative root is irrational. We have to resort to the quadratic formula to obtain the remaining two solutions:
$x = \frac{3 \pm \sqrt{69}}{6}$

Note that one of these is negative and one is positive, verifying our conclusions with Descartes' rule.