# 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)=-17x^3+5x^2+34x-10?

Jan 2, 2018

$x = \frac{5}{17} , x = \pm \sqrt{2}$

#### Explanation:

The pattern of signs of $f \left(x\right)$ is $- + + -$. This has two sign changes, and Descartes' rule of signs thereby tells us that the polynomial has either no or two positive roots.

The sign pattern of $f \left(- x\right)$ is $+ + - -$, meaning it has $1$ sign change and thereby the polynomial has $1$ negative root.

The rational roots theorem says that we can find all the rational roots of a polynomial by taking $\pm$ the factors of the last term divided by $\pm$ the factors of the first term. The factors of $10$ are $1 , 2 , 5 , 10$ and the factors of $17$ are $1 , 17$. This makes for the following possible rational roots:
$\pm 1 , \pm 2 , \pm 5 , \pm 10 , \pm \frac{1}{17} , \pm \frac{2}{17} , \pm \frac{5}{17} , \pm \frac{10}{17}$

I will begin by looking at the negative values, since I know the polynomial has a negative root. Unfortunately none of the negative values are roots, so the negative root must be irrational.

Now we'll have to hope that there are two positive roots, and that at least one of them is rational. If we try all the positives, we find that $f \left(\frac{5}{17}\right)$ is a root. This means we can factor as follows using polynomial long division:
$\left(x - \frac{5}{17}\right) \left(- 17 {x}^{2} + 34\right)$

Now we set the other factor equal to zero to find its roots:
$- 17 {x}^{2} + 34 = 0$

$17 {x}^{2} = 34$

$\frac{17 {x}^{2}}{17} = \frac{34}{17}$

${x}^{2} = 2$

$x = \pm \sqrt{2}$

This matches up with our conclusions using Descartes' rule, since we ended up with $2$ positive roots and one negative root.