# What are all the possible rational zeros for f(x)=x^3-9x^2+15x-2 and how do you find all zeros?

Dec 18, 2017

$x = 2$ and $x = \frac{7 \pm 3 \sqrt{5}}{2}$

#### Explanation:

The rational root theorem tells us that all the possible rational zeroes will be $\pm$ all the factors of the coefficient of the last term divided by $\pm$ all the coefficients of the first term.

In this case, we get: $\pm 1 , \pm 2$ (because $2$ is prime, and because $1$ doesn't have any factors besides itself).

If we plug these into our equation we see that $x = 2$ is a solution.

To find the remaining zeroes, we can divide by $x - 2$ (using polynomial long division) to factor:
$\frac{{x}^{3} - 9 {x}^{2} + 15 x - 2}{x - 2} = {x}^{2} - 7 x + 1$

This means that we can factor like so:
$\left(x - 2\right) \left({x}^{2} - 7 x + 1\right) = 0$

Since we already know $2$ is a root, we want to look at when the other factor equals $0$:
${x}^{2} - 7 x + 1 = 0$

We can solve this by completing the square:
${\left(x - \frac{7}{2}\right)}^{2} - \frac{45}{4} = 0$

${\left(x - \frac{7}{2}\right)}^{2} = \frac{45}{4}$

$x - \frac{7}{2} = \pm \sqrt{\frac{45}{4}}$

$x = \frac{7}{2} \pm \sqrt{\frac{45}{4}}$

$x = \frac{7 \pm 3 \sqrt{5}}{2}$