# What are all the possible rational zeros for f(x)=3x^3-x^2-3x+1 and how do you find all zeros?

Jan 2, 2018

$x = - 1 , x = \frac{1}{3} , x = 1$

#### Explanation:

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

The pattern of signs for $f \left(- x\right)$ is $- - + +$, which has one change, meaning the polynomial has one negative root.

The rational roots theorem states that we can find all possible rational roots to a polynomial by dividing $\pm$ all factors of the last term by $\pm$ all factors of the first. In this case, we get:
$\pm 1 , \pm \frac{1}{3}$.

I'll begin by trying the negative values since I know for a fact the polynomial has a real root. Of these, we find that $f \left(- 1\right)$ is a root.

This means we can bring out a factor of $\left(x + 1\right)$ out of the polynomial using long division:
$\left(x + 1\right) \left(3 {x}^{2} - 4 x + 1\right)$

We can factor this quadratic by grouping:
$\left(x + 1\right) \left(3 {x}^{2} - 3 x - x + 1\right)$

$\left(x + 1\right) \left(3 x \left(x - 1\right) - \left(x - 1\right)\right)$

$\left(x + 1\right) \left(3 x - 1\right) \left(x - 1\right)$

We can use the zero factor principle to find that the remaining roots are $x = 1$ and $x = \frac{1}{3}$, both of which are positive, verifying our conclusions using Descartes' rule of signs.