# How do you use the rational roots theorem to find all possible zeros of f(x) = 2x^3 + 3x^2 – 8x + 3?

Aug 9, 2016

$f \left(x\right)$ has zeros $1$, $\frac{1}{2}$ and $- 3$

#### Explanation:

$f \left(x\right) = 2 {x}^{3} + 3 {x}^{2} - 8 x + 3$

By the rational root theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $3$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational zeros are:

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

We find:

$f \left(1\right) = 2 + 3 - 8 + 3 = 0$

So $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

$2 {x}^{3} + 3 {x}^{2} - 8 x + 3 = \left(x - 1\right) \left(2 {x}^{2} + 5 x - 3\right)$

Substituting $x = \frac{1}{2}$ in the remaining quadratic we find:

$2 {x}^{2} + 5 x - 3 = 2 \left(\frac{1}{4}\right) + 5 \left(\frac{1}{2}\right) - 3 = \frac{1}{2} + \frac{5}{2} - 3 = 0$

So $x = \frac{1}{2}$ is a zero and $\left(2 x - 1\right)$ a factor:

$2 {x}^{2} + 5 x - 3 = \left(2 x - 1\right) \left(x + 3\right)$

So the final zero is $x = - 3$