# How do you find all possible rational zeros of f(x) = 2x^3 - 5x^2 + 3x - 1?

May 27, 2016

Use the rational root theorem to help find that it has no rational zeros.

#### Explanation:

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

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 $- 1$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational zeros are:

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

In addition, note that there are no changes of signs of coefficients in $f \left(- x\right) = - 2 {x}^{3} - 5 {x}^{2} - 3 x - 1$, so $f \left(x\right)$ has no negative zeros.

That leaves possible rational zeros:

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

Then we find:

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

$f \left(1\right) = 2 - 5 + 3 - 1 = - 1$

So this cubic has no rational zeros.