# What are all the possible rational zeros for f(x)=5x^3-31x^2+31x-5 and how do you find all zeros?

Dec 11, 2016

$\frac{1}{5} , 1 \mathmr{and} 5.$

#### Explanation:

graph{y-5x^3+31x^2-31x+5=0 [-10, 10, -5, 5]} The sum of the coefficients is 0. So, 1 is a zero for f.

Now, $f \left(x\right) = \left(x - 1\right) \left(5 {x}^{2} - 26 x + 5\right)$.

The zeros of the quadratic factor are

$\frac{26 \pm \sqrt{{26}^{2} - 100}}{10} = \frac{26 \pm 24}{10} = 5 \mathmr{and} \frac{1}{5}$.

The x-intercepts in the graph give the zeros of f.