# How do you find the zeroes of the function g(x)= 2x^3 - 5x^2 + 4?

Jun 28, 2016

$x = \frac{1 \pm \sqrt{17}}{4} \mathmr{and} x = 2$

#### Explanation:

let's consider the known term 4: it is divided by

$\pm 1 , \pm 2 , \pm 4$

By the remainder theorem, let's substitute each of the dividers in x in the function g(x), so

$g \left(1\right) = 2 - 5 + 4 \ne 0$

$g \left(- 1\right) = - 2 - 5 + 4 \ne 0$

$g \left(2\right) = 16 - 20 + 4 = 0$

So, one of the zeroes is 2

Then we can divide the polynomial g(x) by the bynomial x-2

and obtain:

$2 {x}^{2} - x - 2 = 0$

by which you can find two zeroes

$x = \frac{1 \pm \sqrt{17}}{4}$