# What are all the zeroes of g(x)= 2x^3 - 5x^2 + 4 ?

Aug 8, 2016

$g \left(x\right)$ has zeros $2$ and $\frac{1}{4} \pm \frac{\sqrt{17}}{4}$

#### Explanation:

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

By the rational root theorem, any rational zeros of $g \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $4$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means the only possible rational zeros are:

$\pm \frac{1}{2} , \pm 1 , \pm 2 , \pm 4$

We find:

$g \left(2\right) = 2 \left(8\right) - 5 \left(4\right) + 4 = 16 - 20 + 4 = 0$

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

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

The remaining quadratic is in the form $a {x}^{2} + b x + c$ with $a = 2$, $b = - 1$ and $c = - 2$. This has disrciminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(- 1\right)}^{2} - 4 \left(2\right) \left(- 2\right) = 1 + 16 = 17$

Since this is positive but not a perfect square, the remaining zeros are Real but irrational. They are given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{1 \pm \sqrt{\Delta}}{4}$

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

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