# How do you factor and find the zeroes for g(x)= x^3 -2x^2 + 5x?

May 30, 2015

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

So when $x = 0$, $g \left(x\right) = 0$

The quadratic ${x}^{2} - 2 x + 5$ is of the form $a {x}^{2} + b x + c$ with $a = 1$, $b = - 2$ and $c = 5$. This has discriminant given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(- 2\right)}^{2} - \left(4 \times 1 \times 5\right) = 4 - 20 = - 16$

Since $\Delta < 0$ the quadratic has no real roots. It has two distinct complex roots.

The only real zero of $g \left(x\right)$ is $x = 0$