# How do you find all zeros of g(x)=5x(x^2-2x-1)?

Jan 18, 2017

$0$, $1 + \sqrt{2}$, and $1 - \sqrt{2}$.

#### Explanation:

$0$ is clearly a root, since $g \left(x\right)$ is a product that contains $5 x$.

${x}^{2} - 2 x - 1$ is a polynomial with discriminant:

$D = {b}^{2} - 4 a c$, where $a , b , c$ the coefficients of the polynomial ($a$ for ${x}^{2}$, $b$ for $x$, $c$ for the constant). So,

$D = 4 + 4 = 8 > 0$ and it follows that the aforementioned polynomial has two unique roots given by:

$x = \frac{- b \pm \sqrt{D}}{2 a} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2 \sqrt{2}}{2} = 1 \pm \sqrt{2.}$

So, $g$ has three roots: $x = 0$, $x = 1 + \sqrt{2}$, and $x = 1 - \sqrt{2}$.