# How do you find all the zeros of f(x)=x^4 - x^3 - 20x^2?

Feb 27, 2016

Factor $f \left(x\right) = {x}^{2} \left(x - 5\right) \left(x + 4\right)$, hence find zeros:

$x = 0$ (twice), $x = 5$ and $x = - 4$

#### Explanation:

Separate out the common factor ${x}^{2}$ of the terms, then factor the remaining quadratic using $5 \times 4 = 20$ and $5 - 4 = 1$ as follows:

$f \left(x\right) = {x}^{4} - {x}^{3} - 20 {x}^{2}$

$= {x}^{2} \left({x}^{2} - x - 20\right)$

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

Hence zeros: $x = 0$ (twice), $x = 5$ and $x = - 4$