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

Dec 12, 2017

$x = - 4 , x = 0 , x = 5$

#### Explanation:

$\text{take out a common factor of } {x}^{2}$

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

$\text{the factors of - 20 which sum to - 1 are - 5 and + 4}$

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

$\text{to find zeros set } f \left(x\right) = 0$

$\Rightarrow {x}^{2} \left(x - 5\right) \left(x + 4\right) = 0$

$\text{equate each factor to zero and solve for x}$

${x}^{2} = 0 \Rightarrow x = 0 \text{ with multiplicity 2}$

$x - 5 = 0 \Rightarrow x = 5$

$x + 4 = 0 \Rightarrow x = - 4$