How do you find all the zeros of f(x)=x^3+5x^2-4x-20?

Jul 18, 2016

x = -5 , x = ± 2

Explanation:

The zeros of f(x) are the values of x that make f(x) = 0

solve: ${x}^{3} + 5 {x}^{2} - 4 x - 20 = 0$

Begin by factorising the left side. Group the term in 'pairs'

$\left[{x}^{3} + 5 {x}^{2}\right] + \left[- 4 x - 20\right]$ and factorising each pair gives.

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

$= \left(x + 5\right) \left(x - 2\right) \left(x + 2\right)$

Solving $\left(x + 5\right) \left(x - 2\right) \left(x + 2\right) = 0$

The zeros are x=-5,x=±2