# How do you find all the zeros of f(x) = x^3 - 2x^2- 21x -18 with -3 as a zero?

Mar 22, 2016

First divide the polynom by x+3 and calculate the solutions with the quadratic formula.

#### Explanation:

$\frac{{x}^{3} - 2 {x}^{2} - 21 x - 18}{x + 3} = {x}^{2} - 5 x - 6$

$x = \frac{5 \pm \sqrt{{5}^{2} - 4 \cdot 1 \cdot \left(- 6\right)}}{2}$

$x = \frac{5 \pm \sqrt{49}}{2}$

$x = \frac{5 \pm 7}{2}$

$x = 6 \mathmr{and} x = - 1$