# How do you find all the zeros of f(x)=x^3+x^2-6x?

Jun 17, 2018

Assuming you meant $f \left(x\right) = 0$
$x = 0 \mathmr{and} 2 \mathmr{and} - 3$

#### Explanation:

${x}^{3} + {x}^{2} - 6 x = 0$

$x \left({x}^{2} + x - 6\right) = 0$
$\therefore x = 0 , f \left(x\right) = 0$ (because of the $x$ outside the brackets)

${x}^{2} + x - 6 = 0$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot - 6}}{2}$
$x = 2 \mathmr{and} - 3$

Turning points

$f ' \left(x\right) = 3 {x}^{2} + 2 x - 6$
$3 {x}^{2} + 2 x - 6 = 0$

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 3 \cdot - 6}}{6}$

$x = \frac{- 2 + \sqrt{52}}{6}$ $x = \frac{- 2 - \sqrt{52}}{6}$