# How do you find zeros of y=6x^3+x^2-x?

Sep 17, 2017

$x = 0 , x = - \frac{1}{2} , x = \frac{1}{3}$

#### Explanation:

$y = 6 {x}^{3} + {x}^{2} - x$
to find zeros
$y = 0$
$6 {x}^{3} + {x}^{2} - x = 0$
$x \left(6 {x}^{2} + x - 1\right) = 0$
this implies either $x = 0$ or $6 {x}^{2} + x - 1 = 0$

Now we solve $6 {x}^{2} + x - 1 = 0$
here $a = 6$, $b = 1$, $c = - 1$
using formula $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$x = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \left(6\right) \left(- 1\right)}}{\left(2\right) \left(6\right)}$
$x = \frac{- 1 \pm \sqrt{1 + 24}}{12} = \frac{- 1 \pm \sqrt{25}}{12} = \frac{- 1 \pm 5}{12}$
either $x = \frac{- 1 - 5}{12} \mathmr{and} x = \frac{- 1 + 5}{12}$
either $x = \frac{- 6}{12} \mathmr{and} x = \frac{4}{12}$
either $x = - \frac{1}{2} \mathmr{and} x = \frac{1}{3}$

Hence zeros of given function $y = 6 {x}^{3} + {x}^{2} - x$ are
$x = 0 , x = - \frac{1}{2} , x = \frac{1}{3}$