# How to solve x^3-3x-2=0 ?

Jun 3, 2018

The roots are $- 1 , - 1 , 2$

#### Explanation:

It is easy to see by inspection that $x = - 1$ satisfies the equation :

${\left(- 1\right)}^{3} - 3 \times \left(- 1\right) - 2 = - 1 + 3 - 2 = 0$

To find the other roots let us rewrite ${x}^{3} - 3 x - 2$ keeping in mind that $x + 1$ is a factor:

${x}^{3} - 3 x - 2 = {x}^{3} + {x}^{2} - {x}^{2} - x - 2 x - 2$
$q \quad q \quad = {x}^{2} \left(x + 1\right) - x \left(x + 1\right) - 2 \left(x + 1\right)$
$q \quad q \quad = \left(x + 1\right) \left({x}^{2} - x - 2\right)$
$q \quad q \quad = \left(x + 1\right) \left({x}^{2} + x - 2 x - 2\right)$
$q \quad q \quad = \left(x + 1\right) \left\{x \left(x + 1\right) - 2 \left(x + 1\right)\right\}$
$q \quad q \quad = {\left(x + 1\right)}^{2} \left(x - 2\right)$

Thus, our equation becomes

${\left(x + 1\right)}^{2} \left(x - 2\right) = 0$

which obviously has roots $- 1 , - 1 , 2$

We can also see it in the graph:

graph{x^3-3x-2}

Jun 4, 2018

${x}_{1} = {x}_{2} = - 1$ and ${x}_{3} = 2$

#### Explanation:

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

${x}^{3} + 1 - \left(3 x + 3\right) = 0$

$\left(x + 1\right) \left({x}^{2} - x + 1\right) - 3 \left(x + 1\right) = 0$

$\left(x + 1\right) \left({x}^{2} - x + 1 - 3\right) = 0$

$\left(x + 1\right) \left({x}^{2} - x - 2\right) = 0$

$\left(x + 1\right) \left(x + 1\right) \left(x - 2\right) = 0$

${\left(x + 1\right)}^{2} \cdot \left(x - 2\right) = 0$

Thus ${x}_{1} = {x}_{2} = - 1$ and ${x}_{3} = 2$