# How do you find the roots of x^3-2x^2-5x+6=0?

Dec 30, 2016

$x \in \left\{+ 1 , + 3 , - 2\right\}$

#### Explanation:

If we note that the sum of the coefficients of the terms equals zero
then it follows that $x = 1$ is one solution to this equation.

That is $\left(x - 1\right)$ is a factor of the right side.
Either using synthetic or long division we can obtain
$\textcolor{w h i t e}{\text{XXX}} \left({x}^{3} - 2 {x}^{2} - 5 x + 6\right) \div \left(x - 1\right) = \textcolor{g r e e n}{{x}^{2} - x - 6}$

Using basic principles we can factor
$\textcolor{w h i t e}{\text{XXX}} \left({x}^{2} - x - 6\right) = \left(x - 3\right) \left(x + 2\right)$

So ${x}^{3} - 2 {x}^{2} - 5 x + 6 = 0$

$\rightarrow \left(x - 1\right) \left(x - 3\right) \left(+ 2\right) = 0$

giving the solutions
$\textcolor{w h i t e}{\text{XXX}} x = + 1 \mathmr{and} x = + 3 \mathmr{and} x = - 2$