# How do you solve x^3 - 2x^2 - 5x + 6 = 0??

Jul 28, 2015

First observe that the sum of the coefficients is $0$, so $x = 1$ is a root. Then divide by $\left(x - 1\right)$ to get a quadratic that is easier to factor and thus solve to find two other roots $x = 3$ and $x = - 2$.

#### Explanation:

Let $f \left(x\right) = {x}^{3} - 2 {x}^{2} - 5 x + 6$

First note that the sum of the coefficients is $0$, so $x = 1$ is a zero of $f \left(x\right)$...

$f \left(1\right) = 1 - 2 - 5 + 6 = 0$

Divide $f \left(x\right)$ by $\left(x - 1\right)$ to find:

$f \left(x\right) = {x}^{3} - 2 {x}^{2} - 5 x + 6 = \left(x - 1\right) \left({x}^{2} - x - 6\right)$

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

So the other two roots of $f \left(x\right) = 0$ are $x = 3$ and $x = - 2$

Alternatively, use the rational roots theorem to know that any rational roots of $f \left(x\right) = 0$ must be of the form $\frac{p}{q}$ where $p$, $q$ are integers, $q \ne 0$, $p$ a factor of the constant term $6$ and $q$ a factor of the coefficient $1$ of the term (${x}^{3}$) of highest degree. So the only possible rational roots are:

$\pm 1$, $\pm 2$, $\pm 3$ and $\pm 6$

Try each to find $f \left(1\right) = 0$, $f \left(- 2\right) = 0$ and $f \left(3\right) = 0$.