# How do you factor r^3 - 2r^2 - 5r + 6 = 0?

Oct 10, 2015

$\left(r - 1\right) \left(r + 2\right) \left(r - 3\right)$

#### Explanation:

Using the rational root theorem, you can deduce that the possible roots of the function are $1 , 2 , 3 , 6 , - 1 , - 2 , - 3 ,$ and$- 6$ (which are simply possible factors of the term without $r$).

Now, just plug in some of the possible roots into the equation to see if it equals $0$. Plugging in $1$, we see that ${1}^{3} - 2 \left({1}^{2}\right) - 5 \left(1\right) + 6$ is $1 - 2 - 5 + 6$ which equates to $0$.

Now that we have determined a root of the original equation, we can use synthetic division to divide the polynomial by the binomial $\left(x - 1\right)$. Dividing, we get ${r}^{2} - r - 6$.

Now it is simply factoring quadratic equation. When we factor, we get $\left(r + 2\right) \left(r - 3\right)$.

Combining these roots with our roots determined earlier, we can get the full answer as

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