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

1 Answer
Oct 10, 2015

(r-1)(r+2)(r-3)

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(1^2)-5(1)+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 (x-1). Dividing, we get r^2-r-6.

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

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

r^3-2r^2-5r+6=(r-1)(r+2)(r-3)