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

1 Answer
Oct 10, 2015

Answer:

#(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)#