# Factoring a cubic

• January 30th 2011, 01:58 PM
DarK
Factoring a cubic
I tried factoring by grouping and everything else I could think of but I can't seem to end up with a solution.

$r^3 - 6r^2 + 11r - 6 = 0$

Would there be a systematic way of factoring this or is it just guesswork?

The final factored form is:

$(r-1)(r-2)(r-3)$

Any help is appreciated!
• January 30th 2011, 02:22 PM
e^(i*pi)
You can use the rational root theorem to find the first root since it must be a factor of -6. Since 1 is always easy to check start there: $f(1) = 1 - 6+11-6 = 0$. Hence r = 1 is a root.

By the factor theorem, if 1 is a root then (r-1) must be a factor of the cubic

$r^3-6r^2+11r-6 = (r-1)(Ar^2+Br+C)$ where A, B and C are constants to be found either by comparing coefficients or by long division.

Once you've found the quadratic remember to check if it factors (it obviously does judging by the answer)