Using the rational roots theorem (and polynomial/synthetic division), we find:
Once the rational roots theorem gives you the roots 0, 1, 2, then compute:
This is actually for my ODE class, but I'm stuck on a factorization type question.
r^5 - 3r^4 + 3r^3 - 3r^2 + 2r = 0
First thing I can do is factor out an r, leaving me with:
r ( r^4 - 3r^3 + 3r^2 - 3r^1 + 2) = 0.
One root is obviously 0. Cool. Now how do I factor the expression in the parentheses to find the roots of the equation? The method my book advises involves finding the factors of a0 (the coefficient of r^4 in this case) and an (2, in this case). So, I am left with +/- 1 and +/- 2. I tried plugging these into the equation and found that +1 and +2 were also roots.
BUT apparently there is a way to factor the above parenthetical expression into (r-1)(r-2)(r^2 + 1), in which case we would find the roots listed above, AS WELL AS +/- i. So, what is the method used to factor that equation that way? Since it seems the coefficient method was not completely sufficient.
Yes, the rational root theorem says that IF polynomial equation has a rational root, , then p must evenly divide the constant term, , and q must evenly divide the leading coefficient, .
Here, the constant term is 2 and the leading coefficient is 1. The only divisors or 2 are and . The only divisors of 1 are . Therefore, if there are any rational roots, they must be of the form 1, -1, 2, or -2. It's easy to check that , not 0. . , not 0. .
So we know that 0, 1, and 2 are roots. Dividing by x(x- 1)(x- 2) as MarkFL2 suggests leaves which has roots i and -i.
Of course, it is possible that shuch and equation might NOT have any rational roots. Checking "possible" rational roots would tell us that- but no one could rationally expect you to be able to solve easily a fourth degree equation that had no rational roots.
(Sorry about that- I couldn't resist.)