Factoring an equation to find its roots

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.

Re: Factoring an equation to find its roots

Using the rational roots theorem (and polynomial/synthetic division), we find:

$\displaystyle r^5-3r^4+3r^3-3r^2+2r=r(r-1)(r-2)(r^2+1)$

Once the rational roots theorem gives you the roots 0, 1, 2, then compute:

$\displaystyle \frac{r^4-3r^3+3r^2-3r+2}{r^2-3r+2}=r^2+1$

Re: Factoring an equation to find its roots

Thanks... I understand how to find the roots, but how do we arrive at that factorization though?

Re: Factoring an equation to find its roots

Yes, the rational root theorem says that IF polynomial equation $\displaystyle a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0$ has a rational root, $\displaystyle \frac{p}{q}$, then p must evenly divide the constant term, $\displaystyle a_0$, and q must evenly divide the leading coefficient, $\displaystyle a_n$.

Here, the constant term is 2 and the leading coefficient is 1. The only divisors or 2 are $\displaystyle \pm 1$ and $\displaystyle \pm 2$. The only divisors of 1 are $\displaystyle \pm 1$. Therefore, **if** there are any rational roots, they must be of the form 1, -1, 2, or -2. It's easy to check that $\displaystyle (-1)^4- 3(-1)^3+ 3(-1)^2- 3(-1)+ 2= 12$, not 0. $\displaystyle (1)^4- 3(1)^3+ 3(1)^2- 3(1)+ 2= 0$. $\displaystyle (-2)^4- 3(-2)^3+ 3(-2)^2- 3(-2)+ 2= 60$, not 0. $\displaystyle (2)^4- 3(2)^3+ 3(2)^2- 3(2)+ 2= 0$.

So we know that 0, 1, and 2 are roots. Dividing $\displaystyle r^4- 3r^3+ 3r^2- 3r+ 2$ by x(x- 1)(x- 2) as MarkFL2 suggests leaves $\displaystyle r^2+ 1= 0$ 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.)