Factoring an equation to find its roots

**MN1987** · October 12th 2012, 05:49 PM

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.

**MarkFL** · October 12th 2012, 05:54 PM

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

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

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

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

**MN1987** · October 12th 2012, 06:06 PM

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

**HallsofIvy** · October 12th 2012, 06:26 PM

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

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

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

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