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Math Help - Factoring an equation to find its roots

  1. #1
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    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.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Factoring an equation to find its roots

    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
    Last edited by MarkFL; October 12th 2012 at 05:58 PM.
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  3. #3
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    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?
    Last edited by MN1987; October 12th 2012 at 06:13 PM.
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  4. #4
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    Re: Factoring an equation to find its roots

    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.)
    Last edited by HallsofIvy; October 12th 2012 at 06:29 PM.
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