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Factoring puzzle

  1. Aug 12, 2008 #1
    factoring "puzzle"

    For what values of "a" can the following be factored and reduced:

    [tex] \frac {a^2 - 2x -3}{x^2 - 4x + a} [/tex]

    Ok, so the first thing on the list is to factor the top term.

    [tex] x = \frac {-(-2) \pm \sqrt{(-2)^2 - 4 * 1 * (-3)} } {2 * 1} [/tex]

    The roots are 3 & -1, and the factors then become

    [tex] (x-3)(x+1) [/tex]

    So, I want the denominator to become either both, or one of those factors.

    I start off by rewriting the denominator as an equation

    [tex] x = \frac {4 \pm \sqrt{16 - 4 * 1 * a} } {2} [/tex]

    At this point I can tell that there are only solutions for [tex] a < 5 [/tex]

    I know that I want the roots the polynomial to be either 3 or -1 (or both), so I substitute everything inside the square root with R, and check which values I need from the square root.

    [tex]
    \begin{align*}
    \frac{4 + R}{2} = 3\\
    4 + R = 6\\
    R = 2\\
    \end{align*}
    [/tex]

    and

    [tex]
    \begin{align*}
    \frac{4 + R}{2} = -1\\
    4 + R = -2\\
    R = -6\\
    \end{align*}
    [/tex]

    Since I would be hard pressed to get the square root back as negative 6, I conclude that I want the result of the square root to be 2.

    In other words

    [tex]
    \begin{align*}
    \sqrt{16 - 4 * 1 * a} = 2\\
    16 - 4a = 4\\
    a = 3\\
    \end{align*}
    [/tex]

    Finally there. I factor the polynomial using [tex] a = 3 [/tex]

    [tex] x = \frac {4 \pm \sqrt{16 - 4 * 3} } {2} [/tex]

    The roots are 3 and 1 which gives the factors

    [tex] (x-3)(x-1) [/tex]

    And I can now factor the original rational expression

    [tex]
    \frac {a^2 - 2x -3}{x^2 - 4x + a} = \frac {(x-3)(x+1)} {(x-3)(x-1)} = \frac {x + 1}{ x -1}
    [/tex]


    However, there are two things that bug me to no end. First off the are two solutions listed, 3 and -5. I only found one of them, but I cant see where I went wrong and "missed" the second one?

    The other thing is that I found my approach rather long winded, and I wondered if there was a better way to do this kind of problems?

    Thanks for any feedback.

    k
     
  2. jcsd
  3. Aug 12, 2008 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi kenewbie! :smile:
    (I assume you mean [tex] \frac {x^2 - 2x -3}{x^2 - 4x + a} [/tex]) ?

    erm … yes! :biggrin:

    Two methods (basically the same):

    i] Just use long division, first with (x - 3) and then with (x + 1).

    The remainder will be a + something, which you want to be zero. :smile:

    ii] You want 3 or 1 to be a root of the denominator, so just plug 3 or 1 into it, again you get the remainder!
     
  4. Aug 12, 2008 #3
    Re: factoring "puzzle"

    I'm afraid I don't quite get what you mean. Can I bother you to type out examples how how you would do it?

    k
     
  5. Aug 12, 2008 #4
    Re: factoring "puzzle"

    ooh, I think I see why I missed the -5 solution.

    I used [tex] 4 + R [/tex] instead of [tex] 4 \pm R[/tex], which would allow me to use the -6 as just 6, which would have come out with -5 as a root.

    k
     
  6. Aug 12, 2008 #5

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Well, for example, if you want x2 - 4x + a to be divisible by (x - b),

    then b will be a root,

    and so b2 - 4b + a must be zero ("Remainder Theorem"). :smile:
     
  7. Aug 12, 2008 #6
    Re: factoring "puzzle"

    Aha. I haven't learned "polynomial long division" or "polynomial remainder theorem" yet, but I see they both have neat little articles on wikipedia, so I'll look into them.

    Thanks

    k
     
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