Results 1 to 3 of 3

Thread: Need help on factor theorem.

  1. #1
    Feb 2010

    Unhappy Need help on factor theorem.

    I am supposed to use the Factor Theorem to either factor the polynomial completely or to prove that it has no linear factors with integer coefficients.
    I think I haven't completely understood the whole concept yet, which is why I couldn't get my brain working on the following problems.

    a. P(x)= x^4+4x^3-7x^2-34x-24
    b. P(x)= x^6-6x^5+15x^4-20x^3+15x^2-6x+1

    I think these are harder than the regular ones like P(x)=x^3+2x^2-9x+3, etc.

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Quacky's Avatar
    Nov 2009
    Windsor, South-East England

    A very long method

    The key here is to look at the constant at the end.
    For part a, this is $\displaystyle -24$. Think about a quadratic. When factorizing $\displaystyle x^2-7x-8=0$, you'd look for factors of 8, and come up with:
    $\displaystyle (x+1)(x-8)=0$
    So, $\displaystyle x=8, x=-1$

    To factorize, though the co-efficients of x are important, it's the constant that determines the possible factors, which in this case are (+ or -) 1,2,4,8. The co-efficients just allow you to select which specific factors will work.

    Now in part a, just apply this to a quartic:

    $\displaystyle P(x)= x^4+4x^3-7x^2-34x-24$
    Assume p(x)=0
    $\displaystyle x^4+4x^3-7x^2-34x-24=0$
    Now work out what the factors of 24 are, and you'll get $\displaystyle 1,2,3,4,6,8,12,24$
    (Yuck: Including negatives, there're a lot to choose from. Perhaps someone at a more advanced level will know of an easier way)

    To start, choose $\displaystyle x=1$

    Now think back to the example of the quadratic, which does not apply to your question but I am using as an example:
    $\displaystyle (x+1)(x-8)=0$
    So $\displaystyle x=8, x=-1$

    So, if x=1 gives a result of P(x)=0, then (x-1) is a factor of P(x). Now let's go back to your example.

    So, if when we substitute x=1 into p(x) and the result is equal to 0, then (x-1) is a factor of p(x)

    $\displaystyle P(x)= x^4+4x^3-7x^2-34x-24$
    P(1)= (1)^4+4(1)^3-7(1)^2-34(1)-24$
    This gives a result which clearly does not equal zero, so $\displaystyle (x-1)$ is not a factor. You'd then try x=-1, then x=2, x=-2, etc...

    As P(x) is a quartic, there will be a maximum of four factors. I'm afraid I don't know of a faster approach.

    Once you find a factor, partially factorize the expression. This way, once you have found two factors, you'll be left with an easy quadratic.
    Last edited by Quacky; Feb 9th 2010 at 04:39 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Jul 2009
    So we agree that the roots of P(x)=0 are in $\displaystyle

    \{\pm 1,\pm 2,\pm 3,\pm 4,\pm 6, \pm 8,\pm 12,\pm 24 \}

    $\displaystyle P(1) \neq 0 $
    $\displaystyle P(-1) = 0 $
    $\displaystyle P(2) \neq 0 $
    $\displaystyle P(-2) = 0 $

    this rules out 24, -24 and 12

    $\displaystyle P(3) = 0 $

    this rules out -3, 8,-8,6,-6

    only ones left are 4,-4

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Factor theorem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Nov 9th 2009, 03:59 PM
  2. Factor theorem
    Posted in the Algebra Forum
    Replies: 8
    Last Post: Nov 2nd 2009, 02:10 PM
  3. Factor theorem.
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Jan 19th 2009, 12:22 PM
  4. Factor Theorem and Remainder Theorem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Sep 8th 2007, 10:50 AM
  5. factor theorem.
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: Jun 26th 2006, 08:18 PM

Search Tags

/mathhelpforum @mathhelpforum