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Math Help - polynomial and factor theorem

  1. #1
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    polynomial and factor theorem

    Hi guys could you check over what i have done and point me in the right direction with part ii

    i had to use polinomial division to find the remainder 3x-1 / 9x^5-4x^4

    i got the remainder to be 1x or (x) is this correct?
    part ii

    show using factor theorem that 2x-1 is a factor of 2x^4-x^3-6x^2+5x-1
    and express 2x^4-x^3-6x^2+5x-1 as a linear and cubic factor.

    TBH i dont know where to start with this one

    any guidance appreciated
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  2. #2
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    try using long division... divide the polynomial by (2x-1), and if there is 0 remainder, that means it is a factor. Also, the division should yield a cubic function, so expressing it as a product of a linear factor and a cubic factor should look like this:

    (2x-1)(the cubic function you get from division)
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  3. #3
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    Right i see it now thanks for the help,

    was my first part correct or was i way off?


    regards
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  4. #4
    Newbie I4talent's Avatar
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    Quote Originally Posted by tommoturbo View Post
    Show using factor theorem that 2x-1 is a factor of 2x^4-x^3-6x^2+5x-1
    The factor theorem states that a polynomial p(x) has a factor (ax - k) if and only if p\left(\frac{k}{a}\right) = 0. So to show that 2x-1 is a factor of 2x^4-x^3-6x^2+5x-1, you plug in \frac{1}{2} for x in 2x^4-x^3-6x^2+5x-1 and see if it yields to zero. If it does (which it hopefully will), then by the factor theorem, 2x-1 is a factor of 2x^4-x^3-6x^2+5x-1; but if it doesn't, then 2x-1 is a not a factor of 2x^4-x^3-6x^2+5x-1. Was that clear? If not, see the spoiler:

    Spoiler:
    First, solve 2x-1: 2x-1 = 0 \implies 2x = 1 \implies x = \frac{1}{2}

    Then substitute \left(\frac{1}{2}\right) for x in 2x^4-x^3-6x^2+5x-1:

    2\left(\frac{1}{2}\right)^4-\left(\frac{1}{2}\right)^3-6\left(\frac{1}{2}\right)^2+5\left(\frac{1}{2}\rig  ht)-1 = \frac{1}{8}-\frac{1}{8}-\frac{3}{2}+\frac{5}{2} = 0

    Thus, by the factor theorem, 2x-1 is a factor of 2x^4-x^3-6x^2+5x-1.




    Express 2x^4-x^3-6x^2+5x-1 as a linear and cubic factor.
    It's asking you to express it in the form (ax^3+bx^2+cx+d)(ex+f). Like this:

    2x^4-x^3-6x^2+5x-1 = (ax^3+bx^2+cx+d)(ex+f)

    We know that 2x-1 is a factor 2x^4-x^3-6x^2+5x-1, so we have:

    2x^4-x^3-6x^2+5x-1 = (ax^3+bx^2+cx+d)(2x-1)

    Now, find ax^3+bx^2+cx+d and you are done.

    Hint:
    Spoiler:
     \frac{2x^4-x^3-6x^2+5x-1}{2x-1} = ax^3+bx^2+cx+d

    Last edited by I4talent; November 2nd 2009 at 05:40 AM.
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  5. #5
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    Wow thanks very much that has explained it fully, the notes in my lesson where limited to two examples and i hadnt grasped it at all



    thank you!
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