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Math Help - Factoring Polynomials

  1. #16
    Newbie nickgc's Avatar
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    Re: Factoring Polynomials

    Assuming the third term ought to be

    if you had a typo on the 4th term also, (+16 instead of -16)
    you would have a polynomial which has 3 factors (some repeated)
    and 3 different integer roots for f(x) = 0.
    so that would be:

    x^5 - 6x^4 + 5x^3 + 16x^2 - 12x - 16

    The factors of the first three terms are:
    (x^3)(x - 5)(x - 1)

    I can't factorise the last three terms
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  2. #17
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    Re: Factoring Polynomials

    Quote Originally Posted by nickgc View Post
    so that would be:

    x^5 - 6x^4 + 5x^3 + 16x^2 - 12x - 16

    The factors of the first three terms are:
    (x^3)(x - 5)(x - 1)

    I can't factorise the last three terms
    factorise x^5 - 6x^4 + 5x^3 + 16x^2 - 12x - 16 - Wolfram|Alpha

    To get the factors, guess a root, use the null factor law, use polynomial long division etc. etc.

    eg. Guess x = 1 is a root. Doesn't work. Guess x = -1. It works therefore x + 1 is a factor. Divide x + 1 into x^5 - 6x^4 + 5x^3 + 16x^2 - 12x - 16. Get a quartic factor. Repeat the process.

    Not difficult, just drudge work for you to do (not us).
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  3. #18
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    Re: Factoring Polynomials

    Quote Originally Posted by nickgc View Post
    so that would be:

    x^5 - 6x^4 + 5x^3 + 16x^2 - 12x - 16

    The factors of the first three terms are:
    (x^3)(x - 5)(x - 1)

    I can't factorise the last three terms
    It seldom helps to factor just part of a polynomial in the way you did the first 3 terms.

    I believe Archie Meade suggested that the polynomial: x^5 - 6x^4 + 5x^3 + 16x^2 - 12x - 16 can be factored (with 'nice' results).

    There are a couple of easy numbers you can plug into Archie's polynomial in order to check for some factors.

    Plug 1 in for x. This gives -12 as a result, so the remainder theorem tells you that (x-1) is not a factor.

    Next, plug -1 in for x. This gives -12 as a result, so the remainder theorem tells you that (x-(-1)) is a factor. Of course, (x-(-1)) = (x+1).

    So, x+1 is a factor of Archie's polynomial.

    Use synthetic division (or else long division) to find the other factor. --It's a degree 4 polynomial. See if -1 is also a root of that polynomial.

    A way to find some other possible factors is to graph the polynomial with a graphing calculator or some graphing software. The roots of the polynomial will give information about factors.
    Last edited by SammyS; July 28th 2011 at 08:43 PM. Reason: typo
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  4. #19
    Newbie nickgc's Avatar
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    Re: Factoring Polynomials

    x^5 - 6x^4 + 5x^3 + 16x^2 - 12x - 16

    = ( x + 1 ) ( x^4 - 7x^3 + 12x^2 + 4x - 16 )
    = ( x + 1 ) ( x + 1 ) ( x^3 - 8x^2 + 20x - 16 )
    = ( x + 1 ) ( x + 1 ) ( x - 2) ( x^2 - 6x + 8 )
    = ( x + 1 ) ( x + 1 ) ( x - 2) ( x - 2) ( x - 4 )

    is this correct?
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  5. #20
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    Re: Factoring Polynomials

    Quote Originally Posted by nickgc View Post
    x^5 - 6x^4 + 5x^3 + 16x^2 - 12x - 16

    = ( x + 1 ) ( x^4 - 7x^3 + 12x^2 + 4x - 16 )
    = ( x + 1 ) ( x + 1 ) ( x^3 - 8x^2 + 20x - 16 )
    = ( x + 1 ) ( x + 1 ) ( x - 2) ( x^2 - 6x + 8 )
    = ( x + 1 ) ( x + 1 ) ( x - 2) ( x - 2) ( x - 4 )

    is this correct?
    factorise x^5 - 6x^4 + 5x^3 + 16x^2 - 12x - 16 - Wolfram|Alpha
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