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Math Help - Factoring help

  1. #1
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    Factoring help

    Hey I need some help with this problem

    Factor the polynomial x^8-98x^4+1 into two non-constant polynomials with
    integral coefficients.

    I want to see steps on how to do this problem thanks.
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  2. #2
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    Re: Factoring help

    Start by letting \displaystyle X = x^4, then the equation becomes \displaystyle \begin{align*} X^2 - 98X + 1 \end{align*}. This doesn't factorise over the integers, but does factorise over the real numbers by completing the square.
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  3. #3
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    Re: Factoring help

    I completed the square and got
    (x-49)^2 +2400
    what do I do next?
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  4. #4
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    Re: Factoring help

    That can't possibly be right - expand it out and check...
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  5. #5
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    Re: Factoring help

    sorry it should be -2400
    (x-49)^2 - 2400 = (x-49)(x-49) - 2400 = x^2 -49x - 49x + 2401 -2400 = x^2 - 98x + 1
    ok so when completing the square i get

    (x-49)^2 - 2400
    what do I do next?
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  6. #6
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    Re: Factoring help

    Write it as \displaystyle \begin{align*} (X - 49)^2 - \left( 20\sqrt{6} \right)^2 \end{align*} and then use the Difference of Two Squares rule to factorise it.
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  7. #7
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    Re: Factoring help

    so it is ((x-49)^2 + (20sqrt(6))^2)((x-49)^2 - (20sqrt(6))^2))
    what do I do from here?
    Last edited by gfbrd; November 10th 2012 at 10:59 PM.
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  8. #8
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    Re: Factoring help

    Hello, gfbrd!

    This takes a little Imagination . . .


    Factor x^8-98x^4+1 into two non-constant polynomials with integral coefficients.

    \begin{array}{cc}\text{We are given:} & x^8 - 98x^4 + 1 \\ \text{Add/subtract } 100x^4\!: & x^8 - 98x^4 {\color{blue}+ 100x^4} + 1 {\color{blue}- 100x^4} \\ \text{And we have:} & x^8 + 2x^4 + 1 - 100x^4  \\ \text{Factor:} & (x^4 +1)^2 - (10x^2)^2 \\ \text{Factor:} & (x^4+1-10x^2)(x^4+1+10x^2) \\ \text{Therefore:} & (x^4 - 10x^2+1)(x^4+10x^2+1) \end{array}
    Thanks from gfbrd
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  9. #9
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    Re: Factoring help

    lol sorry my imagination isn't really as good as yours
    thanks a lot soroban
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