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Math Help - Factor completely

  1. #1
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    Factor completely

    Hey guys,

    the task: Factor completely (that is, remove common factors first).

    a)
    x^3 +x^2 -12x

    Could somebody please explain how I can factor completely and remove common factors first?
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Factor completely

    Observe that all terms have x as a factor, so pull that out first, then you are left with a quadratic...to factor this you need two factors of -12 whose sum is 1. Can you finish?
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  3. #3
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    Re: Factor completely

    -3 times 4 = -12 and
    the sum of -3+4 = 1
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Factor completely

    Yes, so what is the factored form of the given expression?
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  5. #5
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    Re: Factor completely

    Factored form should be (x-3)(x+4) = x^2 +4x -3x -12 = x^2 +x -12 = x(x^2 +x -12) = x^3 +x^2 -12x

    So the factored form should be just: x(x^2 +x -12) or not???
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  6. #6
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    Re: Factor completely

    Quote Originally Posted by ford2008 View Post
    Factored form should be (x-3)(x+4) = x^2 +4x -3x -12 = x^2 +x -12 = x(x^2 +x -12) = x^3 +x^2 -12x
    So the factored form should be just: x(x^2 +x -12) or not???
    Completely factored is x(x+4)(x-3)~.
    Thanks from ford2008
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  7. #7
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    Re: Factor completely

    2 other examples:

    1.
    2x^2 +4x -30

    Common factor is 2

    so 2(x-3)(x+5) is completely factored???

    2.
    3x^3 +21x^2 +36x

    Common factor is 3x

    so 3x(x+3)(x+4) is completely factored???
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