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Math Help - Factorising...x^3 + x^2 - 8x - 12

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    Factorising...x^3 + x^2 - 8x - 12

    How would I go about doing this one?

    If I remember correctly I must find two numbers which multiply together to make -12 and add together to make -8x right?

    I don't know what to do after that.
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  2. #2
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    Quote Originally Posted by Jamie88 View Post
    How would I go about doing this one?

    If I remember correctly I must find two numbers which multiply together to make -12 and add together to make -8x right?

    I don't know what to do after that.
    That is true only if you are factoring a quadratic (degree 2) Trinomial (has three terms) and the lead coeffient is 1.

    For degree 3 or larger if factoring by grouping isn't possible we need to use the rational roots theorem.

    The rational roots theorem tells us that all of the zero must be the ratio of the factors constant term and the factors of the lead coeffient.

    ie

    ax^n+bx^{n-1}+...mx+n

    so our constant term n=-12 and our lead term is a=1.

    the factors of negative twelve are (plus or minus) 1,2,3,4,6,12

    So the possiblities for the rational zero's are

    so checking some of them from our list....

    x^3+x^2-8x-12

    if we evaluate at 3 we get

    3^3+3^2-8(3)-12=27+9-24-12=0

    so x=3 must be a zero and (x-3) is a factor

    using polynomial long division or synthetic division you get

    (x-3)(x^2+4x+4) this can be factored again...

    (x-3)(x+2)^2
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    That is true only if you are factoring a quadratic (degree 2) Trinomial (has three terms) and the lead coeffient is 1.

    For degree 3 or larger if factoring by grouping isn't possible we need to use the rational roots theorem.

    The rational roots theorem tells us that all of the zero must be the ratio of the factors constant term and the factors of the lead coeffient.
    I'm afraid that is not what the rational root theorem tells us. What it does tell
    us is that if it does have rational roots then they are the ratio of a factor of
    the constant term to a factor of the coefficient of the highest power of the
    variable.

    There is no guarantee that a polynomial with integer coefficients has any rational roots. (example x^2-2=0).

    RonL
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    Talking Thanks

    Thanks for being clearer. I should have said all of the rational roots.
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    Brilliant, cheers mate.

    (x-3)(x^2+4x+4) this can be factored again...

    I understand this completely, but I'm just curious how to get (x^2+4x+4)?
    I know this works out to form the original polynomial, but I'm just curious as to the method you use too work this out? I've been substituting in various values until I get the original equation.

    Also (x+2) is a factor.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jamie88 View Post
    Brilliant, cheers mate.

    (x-3)(x^2+4x+4) this can be factored again...

    I understand this completely, but I'm just curious how to get (x^2+4x+4)?
    I know this works out to form the original polynomial, but I'm just curious as to the method you use too work this out? I've been substituting in various values until I get the original equation.
    he did (x^3 + x^2 - 8x - 12) \div (x - 3) using long division or synthetic division

    Also (x+2) is a factor.
    yes, you could have done it that way as well. and do long division a second time. that would be harder though. or you could do that for your first long division.
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    LaTex Question

    This may not be the best place to ask but....


    does Latex have a command for the long division or synthetic division symbols?
    Last edited by TheEmptySet; February 24th 2008 at 12:24 PM. Reason: Because I am spelling challanged
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