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Math Help - factorising

  1. #1
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    factorising

    hi, i've done part of a question that i need help finishing. the question is:

    use the factor theorem to show that (x - 2) is a factor of P(x) = x^3 - 3x^2 - 4x + 12 (which i did)
    then it says:
    write P(x) as a product of three linear factors.
    now obviously (x - 2) is one of them so i found out what the equation was that had to be multiplied by (x - 2) to get P(x) = x^3 - 3x^2 - 4x + 12 and i came to (x^2 - x -6) and tried to factorise it further but couldn't. why can't i find two numbers that multiply to give -6 but add to give -1? surely this is the way to do it?

    thanks, mark
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  2. #2
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    Quote Originally Posted by mark View Post
    i came to (x^2 - x -6) and tried to factorise it further but couldn't. why can't i find two numbers that multiply to give -6 but add to give -1? surely this is the way to do it?

    thanks, mark
    Hi

    You mean : find two numbers that multiply to give -6 but add to give +1 ?
    -2 and 3
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  3. #3
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    i looked at it again and i can't see why its two numbers that multiply to give -6 but add to give +1? the equation does come to x^2 - x - 6 to factorise it from there you need to multiply 1 from the x^2 by minus 6 (the constant) and then find two numbers that multiply to give -6 and add to give -1 (from the middle term). can anyone point me in the right direction here please?
    Last edited by mark; September 16th 2009 at 06:40 AM.
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  4. #4
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    Quote Originally Posted by mark View Post
    i can't see why its two numbers that multiply to give -6 but add to give +1?
    You're right: You need factors of -6 that add to -1, not +1. So use the numbers provided, but reverse their signs.
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  5. #5
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    I am sorry but a x^2 + bx + c has 2 roots whose sum is -\frac{b}{a} and whose product is +\frac{c}{a}

    In your case x^2 - x - 6, the sum is +1 and the product is -6
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  6. #6
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    Exclamation

    (x + 3)(x - 2) = x^2 + 1x - 6

    This is not the original quadratic, so adding to +1 would seem to lead to an invalid result.
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