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Math Help - Can someone help me solve my Factorization problem.

  1. #1
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    Can someone help me solve my Factorization problem.

    2*x ^2 - 13*x + 18 = 0

    So far I get :

    General Term : a*x^2 + b*x + c

    Which would be factored into :

    (m*x + p) * ( n*x + q)

    Where

    m*n = a
    p*q = c and
    P*n + m*q = b

    Step 1 -

    m*n = 2
    p*q = 18
    P*n + m*q = -13

    Step 2 -

    Possible integer multiples
    m n (and obverse)
    2 = 1 & 2

    Possible integer multiples
    p q (and obverse)
    18 = 1 & 18
    2 & 9
    3 & 6

    So I get

    (2*x + p)*(x + q) now I think I get a - 13 and an 18 so lets try

    (2*x - 3) * (x - 6)

    I now get lost help anyone. ...
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  2. #2
    Eater of Worlds
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    What 2 numbers when multiplied equal 36 and when added equal -13?.

    Hmmmm....how about -9 and -4

    2x^{2}-9x-4x+18

    Group:

    (2x^{2}-9x)-(4x-18)

    Factor out something common:

    x(2x-9)-2(2x-9)=(x-2)(2x-9)

    Voila!
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  3. #3
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    Thanks.

    Thank you, very quick answer. Button pressed.
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  4. #4
    MHF Contributor red_dog's Avatar
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    If f(x)=ax^2+bx+c and x_1,x_2 are the real roots of f, then f(x)=a(x-x_1)(x-x_2).
    So, the roots of 2x^2-13x+18 are \displaystyle x_1=2,x_2=\frac{9}{2}.
    Then \displaystyle 2x^2-13x+18=2(x-2)\left(x-\frac{9}{2}\right)=(x-2)(2x-9).
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