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Thread: Fatoring polynomials - what am I doing wrong?

  1. Mar 11th 2011, 06:44 PM #1
    Eloise1995
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    Fatoring polynomials - what am I doing wrong?

    Hi! Here is the problem I was trying to solve, and what I did:

    3x^2+7x-6=0
    Since 3*(-6)=-18, I need two numbrers, the product of which is -18, and the sum of which is 7. I chose -9 and 2.
    3x^2-9x+2x-6=0
    3x(x-3)+2(x-3)=0
    (3x+2)(x-3)=0
    3x+2=0 or x-3=0
    x=3 x=-2/3


    However, the back of the textbook gives the answer x=-3 x=2/3
    What is wrong? Thanks if you could help, big time! I didn't do grade 10 math, and in grade 11 they only review factoring for a couple of lessons.

    P.S. Sorry for doubleposting!!
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  2. Mar 11th 2011, 06:52 PM #2
    pickslides
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    I see it like this

    \displaystyle 3x^2+7x-6=0

    \displaystyle 3x^2+9x-2x-6=0

    \displaystyle 3x(x+3)-2(x+3)=0

    \displaystyle (x+3)(3x-2)=0
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  3. Mar 11th 2011, 06:56 PM #3
    Prove It
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    \displaystyle -9 + 2 \neq 7. But \displaystyle 9 + (-2) does...
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