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Math Help - Factor Question

  1. #1
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    Factor Question

    How did this factorization come to be?



    I see where 4x and 2x came from, and I under stand that -3*3=-9 for the C position, but how did the 6 get there in the B position? Isn't it supposed to be what you get when you add -3 +3? That would be 0, maybe something Im missing?
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  2. #2
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    Re: Factor Question

    The coefficients of x inside the brackets are 4 and two - the expansion in the b position is then:
    4x*3+2x*(-3)=6x, as shown.
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  3. #3
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    Re: Factor Question

    Quote Originally Posted by itgl72 View Post
    How did this factorization come to be?



    I see where 4x and 2x came from, and I under stand that -3*3=-9 for the C position, but how did the 6 get there in the B position? Isn't it supposed to be what you get when you add -3 +3? That would be 0, maybe something Im missing?
    Multiply your a and c values, to give 8 x (-9) = -72. You are looking for two numbers that multiply to give -72 and add to give the b value, 6. The numbers are -6 and 12. So we break up the middle term according to these two numbers and factorise by grouping.

    \displaystyle \begin{align*} 8x^2 + 6x - 9 &= 8x^2 + 12x - 6x - 9 \\ &= 4x(2x + 3) - 3(2x + 3) \\ &= (2x + 3)(4x - 3)  \end{align*}
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  4. #4
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    Re: Factor Question

    Im good on this one, thanks all.
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  5. #5
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    Re: Factor Question

    8x^2+6x-9
    Factors of this equations are:
    => (4x-3)(2x+3)

    Since the equation is related to factoring trinomials method i.e. ax^2+bx+c. Solve it by factorizing each term as :
    8 can be factorize as 4 & 2. 6 can be factorize as 2 , 3 & -9 can be factorize as 3, -3. Now them accordingly to make equation balanced.
    8x^2+6x-9
    => 8x^2+12x-6x-9
    =>4x(2x+3)-3(2x+3)
    =>(4x-3)(2x+3)
    Thus the equation get its factors.
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