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Math Help - Factoring polynomials

  1. #1
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    Factoring polynomials

    I'm in a college algebra class and we're getting started with factoring. I'm going through most of them just fine but I'm stuck on this:

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

    I'm guessing I multiply everything out, and I did, and got to this point:

    -x^2(x+7) + 3(7x+9)

    Is that it? Is it even right? I'd appreciate some help Thanks!!
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  2. #2
    MHF Contributor red_dog's Avatar
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    I think you want to factor, not to expand the polynomial.

    2(x+1)(x-3)^2-3(x+1)^2(x-3)=(x+1)(x-3)[2(x-3)-3(x+1)]=

    =(x+1)(x-3)(-x-9)=-(x+1)(x-3)(x+9)
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  3. #3
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    Quote Originally Posted by shenaniganny View Post
    I'm in a college algebra class and we're getting started with factoring. I'm going through most of them just fine but I'm stuck on this:

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

    I'm guessing I multiply everything out, and I did, and got to this point:

    -x^2(x+7) + 3(7x+9)

    Is that it? Is it even right? I'd appreciate some help Thanks!!
    Factor out the common factors:

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

    {\color{white}2(x+1)(x-3)^2 - 3(x+1)^2(x-3)} = (x+1)(x-3)\left( 2(x-3) - 3(x+1) \right)

    {\color{white}2(x+1)(x-3)^2 - 3(x+1)^2(x-3)} =(x+1)(x-3)\left( -x-9)\right)

    {\color{white}2(x+1)(x-3)^2 - 3(x+1)^2(x-3)} =- (x+1)(x-3)\left( x+9)\right)
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  4. #4
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    heh seems really obvious now, thank you both, I appreciate it!
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