Precalculus

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Factor completely

2(3x - 5)3(2x + 1)^3 + (3x - 5)^2 (3)(2x + 1)^2 (2)

I'm not really sure how to do this problem

I started out by rewriting this
6(2x + 1)^3(3x -5) + 6(3x - 5)^2(2x + 1)^2

I am unsure were do go from here...

  • Precalculus -

    this was the answer in the back book. I do not see how to get this though

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

  • Precalculus -

    6(2x + 1)^3(3x -5) + 6(3x - 5)^2(2x + 1)^2

    6(2x+1)^2(2x+1)(3x-5) +6(2x+1)^2 (3x-5)(3x-5)

    6(2x+1)^2(3x-5) [ 2x+1 + 3x-5 ]

    6(2x+1)^2(3x-5)(5x-4)

    So I do not see how to get that answer either.

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