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simpliy the following: (a) {[4x^2-49]/[8x^3+27]} *{ [4x^2+12x+9]/[2x^2-13x+21]} (b) {[x^2+x-6]/[x^2+7x-6]}*{[20x^2-7x-3]/[6x^2-25x+4]}*{[4x^2-11x-3]/[6x^2-19x+3]}

  • Algebra - ,

    Here's (a):

    {[4x^2-49]/[8x^3+27]} *{ [4x^2+12x+9]/[2x^2-13x+21]}

    4x^2-49 = (2x-7)(2x+7)
    8x^3+27 = (2x+3)(4x^2-6x+9)
    4x^2+12x+9 = (2x+3)(2x+3)
    2x^2-13x+21 = (2x-7)(x-3)

    Putting it all together, you can cancel the (2x-7)(2x+3) to get

    (2x+7)(2x+3) / (x-3)(4x^2-6x+9)

    Do (b) likewise. The key is to know how to factor these babies. Recognize differences of square and cubes.

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