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pre cal

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Factor completely: 3x^2(4x2 + 1)^8 + 64x^4(4x2 + 1)^7.

  • pre cal -


  • pre cal -

    how did you solve this??? I am so los.

  • pre cal -

    3x^2(4x2 + 1)^8 + 64x^4(4x2 + 1)^7

    Notice that x^2 is a factor of each term, and so is a power of (4x^2+1)
    So, factoring them out, we have

    x^2 (4x^2+1)^7 (3(4x^2+1)+64x^2)
    3(4x^2+1)+64x^2 = 12x^2+3+64x^2 = 76x^2+3

    and that's Tony's answer

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