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Posted by on Monday, March 8, 2010 at 6:20pm.

Factor by introducing an appropriate substitution.

5y10 - 31y5 - 28

  • Math - , Monday, March 8, 2010 at 6:39pm

    You must have meant to type
    5y^10 - 31y^5 - 28
    then ... let y^5 = x
    5x^ - 31x - 28
    = (x-7)(5x + 4)
    = (y^5 - 7)(5y^5 + 4)

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