Asked by DK
Factor by introducing an appropriate substitution.
5y10 - 31y5 - 28
5y10 - 31y5 - 28
Answers
Answered by
Reiny
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)
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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