SOLUTION: Factor completely {{{(y-1)^4-(y-1)^2}}}
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Question 726670
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Factor completely
Answer by
lwsshak3(11628)
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(y-1)^4-(y-1)^2
This is a difference of 2 squares
[(y-1)^2+(y-1)][(y-1)^2-(y-1)]
(y^2-2y+1+y-1)(y^2-2y+1-y+1)
(y^2-y)(y^2-3y+2)
y(y-1)(y-2)(y-1)