InterviewSolution
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`2y^3+y^2-2y-1` |
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Answer» Here, `p(y) = 2y^3+y^2-2y-1` `=y^2(2y+1)-1(2y+1)` `= (y^2-1)(2y+1)` `=(y+1)(y-1)(2y+1)` So, zeroes of `p(y)` will be `1,-1 and -1/2.` |
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