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2y^3+y^2-2y-1...

`2y^3+y^2-2y-1`

Text Solution

Verified by Experts

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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