The roots of the equation will be (y+1)(y-1)(2y+1)
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By factor theorem,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, Put y = 1 we get 2+1-2-1=0 =>!(y-1) is the factor of the given equation now divide the equ by y-1 we get 2y^2 -3y+1 ,,,,,,,,,,,,,,,, now again using factor theorem put y=1 we get 2 -3+1 =0 => (y-1) is the factor of 2y^2-3y+1 divide it by (y-1) we get 2y-1 so the equ become (y-1)(y-1)(2y-1)