Question

find zeroes and verify the relation between 2y^3+y^2-2y-1​

Answer 1

$2y {}^{3} + y {}^{2} - 2y - 1 = 0 \\ 2y {}^{3} + 2y {}^{2} - y {}^{2} - y - y - 1 = 0 \\ 2y {}^{2} (y + 1) - y(y + 1) - 1(y + 1) = 0 \\ (2y {}^{2} - y - 1)(y + 1) = 0 \\ (2y {}^{2} - 2y + y - 1)(y + 1) = 0 \\ (2y(y - 1) + 1(y - 1))(y + 1) = 0 \\ (2y + 1)(y - 1)(y + 1) = 0 \\ 2y + 1 = 0 \\ 2y = - 1 \\ y - \frac{1}{2} \\ y + 1 = 0 \\ y = - 1 \\ y - 1 = 0 \\ y = 1 \\ hence \: \: \: y = - ( \frac{1}{2} ) (y = - 1) \: \: \: (y = 1) \: are \: the \: three \: zero \: of \: this \: equation \:$