Factorise: 2y^{3} + y^{2} - 2y - 1 | vedclass

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Question

(A) Let $p(y) = 2y^{3} + y^{2} - 2y - 1$.By trial,we check $p(1)$:$p(1) = 2(1)^{3} + (1)^{2} - 2(1) - 1 = 2 + 1 - 2 - 1 = 0$.Since $p(1) = 0$,by the F

Vedclass · Accepted Answer

$(y - 1)(y + 1)(2y + 1)$

Vedclass · Answer

(A) Let $p(y) = 2y^{3} + y^{2} - 2y - 1$.By trial,we check $p(1)$:$p(1) = 2(1)^{3} + (1)^{2} - 2(1) - 1 = 2 + 1 - 2 - 1 = 0$.Since $p(1) = 0$,by the Factor Theorem,$(y - 1)$ is a factor of $p(y)$.Now,divide $p(y)$ by $(y - 1)$:$2y^{3} + y^{2} - 2y - 1 = 2y^{2}(y - 1) + 3y(y - 1) + 1(y - 1)$$= (y - 1)(2y^{2} + 3y + 1)$.Factorize the quadratic expression $2y^{2} + 3y + 1$ by splitting the middle term:$2y^{2} + 2y + y + 1 = 2y(y + 1) + 1(y + 1) = (y + 1)(2y + 1)$.Thus,the factors are $(y - 1)(y + 1)(2y + 1)$.

Factorise: $2y^{3} + y^{2} - 2y - 1$

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