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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

Without actually calculating the cubes,find the value of each of the following: $(28)^{3} + (-15)^{3} + (-13)^{3}$

Determine whether $(x + 1)$ is a factor of the polynomial $p(x) = x^{4} + x^{3} + x^{2} + x + 1$.

Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer: $x^{10} + y^{3} + t^{50}$

Give possible expressions for the length and breadth of each of the following rectangles,in which their areas are given: $\text{Area} = 35y^2 + 13y - 12$.

Check whether the polynomial $q(t) = 4t^3 + 4t^2 - t - 1$ is a multiple of $2t + 1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module