Factorise : x^{3}-3 x^{2}-9 x-5 | vedclass

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

Question

(A) Let $p(x) = x^{3}-3x^{2}-9x-5$.By the factor theorem,we test values to find a root. Let $x = -1$:$p(-1) = (-1)^{3} - 3(-1)^{2} - 9(-1) - 5$$p(-1)

Vedclass · Accepted Answer

$(x+1)(x+1)(x-5)$

Vedclass · Answer

(A) Let $p(x) = x^{3}-3x^{2}-9x-5$.By the factor theorem,we test values to find a root. Let $x = -1$:$p(-1) = (-1)^{3} - 3(-1)^{2} - 9(-1) - 5$$p(-1) = -1 - 3(1) + 9 - 5$$p(-1) = -1 - 3 + 9 - 5 = 0$.Since $p(-1) = 0$,$(x+1)$ is a factor of $p(x)$.Now,divide $p(x)$ by $(x+1)$ using long division or synthetic division:$x^{3}-3x^{2}-9x-5 = (x+1)(x^{2}-4x-5)$.Factorize the quadratic expression $x^{2}-4x-5$ by splitting the middle term:$x^{2}-5x+x-5 = x(x-5) + 1(x-5) = (x-5)(x+1)$.Therefore,the complete factorization is:$x^{3}-3x^{2}-9x-5 = (x+1)(x+1)(x-5) = (x+1)^{2}(x-5)$.

Factorise : $x^{3}-3 x^{2}-9 x-5$

Similar Questions

Use suitable identities to find the product: $\left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right)$

Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer: $y + \frac{2}{y}$

Find the zero of the polynomial: $p(x) = x - 5$.

Find $p(0)$,$p(1)$,and $p(2)$ for the following polynomial: $p(x) = (x - 1)(x + 1)$.

Find the remainder when $x^4+x^3-2x^2+x+1$ is divided by $x-1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module