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

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

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