Factorise x^{3}+8x^{2}+9x-18. | vedclass

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Question

(A) Let $p(x) = x^{3}+8x^{2}+9x-18$.By the factor theorem,we test values to find a root. If $x=1$,then $p(1) = 1^{3} + 8(1)^{2} + 9(1) - 18 = 1 + 8 +

Vedclass · Accepted Answer

$(x-1)(x+3)(x+6)$

Vedclass · Answer

(A) Let $p(x) = x^{3}+8x^{2}+9x-18$.By the factor theorem,we test values to find a root. If $x=1$,then $p(1) = 1^{3} + 8(1)^{2} + 9(1) - 18 = 1 + 8 + 9 - 18 = 0$.Since $p(1) = 0$,$(x-1)$ is a factor of $p(x)$.Now,divide $x^{3}+8x^{2}+9x-18$ by $(x-1)$ using synthetic division or long division:$x^{3}+8x^{2}+9x-18 = (x-1)(x^{2}+9x+18)$.Next,factorize the quadratic expression $x^{2}+9x+18$:$x^{2}+6x+3x+18 = x(x+6) + 3(x+6) = (x+3)(x+6)$.Thus,the complete factorization is $(x-1)(x+3)(x+6)$.

Factorise $x^{3}+8x^{2}+9x-18$.

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