Factorise : x^{3}+x^{2}-26 x+24 | vedclass

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Question

(A) Let $p(x) = x^{3}+x^{2}-26 x+24$.Sum of all the coefficients of the polynomial is $1+1-26+24 = 0$.Since the sum of coefficients is $0$,$(x-1)$ is

Vedclass · Accepted Answer

(x-$1$)(x+$6$)(x-$4$)

Vedclass · Answer

(A) Let $p(x) = x^{3}+x^{2}-26 x+24$.Sum of all the coefficients of the polynomial is $1+1-26+24 = 0$.Since the sum of coefficients is $0$,$(x-1)$ is a factor of $p(x)$.Now,we rewrite the polynomial to factorize it:$x^{3}+x^{2}-26 x+24 = x^{3}-x^{2}+2x^{2}-2x-24x+24$$= x^{2}(x-1)+2x(x-1)-24(x-1)$$= (x-1)(x^{2}+2x-24)$Now,factorize the quadratic expression $(x^{2}+2x-24)$ by splitting the middle term:$x^{2}+6x-4x-24 = x(x+6)-4(x+6) = (x-4)(x+6)$Therefore,the factors are $(x-1)(x-4)(x+6)$.

Factorise $: x^{3}+x^{2}-26 x+24$

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