Factorise x^{3}-11 x^{2}+20 x+32. | vedclass

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Question

(A) Let $p(x) = x^{3}-11 x^{2}+20 x+32$.By the Factor Theorem,we test values to find a root. Let $x = -1$:$p(-1) = (-1)^{3} - 11(-1)^{2} + 20(-1) + 32

Vedclass · Accepted Answer

$(x+1)(x-4)(x-8)$

Vedclass · Answer

(A) Let $p(x) = x^{3}-11 x^{2}+20 x+32$.By the Factor Theorem,we test values to find a root. Let $x = -1$:$p(-1) = (-1)^{3} - 11(-1)^{2} + 20(-1) + 32 = -1 - 11 - 20 + 32 = 0$.Since $p(-1) = 0$,$(x+1)$ is a factor of $p(x)$.Now,divide $x^{3}-11 x^{2}+20 x+32$ by $(x+1)$ using synthetic division or long division:$x^{3}+x^{2} - 12x^{2}-12x + 32x+32 = x^{2}(x+1) - 12x(x+1) + 32(x+1) = (x+1)(x^{2}-12x+32)$.Factor the quadratic $x^{2}-12x+32$:$x^{2}-8x-4x+32 = x(x-8)-4(x-8) = (x-4)(x-8)$.Thus,the factors are $(x+1)(x-4)(x-8)$.

Factorise $x^{3}-11 x^{2}+20 x+32$.

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