Factorise x^{3}+12 x^{2}+39 x+28. | vedclass

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Question

(A) Let $p(x) = x^{3}+12 x^{2}+39 x+28$.By the Factor Theorem,if $(x+a)$ is a factor,then $p(-a) = 0$.Testing $x = -1$: $p(-1) = (-1)^{3} + 12(-1)^{2}

Vedclass · Accepted Answer

$(x+1)(x+4)(x+7)$

Vedclass · Answer

(A) Let $p(x) = x^{3}+12 x^{2}+39 x+28$.By the Factor Theorem,if $(x+a)$ is a factor,then $p(-a) = 0$.Testing $x = -1$: $p(-1) = (-1)^{3} + 12(-1)^{2} + 39(-1) + 28 = -1 + 12 - 39 + 28 = 0$.Since $p(-1) = 0$,$(x+1)$ is a factor of $p(x)$.Now,divide $p(x)$ by $(x+1)$ using synthetic division or long division:$x^{3}+12 x^{2}+39 x+28 = (x+1)(x^{2}+11 x+28)$.Factor the quadratic expression $x^{2}+11 x+28$ by splitting the middle term:$x^{2} + 7x + 4x + 28 = x(x+7) + 4(x+7) = (x+4)(x+7)$.Thus,the complete factorization is $(x+1)(x+4)(x+7)$.

Factorise $x^{3}+12 x^{2}+39 x+28$.

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Factorise:
$3x^{3}-x^{2}-3x+1$

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