Check whether the polynomial p(x) = x^{3} + 9 | vedclass

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Question

(A) The polynomial $p(x)$ will be a multiple of $x + 2$ if and only if $x + 2$ divides $p(x)$ leaving a remainder of $0$.According to the Factor Theor

Vedclass · Accepted Answer

Yes,it is a multiple.

Vedclass · Answer

(A) The polynomial $p(x)$ will be a multiple of $x + 2$ if and only if $x + 2$ divides $p(x)$ leaving a remainder of $0$.According to the Factor Theorem,if $p(a) = 0$,then $(x - a)$ is a factor of $p(x)$.Setting $x + 2 = 0$,we get $x = -2$.Now,we evaluate $p(-2)$:$p(-2) = (-2)^{3} + 9(-2)^{2} + 26(-2) + 24$$p(-2) = -8 + 9(4) - 52 + 24$$p(-2) = -8 + 36 - 52 + 24$$p(-2) = (-8 - 52) + (36 + 24)$$p(-2) = -60 + 60$$p(-2) = 0$Since the remainder is $0$,$(x + 2)$ is a factor of $p(x)$.Therefore,$p(x)$ is a multiple of $x + 2$.

Check whether the polynomial $p(x) = x^{3} + 9x^{2} + 26x + 24$ is a multiple of $x + 2$ or not.

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