Dividing p(x) = x^{2} + 12x + 36 by (x + 5),t | vedclass

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Question

(A) According to the Remainder Theorem,if a polynomial $p(x)$ is divided by $(x - a)$,the remainder is $p(a)$.Here,we are dividing $p(x) = x^{2} + 12x

Vedclass · Accepted Answer

$1$

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(A) According to the Remainder Theorem,if a polynomial $p(x)$ is divided by $(x - a)$,the remainder is $p(a)$.Here,we are dividing $p(x) = x^{2} + 12x + 36$ by $(x + 5)$,which is equivalent to $(x - (-5))$.Therefore,the remainder is $p(-5)$.Substitute $x = -5$ into the polynomial:$p(-5) = (-5)^{2} + 12(-5) + 36$$p(-5) = 25 - 60 + 36$$p(-5) = 61 - 60$$p(-5) = 1$Thus,the remainder is $1$.

Dividing $p(x) = x^{2} + 12x + 36$ by $(x + 5)$,the remainder is.......

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