On dividing x^{3} + ax^{2} + 19x + 20 by (x + | vedclass

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(B) Let $p(x) = x^{3} + ax^{2} + 19x + 20$. According to the Remainder Theorem,when $p(x)$ is divided by $(x + 3)$,the remainder is $p(-3)$. Given tha

Vedclass · Accepted Answer

$8$

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(B) Let $p(x) = x^{3} + ax^{2} + 19x + 20$. According to the Remainder Theorem,when $p(x)$ is divided by $(x + 3)$,the remainder is $p(-3)$. Given that the remainder is $a$,we have $p(-3) = a$. Substituting $x = -3$ into the polynomial: $(-3)^{3} + a(-3)^{2} + 19(-3) + 20 = a$ $-27 + 9a - 57 + 20 = a$ $9a - 64 = a$ $8a = 64$ $a = 8$.

On dividing $x^{3} + ax^{2} + 19x + 20$ by $(x + 3)$,if the remainder is $a$,then find the value of $a$.

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