Find the remainder when p(x) = 40x^2 + 11x - | vedclass

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(B) According to the Remainder Theorem,if a polynomial $p(x)$ is divided by $(x - a)$,the remainder is $p(a)$.Here,the divisor is $x + 2$,which can be

Vedclass · Accepted Answer

$75$

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(B) According to the Remainder Theorem,if a polynomial $p(x)$ is divided by $(x - a)$,the remainder is $p(a)$.Here,the divisor is $x + 2$,which can be written as $x - (-2)$.Therefore,$a = -2$.Now,substitute $x = -2$ into the polynomial $p(x) = 40x^2 + 11x - 63$:$p(-2) = 40(-2)^2 + 11(-2) - 63$$p(-2) = 40(4) - 22 - 63$$p(-2) = 160 - 22 - 63$$p(-2) = 160 - 85$$p(-2) = 75$Thus,the remainder is $75$.

Find the remainder when $p(x) = 40x^2 + 11x - 63$ is divided by $x + 2$.

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