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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$75$

Vedclass · Answer

(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 $x + 2$.

Similar Questions

The difference between the $HCF$ and $LCM$ of two prime numbers $7$ and $11$ is . . . . . . .

Is $\sqrt{2}$ a rational number?

Find the $HCF$ and $LCM$ of the following pair of integers $510$ and $92$ and verify that $HCF \times LCM = \text{product of the two numbers}$.

For the polynomial $p(x) = x^2 - 4x + 3$,the value of $\alpha + \beta$ is . . . . . . .

For any positive integer $n$,the last digit of $5^n$ is always . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module