Question

The polynomials $p\left(x\right)=2x^2+3x+b$ and $q\left(x\right)=x^2+4x+6$ leave the same remainder when divided by $x-2$. What is the value of b?

• A. 4
• B. - 6
• C. 6
• D. - 4

Answer 1

The correct option is A 4

From remainder theorem, we know
The remainder when $p\left(x\right)$ is divided by $x-2isp\left(2\right)$
$\therefore$
$p\left(2\right)=2\left(2^2\right)+3\left(2\right)+b$
= 8 + 6 + b = 14 + b

Also,
The remainder when $q\left(x\right)$ is divided by $x-2isq\left(2\right)$
$\therefore$
$p\left(2\right)=\left(2^2\right)+4\left(2\right)+6$
= 4 + 8 + 6 = 18

Given,
These remainders are equal
$\Rightarrow$ 14 + b = 18
$\Rightarrow$ b = 4