Question

Let $p\left(x\right)$ be a polynomial such that when $p\left(x\right)$ is divided by $(x-19)$, the remainder is $99$ and when $p\left(x\right)$ is divided by $(x-99)$, the remainder is $19$. The remainder when $p\left(x\right)$ is divided by $(x-19)(x-99)$ is?

• A. $(-x+80)$
• B. $(x+80)$
• C. $(-x+118)$
• D. $(x+118)$

Answer 1

Answer: (C)

$(-x+118)$

Let the polynomial be $f\left(x\right)\times (x-19)(x-99)+(ax+b)$

Where, $ax+b$ is the remainder.

Now, according to the remainder theorem,

$p\left(19\right)=99$ or

$\Rightarrow$ $19a+b=99$ ----- ( 1 )

$p\left(99\right)=19$

$\Rightarrow$ $99a+b=19$ ----- ( 2 )

Subtracting ( 2 ) from ( 1 ) we get,

$\Rightarrow$ $-80a=80$

$\Rightarrow$ $a=-1$

Substituting value of $a$ in ( 1 ) we get,

$\Rightarrow$ $-19+b=99$

$\Rightarrow$ $b=118$

The required remainder $=(ax+b)=-x+118$