Question

Let $p\left(x\right)$ be any polynomial. When it is divided by $(x-19)$ and $(x-91)$, then the remainders are $91$ and $19$ respectively. The remainder, when $p\left(x\right)$ is divided by $(x-19)(x-91)$, is:

• A. $x-110$
• B. $110$
• C. $110-x$
• D. $4x+88$

Answer 1

The correct option is C $110-x$

This states that there are polynomials $q$ and $r$ such that

$p\left(x\right)=q\left(x\right)(x-19)+91$

and $p\left(x\right)=r\left(x\right)(x-91)+19$.

since $(x-19)(x-91)$ is second-degree, the remainder must be first-degree, i.e. $ax+b$.

$p\left(x\right)=(x-19)(x-91)+ax+b$

Since $p\left(19\right)=q\left(19\right)(19-19)+91=91$ and

$p\left(19\right)=r\left(19\right)(19-19)+91=91$,

We have a system of equations

$19a+b=91$

$91a+b=19$

$\Rightarrow b=91-19a$ and $b=19-91a$ from above

$\Rightarrow 91-19a=19-91a\Rightarrow -72a=72$

$\therefore a=-1$

Substitute $a=-1$ in $19a+b=91$

$\Rightarrow 19\times -1+b=91$

$\therefore b=91+19=110$

Thus, we concluded that the remainder of $p\left(x\right)$ by $(x-19)(x-91)$ is equal to $-x+110$ or $110-x$.