Let $p(x)$ be a polynomial such that when $p(x)$ is divided by $x - 19$ the remainder is $99$, and when $p(x)$ is divided by $x - 99$ remainder is $19$. Find the remainder when $p(x)$ is divided by $(x-19)(x-99)$.
I tried using the division algorithm but it can't be done that way, I am looking for ways that can be understood by senior high school student.
On
You want a polynomial $p(x)$ such that $$p(x)=(x-19)q_1(x)+99\qquad\text{ and }\qquad p(x)=(x-99)q_2(x)+19.$$ Then it follows that $$(x-99)q_2(x)=p(x)-19=(x-19)q_1(x)+80,$$ and pluging in $x=99$ then shows that $q_1(99)=-1$. A simple candidate for $q_1(x)$ is then $$q_1(x)=x-100,$$ and plugging this in yields $$p(x)-19=(x-19)q_1(x)+80=(x-19)(x-100)+80=x^2-119x+1980.$$
Hint: If $p(x)=(x-19)(x-99)q(x)+ax+b$, what do you get if you put $x=19$? And if you put $x=99$?