Consider a polynomial $f(x)$ when it is divided by $(x-91)$ it gives a remainder of $19$ and when divided by $(x-19)$ it gives a remainder $91$. Find the remainder when $p(x)$ is divided by $(x - 19)(x - 91)$.
\(\displaystyle \frac{f(x)}{x-19}=q(x)+\frac{91}{x-19},\\ \text{so} \\ f(x)=q(x)(x-19)+91, \\ f(19) = 91.\)
Similarly,
\(\displaystyle f(x)=r(x)(x-91)+ 19, \\ f(91) = 19.\)
\(\displaystyle \frac{f(x)}{(x-19)(x-91)}=s(x)+\frac{ax+b}{(x-19)(x-91)}, \\ f(x)=s(x)(x-19)(x-91)+ax+b,\\ f(19)=91=19a+b, \\f(91)=19=91a+b.\)
Solving simultaneously,
\(\displaystyle a=-1, \\b=110.\)
so the remainder is 110 - x.
