Let $P(x)$ be a polynomial such that when $P(x)$ is divided by $x-17$, the remainder is $14$, and when $P(x)$ is divided by $x-13$, the remainder is $6$. What is the remainder when $P(x)$ is divided by $(x-13)(x-17)$?
Here was my process, that I'm not sure if it's right or not:
We can write $P(x)$ in the form of $$P(x)=Q(x)(x-17)(x-13)+cx+d$$
Thus, by the remainder theorem, we have a system of equations:
\begin{align*} 14c+d &=6,\\ 6c+d &=14. \end{align*}
Solving gets $c=-1, d=20.$
Thus, our remainder is $\boxed{-x+20}.$
Did I make any flaws during my process. Thanks in advance for helping. :)
From $$P(x)=Q(x)(x-17)(x-13)+cx+d$$
Now, let $x=17$, then we have $$17c+d=14$$
If we let $x=13$, then we have
$$13c+d=6$$
Now solve for $c$ and $d$.
Subtract the two equations, we ahve $4c=8 \iff c=2$. Proceed on to solve for $d$ to get the remainder.