Let $p$ be a real polynomial ($p\in\mathbb{R}[x])$, if
- 5 is the remainder of the division $\frac{p(x)}{x-2}$, and
- 2 is the remainder of the division $\frac{p(x)}{x-5}$
What is the reaminder of the division $\frac{p(x)}{(x-2)(x-5)}$?
I have tried it, but I have no idea... Thank you
$p(x)=(x−2)(x−5)q(x)+ax+b$, for some $a,b \in \mathbb{R}$
Since we know the remainder of $p(x)$ when divided by $x-2$ and $x-5$
we obtain the following \begin{align} p(2) &= (2-2)(2-5)q(2) + a(2)+b\\ &= 2a+b = 5 \label{a} \end{align}
\begin{align} p(5) &= (5-2)(5-5)q(5) + a(5)+b\\ &= 5a+b = 2 \label{b} \end{align}
Solving the simultaneous equations \ref{a} and \ref{b}, we obtain $a = -1 \text{ and } b = 7$.
Hence, the remainder of the division $\frac{p(x)}{(x−2)(x−5)}$ is $-x + 7$.