\section*{Problem 6B - 151236B}

Let \(P(x)\) be a polynomial with real coefficients and \(p, q, r, s\) be real numbers such that \(p \neq q\).

When this polynomial is divided by \((x-p)\), the remainder is the number \(r\), and when the same polynomial is divided by \((x-q)\), the remainder is the number \(s\).

Under these conditions, what remainder results from the division of the polynomial \(P(x)\) by \((x-p)(x-q)\)?
Let \(R(x)\) be the polynomial that results from dividing the polynomial \(P(x)\) by \((x-p)(x-q)\). Since \((x-p)(x-q)\) has degree 2, \(R\) can have a maximum degree of 1. There are further polynomials \(P_{1}, P_{2}, P_{3}\) such that

\[
P(x)=(x-p) P_{1}(x)+r=(x-q) P_{2}(x)+s=(x-p)(x-q) P_{3}(x)+R(x)
\]

Thus, it follows that

\[
P(p)=r=R(p) \quad \text { and } \quad P(q)=s=R(q)
\]

Since \(R\) has a maximum degree of 1, it is uniquely determined, and it holds that

\[
R(x)=\frac{s-r}{q-p}(x-p)+r
\]