Show that the remainder when the polynomial \(f(x)\) is divided by \((x-a)\) is \(f(a)\).

Show further that, if \(f(x)\) is divided by \((x-a)(x-b)\), where \(a \ne b\), then the remainder is
\[\begin{equation*}
\left( \frac{f(a) - f(b)}{a-b} \right)x + \left( \frac{af(b) - bf(a)}{a-b} \right).
\end{equation*}\]

This time, we can write
\[\begin{equation*}
f(x) = (x-a)(x-b)g(x) + \alpha x + \beta,
\end{equation*}\]
where if \(f(x)\) is of degree \(n, g(x)\) is of degree \(n-2\). By substituting in \(x = a\) and \(x = b\), we have that
\[\begin{align}
f(a) &= \alpha a + \beta, \label{eq:pt-2-linear}\\
f(b) &= \alpha b + \beta. \nonumber
\end{align}\]
By subtracting the second equation from the first, it follows that
\[\begin{equation*}
f(a) - f(b) = \alpha(a-b),
\end{equation*}\]
and hence, as \(a \ne b\),
\[\begin{equation*}
\alpha = \frac{f(a) - f(b)}{a-b}.
\end{equation*}\]
From \(\eqref{eq:pt-2-linear}\), it follows that
\[\begin{equation*}
\beta = f(a) - \alpha a = f(a) - \frac{f(a) - f(b)}{a-b}a,
\end{equation*}\]
so that the remainder term is
\[\begin{align}
\alpha x + \beta &= \left( \frac{f(a) - f(b)}{a-b} \right)x + f(a) - \frac{f(a) - f(b)}{a-b}a\\
&= \frac{f(a) - f(b)}{a-b} x + \frac{(a-b)f(a) - a(f(a) - f(b))}{a-b} \nonumber \\
&= \frac{f(a) - f(b)}{a-b} x + \frac{af(b) - bf(a)}{a-b}. \label{eq:pt-2-final-form}
\end{align}\]
The remainders when \(f(x)\) is divided by \((x-a)(x-b)\) and by \((x-a)(x-c)\), \(a \ne b \ne c\), are equal. Prove that
\[\begin{equation*}
(b-c)f(a) + (c-a)f(b) + (a-b)f(c) = 0.
\end{equation*}\]

From \(\eqref{eq:pt-2-final-form}\) and the assumptions given, we must have that
\[\begin{equation*}
\frac{f(a) - f(b)}{a-b} x + \frac{af(b) - bf(a)}{a-b} = \frac{f(a) - f(c)}{a-c} x + \frac{af(c) - cf(a)}{a-c},
\end{equation*}\]
which implies that
\[\begin{equation*}
\frac{f(a) - f(b)}{a-b} = \frac{f(a) - f(c)}{a-c}.
\end{equation*}\]
By multiplying both sides by \((a-b)(a-c)\), we have that
\[\begin{equation*}
(a-c)f(a) - (a-c)f(b) = (a-b)f(a) - (a-b)f(c),
\end{equation*}\]
which implies that
\[\begin{equation*}
(b-c)f(a) + (c-a)f(b) + (a-b)f(c) = 0.
\end{equation*}\]