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

If \(f(x)\) is a polynomial of degree \(n\), there must be a polynomial \(g(x)\) of degree \(n-1\) and a constant \(r\) such that \[\begin{equation*} f(x) = (x-a)g(x) + r. \end{equation*}\] We say that \(r\) is the remainder on dividing \(f(x)\) by \(x-a\). By taking \(x = a\) in the above, \[\begin{equation*} f(a) = (a-a)g(a) + r = r. \end{equation*}\]
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 \[\begin{equation*} (b-c)f(a) + (c-a)f(b) + (a-b)f(c) = 0. \end{equation*}\]