Review question

# What if dividing $f(x)$ by $(x-a)(x-b)$ and by $(x-a)(x-c)$ gives the same remainder? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8682

## Solution

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 $\begin{equation*} (b-c)f(a) + (c-a)f(b) + (a-b)f(c) = 0. \end{equation*}$