A certain function of \(x\) can be expressed in the form \[a+b(x-1)+c(x-1)(x-2)+2(x-1)(x-2)(x-3),\] where \(a\), \(b\), \(c\), are numerical. When the function is divided in turn by \(x-1\), \(x-2\), \(x-3\) the remainders are \(10\), \(3\), \(4\). Calculate the values of \(a\), \(b\), \(c\).

Let \(f(x)\) denote our function of \(x\). That is, \[\begin{equation} f(x) = a+b(x-1)+c(x-1)(x-2)+2(x-1)(x-2)(x-3). \label{eq:1} \end{equation}\]

As \(f(x)\) is a polynomial, the Remainder Theorem tells us that the remainder when divided by \((x-a)\) is equal to \(f(a)\). Substituting \(1\), \(2\) and \(3\) into \(\eqref{eq:1}\) in turn, we get that \[f(1) = a,\quad f(2)=a+b \quad \text{and} \quad f(3)=a+2b+2c,\] and so \(a=10\), \(a+b=3\) and \(a+2b+2c=4\).

Solving these equations then gives that \(a=10\), \(b=-7\) and \(c=4\).