A calculus of the absurd

4.6 The factor and remainder theorems

The remainder theorem states that for any number \(a\), the remainder when we divide a polynomial (which can be written as \(f(x)\)) by \((x - a)\) is equal to \(f(a)\). 1515 We can prove this without too much trouble. Firstly, by the definition of division, we can write the result of any division as a "quotient" \(Q(x)\) and a remainder \(R(x)\) as a fraction of \(x-a\) - as an equation, that is, that \(\frac {f(x)}{(x-a)} = Q(x) + \frac {R(x)}{x-a}\). If we multiply both sides by \(x-a\), then we obtain that \(f(x) = Q(x)(x-a)+R(x)\). We can then set \(x=a\), which gives that \(f(a) = R(a)\). Note that \(Q(x)(a-a)=0\).

A special case of the remainder theorem is known as the "factor theorem" and it relates the roots of a polynomial to its factors. If \((x-a)\) divides \(f(x)\) with no remainder, then by the remainder theorem, this means that \(f(a)=0\) and so \(a\) is a root as well as a factor. Roots are factors, and factors are roots.

4.6.1 An example

: Given that \((x-1)\) is a factor of \(P(x) = 5x^3 - 9x^2 + 2x + a\), find the value of \(a\).


: We can directly apply the factor theorem here; as \((x-1)\) is a factor of \(P(x)\), it must be the case that \(P(1) = 0\). With this information, we can form an equation where the only variable is \(a\).

\begin{align} & P(1) & = 0 \\ & 5 - 9 + 2 + a & = 0 \\ & -2 + a & = 0 \\ & a & = 0 \end{align}