Remainder and factor theorems: substitute instead of dividing

Cambridge asks this in two ways:

• "Find the remainder when $f(x)$ is divided by $(x - k)$." Work out $f(k)$.
• "Show that $(x - k)$ is a factor." Show $f(k) = 0$.

Both come from the same substitution:

$\text{remainder} = f(k) \quad\text{when dividing by } (x - k)$

$(x - k) \text{ is a factor} \iff f(k) = 0$

Why substitution is enough: dividing by $(x - k)$ always leaves a constant remainder $r$. Substituting $x = k$ makes the divisor zero, so only $r$ survives.

1. Find what makes the divisor zero. For $x - k$ that is $x = k$. For $ax + b$ it is $x = -\dfrac{b}{a}$.
2. Substitute that value into $f(x)$. No division is needed.
3. Read the result. It is the remainder; if it is zero, the divisor is a factor.
4. For a full factorisation: divide out the factor, then factorise the quotient.

The forgotten step: with $ax + b$ the substitution gives the remainder, but the factor to state is still $ax + b$. Writing $x + \dfrac{b}{a}$ instead changes the quotient by a factor of $a$.