Dividing algebraic fractions with factorisation

To divide by an algebraic fraction, multiply by its reciprocal, then factorise and cancel as for multiplication.

Why there is an extra restriction: after turning the divisor over, its numerator becomes a denominator. Dividing by \(\dfrac{2a+6}{a+1}\) is impossible when \(2a+6=0\), so \(a\ne-3\) is needed as well as \(a\ne-1\).

Method

1. Change \(\div\) to \(\times\) and turn over the divisor.
2. Factorise every numerator and denominator.
3. Cancel common factors.
4. State the restrictions: the original denominators and the numerator of the divisor are non-zero.

Example

Simplify \(\dfrac{a^2+3a}{a^2-1}\div\dfrac{2a+6}{a+1}\).

\(\begin{aligned}\dfrac{a^2+3a}{a^2-1}\div\dfrac{2a+6}{a+1}&=\dfrac{a^2+3a}{a^2-1}\times\dfrac{a+1}{2a+6} \\ &=\dfrac{a(a+3)}{(a-1)(a+1)}\times\dfrac{a+1}{2(a+3)} \\ &=\dfrac{a}{2(a-1)},\qquad a\ne1,\ a\ne-1,\ a\ne-3\end{aligned}\)

Check with \(a=2\): \(\dfrac{10}{3}\div\dfrac{10}{3}=1\) and \(\dfrac{2}{2(1)}=1\). Turning over the first fraction instead of the divisor is the most common error.