Multiplying algebraic fractions with factorisation

When the numerators and denominators are sums, factorise every one of them first. In a product, a factor in any numerator may cancel the same factor in any denominator.

Why: \(\dfrac{A}{B}\times\dfrac{C}{D}=\dfrac{AC}{BD}\) is one fraction, so all the numerator factors sit together on top and all the denominator factors sit together underneath. Keeping the brackets unexpanded makes the common factors visible.

Method

1. Factorise each numerator and denominator completely (common factor, difference of two squares, trinomial, reversed bracket).
2. Write the product as one fraction of factors.
3. Cancel common factors from top and bottom.
4. State the restrictions from every original denominator.

Example

Simplify \(\dfrac{x^2-4}{3x+6}\times\dfrac{9}{x^2-2x}\).

\(\begin{aligned}\dfrac{x^2-4}{3x+6}\times\dfrac{9}{x^2-2x}&=\dfrac{(x-2)(x+2)}{3(x+2)}\times\dfrac{9}{x(x-2)} \\ &=\dfrac{9(x-2)(x+2)}{3x(x+2)(x-2)} \\ &=\dfrac{3}{x},\qquad x\ne0,\ x\ne2,\ x\ne-2\end{aligned}\)

Check with \(x=3\): \(\dfrac{5}{15}\times\dfrac{9}{3}=1\) and \(\dfrac33=1\). A common error is to expand \((x-2)(x+2)\) back to \(x^2-4\) before cancelling, which hides the common factor. Another is to drop the restrictions \(x\ne\pm2\) because those factors cancelled.