Multiplying and Dividing Algebraic Fractions
Theory
To multiply algebraic fractions, factorise every numerator and denominator, cancel common factors, then multiply. To divide, multiply by the reciprocal of the second fraction. This Year 10A maths page covers single-term fractions, brackets with a common factor, reversed brackets, chains of operations and fractions inside brackets. Fractions built from trinomials, and adding or subtracting with binomial denominators, are in the next subtopic, Further Simplification of Algebraic Fractions.
Algebraic fractions multiply like number fractions: numerator times numerator, denominator times denominator, \(\dfrac{a}{b}\times\dfrac{c}{d} = \dfrac{ac}{bd}\). Unlike adding, you do not need a common denominator.
The reciprocal of \(\dfrac{c}{d}\) is \(\dfrac{d}{c}\). Dividing by a fraction is the same as multiplying by its reciprocal.
The work is in the cancelling. A factor is something that multiplies the whole numerator or the whole denominator, such as \(4\), \(x\) or a bracket \((x+3)\). Only factors cancel. So the first step is always to factorise every part: take out common factors, then use the difference of two squares or factorise a trinomial.
Cancel across the whole product: a factor on the top of one fraction can cancel with the same factor on the bottom of another. The simplified answer equals the original expression for every value of the pronumeral that does not make any denominator zero.
Factorising toolkit:
| Type | Pattern, with an example |
|---|---|
| Common factor | \(ab+ac = a(b+c)\) e.g. \(5x^2-15x = 5x(x-3)\) |
| Difference of two squares | \(a^2-b^2 = (a-b)(a+b)\) e.g. \(x^2-36 = (x-6)(x+6)\) |
| Perfect square | \(a^2-2ab+b^2 = (a-b)^2\) e.g. \(x^2-10x+25 = (x-5)^2\) |
| Monic trinomial | \(x^2+bx+c\): numbers with product \(c\), sum \(b\) e.g. \(x^2+x-20 = (x+5)(x-4)\) |
| Reversed bracket | \(b-a = -(a-b)\) e.g. \(6-2x = -2(x-3)\) |
How to multiply or divide algebraic fractions
- Change division to multiplication: keep the first fraction, flip the one after the \(\div\) sign.
- Factorise every numerator and denominator fully. Watch for reversed brackets such as \(5-x\).
- Cancel each factor that appears on the top and the bottom, across all the fractions. Cross out one copy from each.
- Multiply what is left: tops together, bottoms together. Simplify any number or power of a pronumeral, such as \(\dfrac{12x^3}{4x} = 3x^2\).
- Check by substituting a simple value, such as \(x = 5\), into the question and your answer.
Other question types
- Chains: in \(A\times B\div C\), flip only \(C\), then cancel across all three fractions.
- Brackets with fractions inside: make each bracket a single fraction first (Example 4). A "show that" question works the same way: simplify one side step by step until it matches the other.
- Two pronumerals: factorise the same way, \(p^2-pq = p(p-q)\), and cancel powers such as \(\dfrac{p^3q^2}{p^2q} = pq\).
- Find a constant: if the product must equal a given fraction, cancel what you can, then decide which factors the unknown expression must contain. Expand those factors and compare coefficients.
- Models: area \(=\) length \(\times\) width, speed \(=\) distance \(\div\) time. Simplify the expression first, then substitute or solve.
(i) \(\dfrac{9p}{4q} \times \dfrac{2q^2}{15p}\)
(ii) \(\dfrac{5x}{6} \div \dfrac{10x^2}{9} \times \dfrac{4x}{3}\)
(i) Nothing needs factorising. Cancel the numbers (\(3\) into \(9\) and \(15\), \(2\) into \(2\) and \(4\)), one \(p\) and one \(q\).
| \(\dfrac{9p}{4q} \times \dfrac{2q^2}{15p}\) | \(=\) | \(\dfrac{3}{2} \times \dfrac{q}{5}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{3q}{10}}\) |
(ii) Flip only the fraction straight after the \(\div\) sign, then cancel across all three fractions.
| \(\) | \(\) | \(\dfrac{5x}{6} \times \dfrac{9}{10x^2} \times \dfrac{4x}{3}\) |
| \(\) | \(=\) | \(\dfrac{5 \times 9 \times 4 \times x \times x}{6 \times 10 \times 3 \times x^2}\) |
| \(\) | \(=\) | \(\dfrac{180x^2}{180x^2} = \mathbf{1}\) |
Every factor cancels, so the answer is \(1\), not \(0\).
Take out the common factor in each bracket. Both become a number times \((x-4)\).
| \(3x-12\) | \(=\) | \(3(x-4)\) |
| \(2x-8\) | \(=\) | \(2(x-4)\) |
| \(\dfrac{3(x-4)}{x^2}\times\dfrac{5x}{2(x-4)}\) | \(=\) | \(\dfrac{3}{x^2}\times\dfrac{5x}{2}\) |
| \(\) | \(=\) | \(\dfrac{15x}{2x^2}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{15}{2x}}\) |
One \(x\) on the top cancels one of the two \(x\)s in \(x^2\). Check with \(x = 5\): \(\dfrac{3}{25}\times\dfrac{25}{2} = \dfrac{3}{2}\), and \(\dfrac{15}{10} = \dfrac{3}{2}\).
Flip the second fraction and multiply. Factorise everything, writing \(2-a = -(a-2)\).
| \(\) | \(\) | \(\dfrac{10-5a}{a^2+3a}\times\dfrac{a+3}{a^2-4}\) |
| \(\) | \(=\) | \(\dfrac{-5(a-2)}{a(a+3)}\times\dfrac{a+3}{(a-2)(a+2)}\) |
| \(\) | \(=\) | \(\boldsymbol{-\dfrac{5}{a(a+2)}}\) |
\(10-5a = 5(2-a) = -5(a-2)\). The minus sign survives the cancelling, so the answer is negative.
Write each bracket as a single fraction first.
| \(x-\dfrac{4}{x}\) | \(=\) | \(\dfrac{x^2-4}{x}\) |
| \(1+\dfrac{2}{x}\) | \(=\) | \(\dfrac{x+2}{x}\) |
Now divide by flipping the second fraction:
| \(\) | \(\) | \(\dfrac{x^2-4}{x}\times\dfrac{x}{x+2}\) |
| \(\) | \(=\) | \(\dfrac{(x-2)(x+2)}{x}\times\dfrac{x}{x+2}\) |
| \(\) | \(=\) | \(\boldsymbol{x-2}\) |
Check with \(x = 7\): \(\left(7-\dfrac{4}{7}\right)\div\left(1+\dfrac{2}{7}\right) = \dfrac{45}{7}\times\dfrac{7}{9} = 5\), and \(7-2 = 5\).
Common pitfalls
Frequently asked questions
How do you multiply algebraic fractions?
Factorise every numerator and denominator, cancel any factor that appears on the top and the bottom, then multiply the remaining numerators and denominators. You do not need a common denominator.
How do you divide algebraic fractions?
Keep the first fraction, change \(\div\) to \(\times\) and flip the second fraction. Then factorise, cancel and multiply as usual.
When can you cancel in an algebraic fraction?
Only when the same factor multiplies the whole numerator and the whole denominator. In \(\dfrac{(x-3)(x+3)}{4(x+3)}\) the bracket \((x+3)\) cancels. In \(\dfrac{x+6}{x+2}\) nothing cancels, because \(x\) is a term, not a factor.
Why did my answer come out negative?
A reversed bracket such as \(3-x\) equals \(-(x-3)\). When it cancels with \(x-3\), a factor of \(-1\) is left behind, so the answer is negative.
How do you divide an algebraic fraction by a whole number?
Write the whole number as a fraction over \(1\), then flip it and multiply: \(\dfrac{14x^2}{9}\div 7 = \dfrac{14x^2}{9}\times\dfrac{1}{7} = \dfrac{2x^2}{9}\).