Multiplying and Dividing Algebraic Fractions
Multiplying and dividing algebraic fractions is mostly about factorising. Once every numerator and denominator is written as a product, common factors cancel, and dividing becomes multiplying by the reciprocal. In NSW Year 10 Advanced maths this uses common factors, differences of two squares and quadratic trinomials, as in \(\dfrac{x^2-25}{2x}\times\dfrac{4x^2}{x^2+2x-15}\) (NESA outcome MA5‑ALG‑P‑02).
You will factorise before you cancel, look out for a factor such as \(5-x\) that is the negative of \(x-5\), work through chains of three fractions, simplify brackets that contain fractions, and find a missing coefficient. Worded questions build the area of a rectangle and an average speed from algebraic lengths and times, then substitute a value. It is part of Year 10 Advanced Algebra Review.
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 10 Advanced maths page covers quadratic factors, reversed brackets, chains of operations and fractions inside brackets.
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 | Example |
|---|---|---|
| Common factor | \(ab+ac = a(b+c)\) | \(5x^2-15x = 5x(x-3)\) |
| Difference of two squares | \(a^2-b^2 = (a-b)(a+b)\) | \(x^2-36 = (x-6)(x+6)\) |
| Perfect square | \(a^2-2ab+b^2 = (a-b)^2\) | \(x^2-10x+25 = (x-5)^2\) |
| Monic trinomial | \(x^2+bx+c\): numbers with product \(c\), sum \(b\) | \(x^2+x-20 = (x+5)(x-4)\) |
| Non-monic trinomial | \(ax^2+bx+c\): numbers with product \(ac\), sum \(b\) | \(3x^2+7x+2 = (3x+1)(x+2)\) |
| Reversed bracket | \(b-a = -(a-b)\) | \(6-2x = -2(x-3)\) |
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).
- 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.
Factorise both quadratics, then cancel every factor that appears on the top and the bottom.
| \(x^2+2x-15\) | \(=\) | \((x+5)(x-3)\) |
| \(x^2-9\) | \(=\) | \((x-3)(x+3)\) |
| \(\dfrac{(x+5)(x-3)}{6x}\times\dfrac{9x^2}{(x-3)(x+3)}\) | \(=\) | \(\dfrac{x+5}{x+3}\times\dfrac{9x^2}{6x}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{3x(x+5)}{2(x+3)}}\) |
\(\dfrac{9x^2}{6x} = \dfrac{3x}{2}\). The \(x+5\) and \(x+3\) are different factors, so nothing more cancels.
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.
Factorise each of the four parts. For \(2x^2-x-3\), find two numbers with product \(2\times(-3) = -6\) and sum \(-1\): \(-3\) and \(2\).
| \(2x^2-x-3\) | \(=\) | \((2x-3)(x+1)\) |
| \(x^2-1\) | \(=\) | \((x-1)(x+1)\) |
| \(4x-6\) | \(=\) | \(2(2x-3)\) |
| \(x^2+x\) | \(=\) | \(x(x+1)\) |
| \(\) | \(\) | \(\dfrac{(2x-3)(x+1)}{(x-1)(x+1)}\times\dfrac{x(x+1)}{2(2x-3)}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{x(x+1)}{2(x-1)}}\) |
One \(x+1\) cancels; the other one stays. Now substitute \(x = 3\):
| \(\dfrac{3\times 4}{2\times 2}\) | \(=\) | \(\boldsymbol{3}\) |
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 factorise 2x squared minus x minus 3?
Find two numbers that multiply to \(2\times(-3) = -6\) and add to \(-1\): they are \(-3\) and \(2\). Split the middle term and group: \(2x^2-3x+2x-3 = (2x-3)(x+1)\).