Factorising Algebraic Fractions
An algebraic fraction can be simplified only by cancelling whole factors, so the numerator and denominator are factorised first. In NSW Year 9 Advanced maths this uses every factorising method in the topic (NESA outcome MA5‑ALG‑P‑02).
You will simplify fractions such as \(\frac{x^2 - x - 20}{x^2 + 7x + 12}\) and \(\frac{9 - x^2}{3x - x^2}\), watch for factors like \(1 - x = -(x - 1)\), and cancel non-monic factors, as in \(\frac{3y^2 + 7y - 6}{2y^2 + 3y - 9}\). You will also divide and subtract fractions after factorising, and show that an expression has the same value for every \(n\) where it is defined. It is part of Year 9 Advanced Algebra: Expanding and Factorising.
Every question with a fully worked solution.
- Factorising Algebraic Fractions - Video - Simplifying Algebraic Fractions Watch
Theory
To simplify an algebraic fraction, factorise the numerator and the denominator, then cancel any common factor. This Year 9 Advanced maths page covers content that leads into Advanced and Extension maths: trinomials, differences of squares, back-to-front brackets and fractions you multiply, divide, add or subtract.
An algebraic fraction is in simplest form when its numerator and denominator have no common factor.
You can cancel only a factor of the whole numerator and the whole denominator. In \(\dfrac{(k + 4)(k - 1)}{3(k - 1)}\) the bracket \((k - 1)\) is a factor of both, so it cancels to leave \(\dfrac{k + 4}{3}\).
In \(\dfrac{k + 4}{k + 7}\) nothing cancels. The \(k\)s are terms, not factors.
Two brackets such as \((5 - y)\) and \((y - 5)\) are negatives of each other: \(5 - y = -(y - 5)\). Their ratio is \(-1\).
The simplified fraction equals the original except where the original denominator is zero; those values are excluded.
Cancelling a common factor \(k \neq 0\):
Factorising tools you will need:
How to simplify an algebraic fraction
- Factorise the numerator and the denominator fully: HCF, then special products or trinomials.
- Match brackets, taking out \(-1\) from any that are back to front.
- Cancel common factors and write what is left.
Operations with algebraic fractions
- Multiply or divide: flip any fraction you divide by, factorise everything, then cancel across all the fractions. For example \(\dfrac{x^2 - 16}{x^2 + x - 12} \times \dfrac{x^2 - 9}{x^2 - x - 12}\) \(= \dfrac{(x - 4)(x + 4)}{(x + 4)(x - 3)} \times \dfrac{(x - 3)(x + 3)}{(x - 4)(x + 3)}\) \(= 1\).
- Add or subtract: factorise each denominator, build the LCD from every different factor (each to its highest power), rewrite each fraction over it and combine the numerators.
- Same value for every \(n\): simplify each fraction first. If everything with \(n\) cancels, the expression is a constant wherever it is defined.
Factorise the numerator (product \(-10\), sum \(3\): \(5\) and \(-2\)) and the denominator (difference of two squares).
| \(\dfrac{x^2 + 3x - 10}{x^2 - 4}\) | \(=\) | \(\dfrac{(x + 5)(x - 2)}{(x - 2)(x + 2)}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{x + 5}{x + 2}}\) |
The answer holds for \(x \neq 2\) and \(x \neq -2\), the values that make the original denominator zero.
Take out the HCF on top, then notice that \(2 - a = -(a - 2)\).
| \(\dfrac{6a - 3a^2}{a^2 - 4}\) | \(=\) | \(\dfrac{3a(2 - a)}{(a - 2)(a + 2)}\) |
| \(\) | \(=\) | \(\dfrac{-3a(a - 2)}{(a - 2)(a + 2)}\) |
| \(\) | \(=\) | \(\boldsymbol{-\dfrac{3a}{a + 2}}\) |
Check with \(a = 1\): \(\dfrac{6 - 3}{1 - 4} = -1\) and \(-\dfrac{3}{3} = -1\).
Numerator: \(ac = -12\), and \(4 + (-3) = 1\). Denominator: \((2x)^2 - 3^2\).
| \(2x^2 + x - 6\) | \(=\) | \(2x^2 + 4x - 3x - 6\) |
| \(\) | \(=\) | \(2x(x + 2) - 3(x + 2)\) |
| \(\) | \(=\) | \((x + 2)(2x - 3)\) |
| \(4x^2 - 9\) | \(=\) | \((2x - 3)(2x + 3)\) |
| \(\dfrac{2x^2 + x - 6}{4x^2 - 9}\) | \(=\) | \(\dfrac{(x + 2)(2x - 3)}{(2x - 3)(2x + 3)}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{x + 2}{2x + 3}}\) |
The \(x\) and \(2x\) left in the answer cannot cancel: they are terms, not factors.
Factorise each denominator. The LCD uses every different factor once: \(x(x - 3)(x + 3)\).
| \(\) | \(\) | \(\dfrac{2}{(x - 3)(x + 3)} + \dfrac{1}{x(x + 3)}\) |
| \(\) | \(=\) | \(\dfrac{2x + (x - 3)}{x(x - 3)(x + 3)}\) |
| \(\) | \(=\) | \(\dfrac{3x - 3}{x(x - 3)(x + 3)}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{3(x - 1)}{x(x - 3)(x + 3)}}\) |
Using the product of the two denominators would give \((x + 3)\) twice, and a harder fraction to simplify.
Common pitfalls
Frequently asked questions
How do you simplify algebraic fractions with quadratics?
Factorise the numerator and the denominator, then cancel any bracket that appears in both. For example (x squared minus 1) over (x squared + 3x + 2) is (x minus 1)(x + 1) over (x + 1)(x + 2), which simplifies to (x minus 1) over (x + 2).
Why can't you cancel the x in (x + 3) over (x + 5)?
Because x is a term, not a factor. Cancelling only works when something multiplies the whole numerator and the whole denominator. Substitute x = 1 to see that 4 over 6 is not 3 over 5.
What is (a minus b) divided by (b minus a)?
It is negative 1, because b minus a is the negative of a minus b. Take out negative 1 from one bracket to make them match before cancelling.
How do you find the common denominator of algebraic fractions with quadratics?
Factorise each denominator first. The lowest common denominator is the product of every different factor, each used as many times as it appears in any one denominator.
Do you need to state restrictions when simplifying algebraic fractions?
The simplified form equals the original only where the original denominator is not zero. If a question asks, list the values of the pronumeral that make any original denominator zero.