01 · Termwise division
(9x³ − 6x + 12)/(3x)
Hint
Divide all three terms by 3x.
Worked solution
3x² − 2 + 4/x, x ≠ 0
Understand · explore · practise
Algebraic fractions
Simplify algebraic fractions by factorising, divide a polynomial term by term and retain excluded values. Worked examples, a missing-point model and practice questions.
Before you startExpanding, factorising and laws of indices
01 / Cancel factors
Cancel a factor shared by the entire numerator and denominator.
A fraction divides its whole numerator by its whole denominator. Cancellation removes a common multiplying factor. It cannot remove one term from a sum.
6x(x + 2) / [3x(x − 1)]
= 2(x + 2)/(x − 1)
This equality holds for x ≠ 0, 1, the exclusions from the original denominator.
In (x + 2)/x you cannot cancel the two x symbols: x is not a factor of the entire numerator. Instead, divide both terms to get 1 + 2/x, for x ≠ 0.
Factor first, then cancel complete factors. Keep brackets around sums so their role is clear.
02 / Divide each term
A single-term denominator divides every numerator term.
Split the numerator into separate fractions. Apply the index laws to each quotient, including any constant term.
The last term below remains a fraction. A constant divided by x does not become a constant.
Divide by 2xWorked example
(4x⁴ − 6x² + 8x − 10)/(2x)
The original expression requires x ≠ 0.
= 4x⁴/(2x) − 6x²/(2x)
+ 8x/(2x) − 10/(2x)
Divide every term.
= 2x³ − 3x + 4 − 5/x
Subtract indices only for powers of the same base.
03 / Factorise both
Look for a common bracket, including non-unit coefficients.
Factorise the numerator and denominator separately. Difference-of-squares denominators often reveal a bracket shared with a quadratic numerator.
The simplified form must retain every excluded value from the original expression, even if the factor causing an exclusion disappears.
Two factorisations, one cancellationWorked example
(3x² − 10x − 8)/(x² − 16)
The denominator is zero at x = 4 and x = −4.
= (3x + 2)(x − 4)
/ [(x − 4)(x + 4)]
Check the numerator by expanding.
= (3x + 2)/(x + 4)
x ≠ 4, −4
Cancel x − 4 only where it is non-zero.
04 / Excluded values
Simplifying does not fill a missing point.
For x ≠ 3, (x² − 9)/(x − 3) = x + 3. At x = 3, the original fraction is 0/0 and is undefined. The straight line x + 3 has value 6 there, but the original expression does not.
The open circle marks the excluded point. Select x = 3 and compare the two values. The algebra agrees everywhere in the original domain.
Two original exclusions
(x² + x − 6)/(x² − x − 2)
= (x + 3)(x − 2)/[(x − 2)(x + 1)]
= (x + 3)/(x + 1), x ≠ 2, −1
The simplified denominator still excludes −1. The cancelled factor leaves the additional exclusion x ≠ 2.
(x² − 9)/(x − 3)Explore at your pace
The graph follows y = x + 3 except for an open hole at (3,6). At x = 2, both expressions equal 5.
Watch cancellation preserve a missing point
Pause, replay or seek freely. The notes explain the same idea and stay in view.
05 / Expression or equation?
Check candidate solutions against the original denominator.
Simplifying rewrites an expression. Solving asks which permitted values make an equation true. An excluded value can never become a solution.
(x² − 4)/(x − 2) = 4, x ≠ 2
x + 2 = 4 ⇒ x = 2
The only candidate is excluded. The original equation therefore has no solution. Substitution would require division by zero.
Multiplying an equation by its denominator is valid only after restricting the domain so that denominator is non-zero.
06 / Unknown constants
Factorisation can reveal a requested form.
To write a rational expression in a given form, factor completely and compare the result. State the original exclusions as well as the constants.
Find a, b and cWorked example
(6x³ + 18x² − 60x)/(3x² − 3x − 6)
= ax(x + b)/(x + c)
Factor out 6x above and 3 below.
= 6x(x + 5)(x − 2)
/ [3(x − 2)(x + 1)]
Both quadratics factor.
= 2x(x + 5)/(x + 1)
a = 2, b = 5, c = 1
The original expression excludes x = 2 and x = −1.
07 / Your turn
Factor carefully and record the domain.
Give excluded values with each simplified answer. An equality between expressions is understood on their common permitted domain.
02 · A shared linear factor
(x² − 25)/(x + 5)
Hint
Use a difference of squares.
Worked solution
(x − 5)(x + 5)/(x + 5)
= x − 5, x ≠ −5
03 · Factor both parts
(2x² + 7x + 3)/(x² − 1)
Hint
The numerator is (2x + 1)(x + 3).
Worked solution
(2x + 1)(x + 3)/[(x − 1)(x + 1)]
No factor cancels. The original expression excludes x = ±1. Similar-looking terms are not common factors.
04 · Opposite brackets
(x² − 4)/(2 − x)
Hint
2 − x = −(x − 2).
Worked solution
−(x + 2), x ≠ 2
05 · Two exclusions
(x² − 3x − 10)/(x² − 4)
Hint
Factor the numerator as (x − 5)(x + 2).
Worked solution
(x − 5)/(x − 2), x ≠ −2, 2
06 · Solve within the domain
(x² − 9)/(x − 3) = 8
Hint
Simplify with x ≠ 3.
Worked solution
x + 3 = 8 ⇒ x = 5
5 is permitted; the original fraction gives (25 − 9)/(5 − 3) = 8.
07 · Reject an excluded root
(x² − 1)/(x + 1) = −2
Hint
The domain excludes x = −1.
Worked solution
x − 1 = −2 ⇒ x = −1
This is excluded, so there is no solution.
08 · Recover constants
Write (4x² − 16)/(2x² + 10x + 12) as a(x + b)/(x + c). Find the constants and exclusions.
Hint
Factor into (x − 2), (x + 2) and (x + 3).
Worked solution
4(x − 2)(x + 2)/[2(x + 2)(x + 3)]
= 2(x − 2)/(x + 3)
a = 2, b = −2, c = 3
x ≠ −2, −3
08 / Recap
A simpler form keeps the original restrictions.
- Cancel common factors, not individual terms in sums.
- Divide every term by a single-term denominator.
- Factorise before cancelling brackets.
- Retain all original denominator exclusions.
- Reject any proposed solution outside the original domain.
Section 1 of 8 · Cancel factors