To simplify an algebraic fraction, factorise the numerator and the denominator completely, cancel factors that appear in both, and state the values that make the original denominator zero. Cancelling only works on multiplied factors, never on terms that are added or subtracted.
It uses everything earlier in algebraic structure, particularly common factors and factorising quadratics.
Why do exclusions matter?
A fraction is not defined when its denominator is zero. The original fraction (x − 1)/((x − 1)(x + 1)) cannot be evaluated at x = 1 or x = −1, even though the simplified version 1/(x + 1) could be evaluated at x = 1.
So the simplified form is equal to the original only where the original exists. Stating the exclusions records that honestly.
How to simplify, step by step
- Factorise the numerator completely.
- Factorise the denominator completely.
- Read the exclusions from the factors of the original denominator: set each to zero.
- Cancel factors that appear in both top and bottom.
- Write the simplified fraction with the exclusions stated.
- Check by substituting a value that is allowed.
Worked example
Simplify (x² − 4) / (x² + 5x + 6) and state the values x cannot take.
Step 1, numerator: x² − 4 is a difference of two squares, so it is (x − 2)(x + 2).
Step 2, denominator: two numbers with product 6 and sum 5 are 2 and 3, so x² + 5x + 6 = (x + 2)(x + 3).
Step 3, exclusions: the original denominator is zero when x + 2 = 0 or x + 3 = 0, which gives x = −2 or x = −3.
Step 4, cancel: the factor (x + 2) appears on both sides.
Step 5, answer:
(x − 2)/(x + 3), where x ≠ −2 and x ≠ −3
Check: with x = 1, the original is (1 − 4)/(1 + 5 + 6) = −3/12 = −1/4. The answer is (−1)/4 = −1/4. With x = 0, the original is −4/6 = −2/3 and the answer is −2/3. Both agree.
Notice that x = −2 is excluded even though the simplified fraction could accept it. The exclusion comes from the original.
The mistake to watch for
A common slip is to cancel a term instead of a factor.
Mistaken working: (x + 6)/6 = x + 1
The student cancelled the 6 in the denominator with the 6 in the numerator. But the 6 in the numerator is added to x, not multiplying the whole expression.
The correction is to ask whether the thing you are cancelling multiplies the whole numerator. Here it does not, so (x + 6)/6 is already as simple as it can be. Substitute x = 6 to see the problem: (6 + 6)/6 = 2, but the mistaken answer gives 7.
A second slip is cancelling correctly but leaving out the exclusions. That costs marks when the question asks for them, so make it part of the routine to write them before you cancel.
Check yourself
Try these without a calculator, then open each answer.
1. Simplify (4x + 8)/(x² − 4) and state the exclusions
Show answer
Top: 4(x + 2). Bottom: (x − 2)(x + 2). The bottom is zero at x = 2 and x = −2. Cancel (x + 2).
4/(x − 2), where x ≠ 2 and x ≠ −2
Check with x = 3: the original is 20/5 = 4, and 4/1 = 4.
2. Simplify (x² − 9)/(x² − 3x) and state the exclusions
Show answer
Top: (x − 3)(x + 3). Bottom: x(x − 3). The bottom is zero at x = 0 and x = 3. Cancel (x − 3).
(x + 3)/x, where x ≠ 0 and x ≠ 3
Check with x = 6: the original is 27/18 = 3/2, and 9/6 = 3/2.
3. Simplify (x − 1)/(x² − 1) and state the exclusions
Show answer
Bottom: (x − 1)(x + 1). The bottom is zero at x = 1 and x = −1. Cancel (x − 1), leaving 1 on top.
1/(x + 1), where x ≠ 1 and x ≠ −1
Check with x = 3: the original is 2/8 = 1/4, and 1/4.
Where this leads next
Try everything together in the algebraic structure practice set. Then equations and formulas uses the same factorising habits to solve for x. The non-calculator working trainer is useful for testing allowed values by hand.
If cancelling and exclusions still cause errors under exam conditions, a teacher can look at your written working and find the step where the structure was missed. That is the kind of work we do in online one-to-one Mathematics tuition.