Simplify (x² − 4)/(x − 2).

x + 2. Factorising gives (x − 2)(x + 2)/(x − 2) = x + 2, valid for x ≠ 2. The restriction remains even though x + 2 alone is defined everywhere.

For which value is (x + 1)/(x − 6) undefined?

x = 6. Only the denominator matters. It is zero at x = 6. A numerator of zero simply makes the fraction zero, which is perfectly legal.

(x + y)/(xy). Common denominator xy gives y/xy + x/xy = (x + y)/xy. Option A is the standard wrong answer and can be disproved with x = y = 1: 1 + 1 = 2, not 1/2.

(a/b) ÷ (c/d) equals:

ad/bc. Invert the divisor and multiply: (a/b) × (d/c) = ad/bc. Inverting the first fraction instead is the usual slip.

Which cancels correctly?

(2x + 4)/(x + 2) → 2. 2x + 4 = 2(x + 2), so the whole factor (x + 2) cancels and 2 is left. Each other option cancels a term rather than a factor.

The LCM of the denominators in 1/(x − 1) + 1/(x² − 1) is:

x² − 1. x² − 1 = (x − 1)(x + 1), which already contains x − 1 as a factor, so the LCM is just x² − 1. Multiplying the two denominators together would give an unnecessarily large fraction.

After cancelling (x − 3) from (x − 3)(x + 1)/[(x − 3)(x − 5)], the restrictions are:

x ≠ 3 and x ≠ 5. Both values came from the original denominator, and cancelling does not undo that. x = 3 gives a hole in the graph and x = 5 gives an asymptote, but neither is in the domain.

The square root of x² + 6x + 9 is:

x + 3. The expression is the perfect square (x + 3)², since 2 × x × 3 = 6x matches the middle term exactly.

Algebraic Fractions — Mathematics Notes, Worked Examples and Quiz

A fraction with letters obeys the same rules as one with numbers

Definition

Excluded value — Any value of the variable that makes a denominator zero. It must be excluded from the domain, and it stays excluded even after the offending factor has been cancelled.

An algebraic fraction is a quotient of two polynomials, such as (x² − 4)/(x + 2). Everything you already know about 6/8 transfers directly: cancel common factors to simplify, use a common denominator to add, invert and multiply to divide.

One thing does not transfer, and it is the whole difficulty of the chapter. With numbers you know instantly that 6/0 is nonsense. With letters, the denominator is only zero for certain values of x, and you have to find them and say so.

You may only cancel factors, never terms

In (x + 3)/(x + 5) nothing cancels — the x values are terms inside a sum, not factors of the whole. Crossing them out to get 3/5 is the single most common error in all of algebra. Cancelling is legal only when the thing you remove multiplies the entire numerator and the entire denominator, which is why you factorise first.

Simplifying: factorise, then cancel, then state the restriction

The routine never changes. Factorise the numerator completely. Factorise the denominator completely. Note the values that make the denominator zero. Cancel any factor appearing in both. Write the answer together with the restriction.

Worked example

Simplify (x² − 5x + 6) / (x² − 9) and state the values of x for which it is undefined.

Numerator: x² − 5x + 6 = (x − 2)(x − 3).Two numbers multiplying to 6 and adding to −5.
Denominator: x² − 9 = (x − 3)(x + 3).Difference of two squares.
The expression is undefined when the denominator is zero, so x ≠ 3 and x ≠ −3.Record this before cancelling, because the next step hides one of them.
Cancel the common factor (x − 3): the fraction becomes (x − 2)/(x + 3).A factor of the whole numerator and the whole denominator — cancelling is legitimate here.
Final answer (x − 2)/(x + 3), with x ≠ 3, −3.The simplified form looks perfectly happy at x = 3, but the original was not, so the restriction stays.

(x − 2)/(x + 3), x ≠ 3 and x ≠ −3

The first option cancels, leaving a straight line with a single point punched out. The second cannot cancel, so the graph blows up into a vertical asymptote instead. Cancelling removes a hole, never a restriction.

Adding and subtracting

You need a common denominator, and the efficient choice is the LCM of the factorised denominators — not their product, which usually leaves you simplifying a much bigger fraction at the end.

The commonest mistake is a sign. When subtracting, the minus applies to every term of the second numerator, so bracket it before you expand.

Worked example

Simplify 3/(x − 2) − 2/(x + 1).

The denominators share no factor, so the LCM is (x − 2)(x + 1).Check for a common factor first; when there is one, the LCM is smaller than the product.
Rewrite: [3(x + 1) − 2(x − 2)] / [(x − 2)(x + 1)].Each numerator is multiplied by whatever its denominator was missing. Keep the subtraction inside the bracket.
Expand the numerator: 3x + 3 − 2x + 4.The −2 multiplies both terms of (x − 2), so −2 × −2 = +4. This is the sign that costs marks.
Collect: (x + 7) / [(x − 2)(x + 1)], with x ≠ 2, −1.Leave the denominator factorised — it shows the restrictions and makes any later cancelling visible.

(x + 7) / [(x − 2)(x + 1)], x ≠ 2, −1

Multiplying and dividing

Multiplication is the easy case: factorise everything, cancel across the two fractions, then write what is left. There is no need for a common denominator.

Division is multiplication by the reciprocal — invert the second fraction and multiply. But note carefully: the excluded values of the original divisor still apply, because you were never allowed to divide by zero in the first place.

(a/b) × (c/d) = ac / bd(a/b) ÷ (c/d) = (a/b) × (d/c)(c ≠ 0 and d ≠ 0)invert the divisor, not the dividend — and both c and d generate restrictions

Where the restrictions come from in a division

For (x+1)/(x−3) ÷ (x+1)/(x+2), three values are excluded: x ≠ 3 from the first denominator, x ≠ −2 from the second, and x ≠ −1 because the divisor itself must not be zero. That third one survives even though the factor cancels, and it is the one students miss.

Square roots of algebraic expressions

The syllabus also asks for square roots of expressions that turn out to be perfect squares. The method is to recognise the pattern a² ± 2ab + b² and write down (a ± b).

For an expression like x² + 1/x² + 2, notice that the middle term 2 is 2 × x × (1/x), so the whole thing is (x + 1/x)² and its square root is x + 1/x. Checking the middle term against 2ab is what confirms the guess.

Before you leave this chapter

Factorise numerator and denominator completely before touching anything else.
Cancel factors only — never terms inside a sum.
Note the excluded values from the original expression, and keep them after cancelling.
To add or subtract, use the LCM of the factorised denominators and bracket the second numerator when subtracting.
To divide, invert the divisor and multiply — and remember the divisor itself must not be zero.

LQ1[6 marks]

Consider the expression E = [1/(x − 2) − 1/(x + 2)] ÷ [4/(x² − 4)].

Simplify the expression inside the first bracket.
Hence simplify E completely.
State all the values of x for which E is undefined.

Mark scheme

Common denominator (x − 2)(x + 2) = x² − 4; numerator = (x + 2) − (x − 2) = 4the x terms cancel — that is the design of the question[1]
First bracket = 4/(x² − 4)[1]
Division becomes multiplication by the reciprocal: [4/(x² − 4)] × [(x² − 4)/4][1]
E = 1[1]
Denominators vanish at x = 2 and x = −2[1]
Those are the only exclusions; the divisor 4/(x² − 4) is never zero, since its numerator is the constant 4checking whether the divisor can be zero is the step being tested[1]

(a) 4/(x² − 4) (b) E = 1 (c) undefined only at x = 2 and x = −2

Examiner tip. Part (c) asks you to check whether the divisor can be zero. Here it cannot, because its numerator is a non-zero constant — say so explicitly rather than leaving the examiner to assume you checked.

Algebraic Fractions