Algebraic Fractions

An algebraic fraction is a fraction whose numerator and/or denominator is a polynomial in some variable, most commonly $x$x. They are the algebraic analogue of rational numbers, and the same arithmetic principles — common denominators for addition, invert-and-multiply for division, simplification by cancelling shared factors — carry across exactly. What changes is the technical demand: every step rests on confident polynomial factorisation, careful tracking of excluded values of $x$x, and disciplined sign work.

This lesson covers the four arithmetic operations on algebraic fractions, the simplification of compound (nested) fractions, and the rewriting of an improper rational expression as a polynomial plus a proper remainder via polynomial long division. The material is foundational: it underwrites partial fractions (lesson 8), the integration of rational functions in the Year 2 calculus chapter, the binomial expansion of $(1+x)^{-1}$(1+x)−1 and its variants, and the algebraic manipulation of differentiated quotients via the quotient rule.

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Spec mapping

This lesson is aligned with the Edexcel 9MA0 specification, Pure Mathematics Paper 1 — Section 2.5: "Simplify rational expressions including by factorising and cancelling, and algebraic division (by linear expressions only)." It draws synoptically on Section 2.3 (factorisation, the factor and remainder theorems — see lesson 6) and feeds directly into Section 2.6 (partial fractions — lesson 8) and Section 8 (integration of rational functions).

Specification reference	Skill assessed
2.5(a)	Simplify a rational expression by factorising numerator and denominator and cancelling common factors
2.5(b)	Add, subtract, multiply and divide algebraic fractions, presenting the answer in fully simplified form
2.5(c)	Perform algebraic long division by a linear expression to obtain quotient and remainder
2.5(d)	Express an improper algebraic fraction in the mixed form quotient + (remainder)/(divisor)

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Key terminology

Term	Meaning
Algebraic fraction (rational expression)	A ratio $P(x)/Q(x)$P(x)/Q(x) of two polynomials with $Q(x) \not\equiv 0$Q(x)≡0
Proper fraction	$\deg P < \deg Q$degP<degQ
Improper fraction	$\deg P \geq \deg Q$degP≥degQ
Excluded value	Any $x$x for which the original denominator is zero; must be stated even after cancellation
Common denominator	A polynomial divisible by every denominator in the sum; the lowest such polynomial is the LCM
Mixed form	$P(x)/Q(x) = \text{quotient}(x) + \text{remainder}(x)/Q(x)$P(x)/Q(x)=quotient(x)+remainder(x)/Q(x)

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Section 1 — Simplification: factor first, then cancel

The single most important principle is this: you may only cancel factors, never individual terms in a sum. The cancellation rule reads $\frac{a \cdot c}{b \cdot c} = \frac{a}{b} \quad (c \neq 0).$b⋅ca⋅c​=ba​(c=0). Both numerator and denominator must be expressed as products before any cancellation takes place. If either is a sum, it must be factorised first. The expression $\frac{x^2 - 9}{x^2 + 5x + 6}$x2+5x+6x2−9​ cannot be simplified by striking out the $x^2$x2 symbols. Instead, factorise: $\frac{x^2 - 9}{x^2 + 5x + 6} = \frac{(x-3)(x+3)}{(x+2)(x+3)} = \frac{x-3}{x+2}, \quad x \neq -3, -2.$x2+5x+6x2−9​=(x+2)(x+3)(x−3)(x+3)​=x+2x−3​,x=−3,−2. Note the two excluded values: $x = -3$x=−3 is excluded because the original denominator vanishes there, even though the simplified expression $(x-3)/(x+2)$(x−3)/(x+2) is perfectly defined at $x=-3$x=−3. Forgetting this is a routine source of mark loss.

Cancel-only-factors warning. Writing $\dfrac{x+2}{x+3} = \dfrac{2}{3}$x+3x+2​=32​ by "cancelling the $x$x" is the most common error in the entire chapter. The $x$x in the numerator is added to $2$2, not multiplied by it. There is no factor of $x$x to cancel.

Worked Example 1 — Simplification

Simplify $\dfrac{2x^2 + 7x + 3}{4x^2 - 1}$4x2−12x2+7x+3​ and state the values of $x$x for which the original expression is undefined.

Solution.

• Factorise the numerator: $2x^2 + 7x + 3 = (2x+1)(x+3)$2x2+7x+3=(2x+1)(x+3).
• Factorise the denominator as a difference of two squares: $4x^2 - 1 = (2x-1)(2x+1)$4x2−1=(2x−1)(2x+1).
• Cancel the common factor $(2x+1)$(2x+1): $\frac{(2x+1)(x+3)}{(2x-1)(2x+1)} = \frac{x+3}{2x-1}.$(2x−1)(2x+1)(2x+1)(x+3)​=2x−1x+3​.
• The original denominator $4x^2 - 1$4x2−1 vanishes at $x = \pm \tfrac{1}{2}$x=±21​, so the original expression is undefined for $x = \tfrac{1}{2}$x=21​ and $x = -\tfrac{1}{2}$x=−21​.

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Section 2 — Addition and subtraction: lowest common denominator

To add or subtract algebraic fractions, you must rewrite them over a common denominator. The lowest common denominator is the LCM of the individual denominators, and to find it you must factorise each denominator first. Skipping that factorisation almost always inflates the denominator unnecessarily and creates an expression that requires re-cancellation at the end.

The general procedure:

1. Factorise every denominator completely.
2. Build the LCM by taking each distinct factor to the highest power in which it appears.
3. Multiply each fraction (top and bottom) by whatever is missing.
4. Combine numerators over the single common denominator.
5. Expand the numerator, collect like terms, and factorise the result to check for any cancellation against the denominator.
6. State excluded values.

Worked Example 2 — Subtraction with a common denominator

Express as a single fraction in fully simplified form: $\frac{3}{x-2} - \frac{x+1}{x^2 - 4}.$x−23​−x2−4x+1​.

Solution.

• Factorise: $x^2 - 4 = (x-2)(x+2)$x2−4=(x−2)(x+2). The LCM is $(x-2)(x+2)$(x−2)(x+2).
• Rewrite the first fraction: $\dfrac{3}{x-2} = \dfrac{3(x+2)}{(x-2)(x+2)}$x−23​=(x−2)(x+2)3(x+2)​.
• Combine: $\frac{3(x+2) - (x+1)}{(x-2)(x+2)}.$(x−2)(x+2)3(x+2)−(x+1)​.
• Sign care. The minus sign distributes over the whole numerator $x+1$x+1: $3(x+2) - (x+1) = 3x + 6 - x - 1 = 2x + 5.$3(x+2)−(x+1)=3x+6−x−1=2x+5.
• Final answer: $\frac{2x+5}{(x-2)(x+2)}, \quad x \neq \pm 2.$(x−2)(x+2)2x+5​,x=±2.

The numerator $2x+5$2x+5 does not factorise to share anything with the denominator, so no further cancellation is possible.

Sign discipline on subtraction. Always bracket the subtracted numerator before distributing the minus sign. The single most common error in this section is to subtract only the first term, leaving the others with the wrong sign.

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Section 3 — Multiplication and division: invert and multiply

Multiplication of algebraic fractions follows $\frac{A}{B} \times \frac{C}{D} = \frac{AC}{BD},$BA​×DC​=BDAC​, and division uses the invert-and-multiply rule $\frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \times \frac{D}{C} = \frac{AD}{BC}, \quad C \neq 0.$BA​÷DC​=BA​×CD​=BCAD​,C=0.

In practice it is almost always cheaper to factorise all four polynomials before multiplying anything out. Cancelling early keeps the algebra small; multiplying first and then trying to factorise a degree-four numerator and denominator is hard work and invites errors.

Worked Example 3 — Multiplication

Simplify $\frac{x^2 + 5x + 6}{x^2 - 1} \times \frac{x^2 - x}{x^2 + 2x}.$x2−1x2+5x+6​×x2+2xx2−x​.

Solution. Factorise everything first:

• $x^2 + 5x + 6 = (x+2)(x+3)$x2+5x+6=(x+2)(x+3)
• $x^2 - 1 = (x-1)(x+1)$x2−1=(x−1)(x+1)
• $x^2 - x = x(x-1)$x2−x=x(x−1)
• $x^2 + 2x = x(x+2)$x2+2x=x(x+2)

Hence $\frac{(x+2)(x+3)}{(x-1)(x+1)} \times \frac{x(x-1)}{x(x+2)}.$(x−1)(x+1)(x+2)(x+3)​×x(x+2)x(x−1)​. Cancel $(x-1)$(x−1), $(x+2)$(x+2) and $x$x: $\frac{x+3}{x+1}, \quad x \neq 0, \pm 1, -2.$x+1x+3​,x=0,±1,−2.

Worked Example 4 — Division

Simplify $\frac{x^2 - 9}{x^2 + 4x} \div \frac{x - 3}{2x + 8}.$x2+4xx2−9​÷2x+8x−3​.

Solution. Invert and multiply: $\frac{x^2 - 9}{x^2 + 4x} \times \frac{2x + 8}{x - 3}.$x2+4xx2−9​×x−32x+8​. Factorise:

• $x^2 - 9 = (x-3)(x+3)$x2−9=(x−3)(x+3)
• $x^2 + 4x = x(x+4)$x2+4x=x(x+4)
• $2x + 8 = 2(x+4)$2x+8=2(x+4)

So $\frac{(x-3)(x+3)}{x(x+4)} \times \frac{2(x+4)}{x-3} = \frac{2(x+3)}{x}, \quad x \neq 0, -4, 3.$x(x+4)(x−3)(x+3)​×x−32(x+4)​=x2(x+3)​,x=0,−4,3.

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Operation method summary

Operation	Standard procedure	Critical step
Simplification	Factorise numerator and denominator; cancel common factors	Both must be in product form before any cancellation
Addition	Factorise denominators; form LCM; rewrite each fraction over LCM; combine numerators; simplify	Re-factorise the resulting numerator to check for further cancellation
Subtraction	As for addition	Bracket the subtracted numerator before distributing the minus sign
Multiplication	Factorise everything; cancel any common factors across both fractions; multiply remaining factors	Cancel before multiplying out, never after
Division	Replace $\div$÷ by multiplication by the reciprocal of the second fraction; then proceed as for multiplication	Invert only the divisor, never the dividend
Long division	Divide leading term, multiply back, subtract, bring down; repeat until remainder has degree below divisor	Bracket the multiplied divisor before subtracting

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Section 4 — Compound (nested) fractions

A compound fraction is a fraction whose numerator or denominator (or both) is itself a fraction. The standard strategy is to multiply the top and the bottom of the outer fraction by a common quantity that clears the inner denominators.

Compound Example

Simplify $\frac{1 + \dfrac{1}{x}}{1 - \dfrac{1}{x^2}}.$1−x21​1+x1​​.

Multiply top and bottom by $x^2$x2: $\frac{x^2 + x}{x^2 - 1} = \frac{x(x+1)}{(x-1)(x+1)} = \frac{x}{x-1}, \quad x \neq 0, \pm 1.$x2−1x2+x​=(x−1)(x+1)x(x+1)​=x−1x​,x=0,±1.

Compound fractions appear naturally in the simplification of derivatives obtained from the quotient rule and in the algebraic manipulation of difference quotients used to define the derivative from first principles.

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Section 5 — Polynomial long division and the mixed form

For Paper 1 you must be able to divide a polynomial by a linear expression $x - a$x−a (the spec restricts division to the linear case, although the same algorithm generalises). Long division converts an improper fraction into the mixed form $\frac{P(x)}{x - a} = q(x) + \frac{r}{x - a},$x−aP(x)​=q(x)+x−ar​, where $q(x)$q(x) is the quotient polynomial and $r$r is the constant remainder (a constant because the divisor is linear).

The relationship between the remainder and the value of the polynomial at $x = a$x=a is the remainder theorem (treated fully in lesson 6): $r = P(a)$r=P(a). This gives an instant arithmetic check on every long division you perform.

Worked Example 5 — Division to mixed form

Express $\dfrac{x^2 + 3x + 5}{x + 1}$x+1x2+3x+5​ in the form $x + b + \dfrac{c}{x+1}$x+b+x+1c​.

Long division.

• Divide $x^2$x2 by $x$x: quotient term $x$x. Multiply back: $x(x+1) = x^2 + x$x(x+1)=x2+x. Subtract: $(x^2 + 3x + 5) - (x^2 + x) = 2x + 5$(x2+3x+5)−(x2+x)=2x+5.
• Divide $2x$2x by $x$x: quotient term $+2$+2. Multiply back: $2(x+1) = 2x + 2$2(x+1)=2x+2. Subtract: $(2x + 5) - (2x + 2) = 3$(2x+5)−(2x+2)=3.
• No more terms to bring down. Quotient: $x + 2$x+2. Remainder: $3$3.

So the quotient is $x + 2$x+2 and the remainder is $3$3, giving $\frac{x^2 + 3x + 5}{x + 1} = x + 2 + \frac{3}{x + 1}.$x+1x2+3x+5​=x+2+x+13​.

Check via the remainder theorem. $P(x) = x^2 + 3x + 5$P(x)=x2+3x+5, so $P(-1) = 1 - 3 + 5 = 3$P(−1)=1−3+5=3. This matches the remainder, confirming the division is correct.

The mixed form is the natural input to integration: $\int (x + 2 + 3/(x+1))\,dx = \tfrac{1}{2}x^2 + 2x + 3\ln|x+1| + C$∫(x+2+3/(x+1))dx=21​x2+2x+3ln∣x+1∣+C, where the polynomial part is integrated termwise and the proper-fraction tail integrates to a logarithm.

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Section 6 — Proper, improper, and the role of the mixed form

Type	Condition on degrees	Example	Standard treatment
Proper	$\deg P < \deg Q$degP<degQ	$\dfrac{3}{x+1}$x+13​, $\dfrac{2x-1}{x^2 - 4}$x2−42x−1​	Decompose into partial fractions (lesson 8)
Improper, $\deg P = \deg Q$degP=degQ	equal degrees	$\dfrac{x+5}{x+1}$x+1x+5​	Long division yields a constant quotient plus a proper tail
Improper, $\deg P > \deg Q$degP>degQ	strictly greater	$\dfrac{x^2 + 3x + 5}{x+1}$x+1x2+3x+5​	Long division yields a non-constant quotient plus a proper tail

The exam convention is that an answer should be left in the form most useful for the next step: as a single fraction when the question asks for one; in mixed form when the question continues to integration or asymptotic analysis; in partial-fraction form when binomial expansion or Laplace-style work follows.

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Worked Example 6 — A harder long division check

Express $\dfrac{2x^3 - x^2 + 4x - 7}{x - 2}$x−22x3−x2+4x−7​ in the form $ax^2 + bx + c + \dfrac{d}{x-2}$ax2+bx+c+x−2d​, stating the values of the constants $a$a, $b$b, $c$c, $d$d.

Solution.

• Divide $2x^3$2x3 by $x$x: quotient term $2x^2$2x2. Multiply back: $2x^2(x-2) = 2x^3 - 4x^2$2x2(x−2)=2x3−4x2. Subtract: $(2x^3 - x^2) - (2x^3 - 4x^2) = 3x^2$(2x3−x2)−(2x3−4x2)=3x2. Bring down $+4x$+4x: running polynomial is $3x^2 + 4x$3x2+4x.
• Divide $3x^2$3x2 by $x$x: quotient term $3x$3x. Multiply back: $3x(x-2) = 3x^2 - 6x$3x(x−2)=3x2−6x. Subtract: $(3x^2 + 4x) - (3x^2 - 6x) = 10x$(3x2+4x)−(3x2−6x)=10x. Bring down $-7$−7: running polynomial is $10x - 7$10x−7.
• Divide $10x$10x by $x$x: quotient term $10$10. Multiply back: $10(x-2) = 10x - 20$10(x−2)=10x−20. Subtract: $(10x - 7) - (10x - 20) = 13$(10x−7)−(10x−20)=13.

Quotient $2x^2 + 3x + 10$2x2+3x+10, remainder $13$13, so $\frac{2x^3 - x^2 + 4x - 7}{x - 2} = 2x^2 + 3x + 10 + \frac{13}{x - 2},$x−22x3−x2+4x−7​=2x2+3x+10+x−213​, giving $a = 2$a=2, $b = 3$b=3, $c = 10$c=10, $d = 13$d=13.

Remainder-theorem check. $P(x) = 2x^3 - x^2 + 4x - 7$P(x)=2x3−x2+4x−7, so $P(2) = 16 - 4 + 8 - 7 = 13$P(2)=16−4+8−7=13. Match.

This kind of question is common in the second half of Paper 1, where the algebraic-fraction skill is folded into a broader question on integration, asymptotes, or curve sketching. The mixed form gives an immediate reading of the curve's behaviour at infinity ($y \sim 2x^2$y∼2x2) and at the vertical asymptote $x = 2$x=2.

Worked Example 7 — A compound fraction that simplifies dramatically

Simplify $\frac{\dfrac{1}{x+h} - \dfrac{1}{x}}{h}.$hx+h1​−x1​​.

This is the difference quotient that defines the derivative of $1/x$1/x from first principles, and the algebra is a standard A-Level exercise in compound-fraction manipulation.