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Pure 03 L01 · Algebraic Fractions
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Pure Mathematics A2 · Unit 01

Algebraic
Fractions

Division, addition and subtraction, improper fractions, and algebraic long division — all in one place.

📖 3 Concepts ✏️ 2 Quizzes 📄 4 Past Paper Qs
01
Division of Algebraic Fractions
Core Rule Dividing by a fraction is the same as multiplying by its reciprocal — swap the numerator and denominator of the second fraction, then multiply.

Division Rule

$$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$$

Key Identity

$$a^2 - b^2 = (a+b)(a-b)$$
// Worked Example — Division
// Simplify fully
$$\frac{x+2}{x+4} \div \frac{3x+6}{x^2-16}$$
Step 1 — Flip & Multiply ▼
$$= \frac{x+2}{x+4} \times \frac{x^2-16}{3x+6}$$
Step 2 — Factorise both expressions ▼
Using \(a^2-b^2=(a+b)(a-b)\): $$x^2-16=(x+4)(x-4)\qquad 3x+6=3(x+2)$$ $$= \frac{x+2}{x+4} \times \frac{(x+4)(x-4)}{3(x+2)}$$
Always factorise fully before cancelling — never cancel terms across a + or −
Step 3 — Cancel common factors ▼
Cancel \((x+2)\) and \((x+4)\) from top and bottom: $$\boxed{\dfrac{x-4}{3}}$$
02
Adding & Subtracting Algebraic Fractions
Core Rule Find a common denominator first — identical to working with numeric fractions. Multiply each numerator by the other fraction's denominator.
// Worked Example — Subtraction
// Express as a single fraction
$$\frac{2}{x+3} - \frac{1}{x+1}$$
Step 1 — Common denominator = (x+3)(x+1) ▼
$$= \frac{2(x+1) - 1(x+3)}{(x+3)(x+1)}$$
Step 2 — Expand & simplify numerator ▼
$$= \frac{2x+2-x-3}{(x+3)(x+1)} = \boxed{\frac{x-1}{(x+3)(x+1)}}$$
Always check: does the numerator share a factor with the denominator? If not, the fraction is fully simplified.
03
Improper Algebraic Fractions
What makes a fraction "improper"? When the degree (highest power) of the numerator \(\geq\) degree of the denominator — just like \(\tfrac{10}{7}\) in arithmetic. Convert to a mixed form using long division.

The polynomial division identity:

$$F(x) \;=\; Q(x)\cdot d(x) \;+\; R \qquad\Longleftrightarrow\qquad \frac{F(x)}{d(x)} = Q(x)+\frac{R}{d(x)}$$
// Worked Example — Algebraic Long Division
// Express in the form Q(x) + R/(x−2)
$$\frac{x^2+5x+8}{x-2}$$
Step 1 — Leading term: x² ÷ x = x ▼
$$x\cdot(x-2)=x^2-2x \qquad \text{Subtract: }(x^2+5x+8)-(x^2-2x)=7x+8$$
Step 2 — Next term: 7x ÷ x = 7 ▼
$$7\cdot(x-2)=7x-14 \qquad \text{Subtract: }(7x+8)-(7x-14)=22\;\leftarrow\text{remainder}$$
Step 3 — Write the final mixed form ▼
$$\frac{x^2+5x+8}{x-2}=(x+7)+\frac{22}{x-2}$$ Or equivalently: \(\;x^2+5x+8=(x+7)(x-2)+22\)
⚡ Remainder Shortcut To find only the remainder: substitute \(x\) = root of the divisor.
For \((x-2)\): set \(x=2\;\Rightarrow\;4+10+8=22\;\checkmark\)
Theory Quiz
Quiz 01 — Core Concepts
Four questions to lock in the theory before past papers.
Q1. To divide two algebraic fractions you should:
Q2. How do you factorise \(x^2-25\)?
Q3. Is \(\dfrac{x^3+2x}{x+1}\) an improper fraction?
Q4. Using the remainder shortcut, find the remainder when \(x^2+5x+8\) is divided by \((x-2)\).
Edexcel Past Paper Questions
202405-02
Improper Fraction → Mixed Form
3 marks
Question
$$g(x)=\frac{2x^2-5x+8}{x-2}$$ Write \(g(x)\) in the form \(\;Ax+B+\dfrac{C}{x-2}\;\) where \(A,B,C\) are integers to be found.
// Model Solution
Set up
$$2x^2-5x+8=(Ax+B)(x-2)+C$$
x = 2
$$C=2(4)-5(2)+8=8-10+8=6$$
Long div
\(2x\cdot(x-2)=2x^2-4x\;\Rightarrow\;\) rem \(-x+8\)
\(-1\cdot(x-2)=-x+2\;\Rightarrow\;\) rem \(6\;\checkmark\)
$$g(x)=2x-1+\frac{6}{x-2}\qquad A=2,\quad B=-1,\quad C=6$$
202210-01
Long Division with Remainder
4 marks
Question
$$f(x)=\frac{2x^3-4x-15}{x^2+3x+4}$$ Show that \(\;f(x)\equiv Ax+B+\dfrac{C(2x+3)}{x^2+3x+4}\;\) finding integers \(A,B,C\).
// Model Solution
Step 1
Divide \(2x^3+0x^2-4x-15\) by \((x^2+3x+4)\):$$2x\cdot(x^2+3x+4)=2x^3+6x^2+8x\;\Rightarrow\;\text{rem: }-6x^2-12x-15$$
Step 2
$$-6\cdot(x^2+3x+4)=-6x^2-18x-24\;\Rightarrow\;\text{rem: }6x+9=3(2x+3)$$
$$f(x)=2x-6+\frac{3(2x+3)}{x^2+3x+4}\qquad A=2,\quad B=-6,\quad C=3$$
202401-04
Add & Simplify Fractions
3 marks
Question  (x > 2)
$$f(x)=\frac{2x^2-32}{3x^2+7x-20}+\frac{8}{3x-5}$$ Show that \(\;f(x)=\dfrac{2x}{3x-5}\).
// Model Solution
Factorise
$$3x^2+7x-20=(3x-5)(x+4)\qquad 2x^2-32=2(x+4)(x-4)$$
Rewrite
$$=\frac{2(x+4)(x-4)}{(3x-5)(x+4)}+\frac{8}{3x-5}=\frac{2(x-4)}{3x-5}+\frac{8}{3x-5}$$
Combine
$$=\frac{2(x-4)+8}{3x-5}=\frac{2x-8+8}{3x-5}$$
$$=\frac{2x}{3x-5}\qquad\checkmark$$
202501-04
Identify Constants from Long Division
4 marks
Question
$$\frac{4x^3+2x^2+3x+8}{x^2+4}\;\equiv\;Ax+B+\frac{Cx+D}{x^2+4}$$ (i) Find \(A\), \(B\) and \(C\).   (ii) Show that \(D=0\).
// Model Solution
Step 1
$$4x\cdot(x^2+4)=4x^3+16x\;\Rightarrow\;\text{rem: }2x^2-13x+8$$
Step 2
$$2\cdot(x^2+4)=2x^2+8\;\Rightarrow\;\text{rem: }-13x+0$$
D = 0
Remainder \(=-13x+0\;\Rightarrow\;D=0\;\checkmark\)
$$\frac{4x^3+2x^2+3x+8}{x^2+4}=4x+2+\frac{-13x}{x^2+4}\qquad A=4,\;B=2,\;C=-13,\;D=0$$
Practice Quiz
Quiz 02 — Past Paper Style
Apply the techniques to exam-style problems.
P1. Simplify fully:
$$\frac{x+2}{x+4}\div\frac{3x+6}{x^2-16}$$
P2. Using the remainder shortcut, find the remainder when \(2x^2-5x+8\) is divided by \((x-2)\).
P3. What is the numerator when expressed as a single fraction?
$$\frac{2}{x+3}-\frac{1}{x+1}$$
P4. In the identity below, find the value of \(B\):
$$\frac{2x^3-4x-15}{x^2+3x+4}\equiv 2x+B+\frac{C(2x+3)}{x^2+3x+4}$$
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Key Takeaways