Algebra (Pure 3) — A-Level Mathematics Pure 3 Study Guide

For: A-Level Mathematics candidates sitting Paper 3 (Pure Mathematics 3).

Covers: Modulus function graphs and equations, polynomial division with factor/remainder theorems, partial fractions (proper, improper, repeated and quadratic factors), binomial expansion for non-integer/negative n with convergence rules, and solving inequalities involving algebraic fractions.

You should already know: A-Level Mathematics Pure 1 (functions, calculus, trig).

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Mathematics style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Algebra (Pure 3)?

Pure 3 Algebra extends foundational algebraic techniques from Pure 1 to solve complex non-linear, fractional, and multi-term problems that form the backbone of all other Paper 3 topics, including integration, differential equations, and complex numbers. It makes up 15-20% of total Paper 3 marks, often tested in combination with other topics in 7-10 mark long-form questions. Unlike Pure 1 Algebra, it focuses on non-terminating series, piecewise function operations, and safe manipulation of expressions with unknown sign.

2. Modulus function $|x|$ — graphs and equations

The modulus (or absolute value) function $|x|$ measures the distance of a real number $x$ from 0 on the number line, so its output is always non-negative. It is defined piecewise: $$|x|=\{\begin{array}{ll}x& x\ge 0\\ -x& x<0\end{array}$$

Key Graph Transformations

• The base graph $y=|x|$ is a V-shape with vertex at $(0,0)$, gradient 1 for $x>0$ and -1 for $x<0$.
• $y=|ax+b|$ is a scaled V-shape with vertex at $x=-\frac{b}{a}$.
• $y=|f(x)|$ reflects all segments of $f(x)$ below the x-axis above the x-axis.
• $y=f(|x|)$ reflects the right half of $f(x)$ across the y-axis, so it is always symmetric about the y-axis.

Solving Modulus Equations

To solve $|f(x)|=g(x)$:

1. Split into two cases: $f(x)=g(x)$ and $f(x)=-g(x)$
2. Check all solutions: since $|f(x)|\ge 0$, $g(x)$ must be non-negative for the solution to be valid. Discard any solutions where $g(x)<0$.

Worked Example: Solve $|3x-5|=x+2$

• Case 1: $3x-5=x+2⟹2x=7⟹x=3.5$. Check RHS: $3.5+2=5.5\ge 0$, valid.
• Case 2: $3x-5=-(x+2)⟹4x=3⟹x=0.75$. Check RHS: $0.75+2=2.75\ge 0$, valid.
• Solutions: $x=0.75,x=3.5$

Exam tip: Examiners regularly set modulus equations with invalid solutions to test if you complete the validity check, which is worth 1-2 marks per question.

3. Polynomial division and the factor / remainder theorem

When dividing a higher-degree polynomial $P(x)$ (dividend) by a lower-degree polynomial $D(x)$ (divisor), you get a quotient polynomial $Q(x)$ and remainder polynomial $R(x)$, such that: $$P(x)=D(x)Q(x)+R(x)$$ where the degree of $R(x)$ is always less than the degree of $D(x)$. If $D(x)$ is linear (degree 1), $R(x)$ is a constant.

Remainder Theorem

If $P(x)$ is divided by a linear factor $(ax+b)$, the remainder is equal to $P\left(-\frac{b}{a}\right)$. This eliminates the need for long division to find remainders for linear divisors.

Factor Theorem

A special case of the remainder theorem: if $P\left(-\frac{b}{a}\right)=0$, then $(ax+b)$ is a factor of $P(x)$.

Worked Example: Find the remainder when $P(x)=2x^3-4x^2+7x-1$ is divided by $(2x+1)$

• Using the remainder theorem: $a=2,b=1$, so evaluate $P\left(-\frac{1}{2}\right)$: $$P\left(-\frac{1}{2}\right)=2{\left(-\frac{1}{2}\right)}^3-4{\left(-\frac{1}{2}\right)}^2+7\left(-\frac{1}{2}\right)-1=-\frac{1}{4}-1-\frac{7}{2}-1=-\frac{23}{4}$$
• The remainder is $-\frac{23}{4}$.

4. Partial fractions — proper, improper, repeated and quadratic factors

Partial fraction decomposition breaks a single rational function into a sum of simpler fractions that are easier to integrate, differentiate, or expand binomially.

First Step: Check if the fraction is proper

A fraction is proper if the degree of the numerator is less than the degree of the denominator. If it is improper, first perform polynomial division to rewrite it as a polynomial plus a proper fraction.

Decomposition Rules by Denominator Factor Type

1. Distinct linear factor $(ax+b)$: Assign a constant numerator: $\frac{A}{ax+b}$
2. Repeated linear factor $(ax+b{)}^k$: Assign terms for each power from 1 to $k$: $\frac{A_1}{ax+b}+\frac{A_2}{(ax+b{)}^2}+...+\frac{A_k}{(ax+b{)}^k}$
3. Irreducible quadratic factor $(a{x}^2+bx+c)$ (discriminant $<0$): Assign a linear numerator: $\frac{Ax+B}{a{x}^2+bx+c}$

Worked Example: Decompose $\frac{4x+3}{(x-1{)}^2(x+2)}$ into partial fractions

• Let $\frac{4x+3}{(x-1{)}^2(x+2)}=\frac{A}{x-1}+\frac{B}{(x-1{)}^2}+\frac{C}{x+2}$
• Multiply both sides by the denominator: $4x+3=A(x-1)(x+2)+B(x+2)+C(x-1{)}^2$
• Substitute $x=1$: $4(1)+3=0+B(3)+0⟹B=\frac{7}{3}$
• Substitute $x=-2$: $4(-2)+3=0+0+C(9)⟹C=-\frac{5}{9}$
• Equate coefficients of $x^2$: $0=A+C⟹A=\frac{5}{9}$
• Final decomposition: $\frac{5}{9(x-1)}+\frac{7}{3(x-1{)}^2}-\frac{5}{9(x+2)}$

5. Binomial expansion for non-integer or negative $n$ — convergence

The Pure 1 binomial expansion for positive integer $n$ is a finite terminating series. For negative or non-integer $n$, the expansion is infinite, and only valid for values of $x$ that make the series converge.

General Formula

For any real $n$, the expansion of $(1+x{)}^n$ is: $$(1+x{)}^n=1+nx+\frac{n(n-1)}{2!}{x}^2+\frac{n(n-1)(n-2)}{3!}{x}^3+...$$ This is valid only when $|x|<1$, so that higher-order terms get smaller and the series converges to a finite value. For expansions of the form $(a+bx{)}^n$, rewrite it as $a^n{\left(1+\frac{b}{a}x\right)}^n$ to use the formula above. The convergence condition becomes $\left|\frac{b}{a}x\right|<1$.

Worked Example: Expand $(2-x{)}^{-2}$ up to the $x^2$ term, and state the valid range of $x$

• Rewrite: $(2-x{)}^{-2}=2^{-2}{\left(1-\frac{x}{2}\right)}^{-2}=\frac{1}{4}{\left(1+\left(-\frac{x}{2}\right)\right)}^{-2}$
• Apply the formula with $n=-2$ and $x$ term = $-\frac{x}{2}$: $$\frac{1}{4}\left[1+(-2)\left(-\frac{x}{2}\right)+\frac{(-2)(-3)}{2!}{\left(-\frac{x}{2}\right)}^2+...\right]$$
• Simplify: $\frac{1}{4}\left[1+x+\frac{3x^2}{4}+...\right]=\frac{1}{4}+\frac{x}{4}+\frac{3x^2}{16}$
• Convergence condition: $\left|-\frac{x}{2}\right|<1⟹|x|<2$

6. Solving inequalities involving algebraic fractions

The cardinal rule for solving these inequalities: never multiply both sides by an expression whose sign you do not know, as multiplying by a negative number reverses the inequality sign. Use one of two safe methods:

1. Multiply by the square of the denominator: Since squares are always non-negative, this does not change the inequality direction.
2. Rearrange to a single fraction: Move all terms to the left-hand side, combine into a single fraction, then test intervals around critical points where the numerator or denominator equals zero.

Worked Example: Solve $\frac{x+1}{x-4}>2$

• Method 1: Multiply both sides by $(x-4{)}^2$ ($x\ne 4$): $$(x+1)(x-4)>2(x-4{)}^2$$
• Expand and rearrange: $x^2-3x-4>2x^2-16x+32⟹x^2-13x+36<0$
• Factor: $(x-4)(x-9)<0$
• Test intervals: Only $4<x<9$ satisfies the inequality.
• Method 2: Rearrange: $\frac{x+1}{x-4}-2>0⟹\frac{x+1-2x+8}{x-4}>0⟹\frac{9-x}{x-4}>0$, same result.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Forgetting to check the validity of modulus equation solutions where the RHS is negative. Why students do it: Rushing through case-solving without verification. Correct move: Always substitute solutions back to confirm the RHS is non-negative, discard invalid ones.
• Wrong move: Decomposing improper partial fractions without first doing polynomial division. Why students do it: Skipping the degree comparison step. Correct move: Compare numerator and denominator degrees first, divide if needed to get a proper fraction before decomposition.
• Wrong move: Using the positive integer binomial formula for non-integer/negative $n$, and omitting the convergence condition. Why students do it: Confusing Pure 1 and Pure 3 binomial rules. Correct move: For non-integer/negative $n$, always rewrite the expression in $(1+x{)}^n$ form and explicitly state the valid $x$ range.
• Wrong move: Multiplying both sides of an algebraic fraction inequality by the unsquared denominator. Why students do it: Assuming the denominator is positive. Correct move: Either multiply by the squared denominator, or rearrange to a single fraction and test intervals.
• Wrong move: Writing repeated linear factors as two identical constant-numerator terms in partial fractions. Why students do it: Treating repeated factors as distinct. Correct move: For $(ax+b{)}^k$, include terms with powers from 1 to $k$ in the decomposition.

8. Practice Questions (A-Level Mathematics Paper 3 Style)

Question 1

(a) Sketch the graph of $y=|2x+2|-3$, labeling the vertex and all axis intercepts. [3 marks] (b) Solve the equation $|2x+2|-3=2x+1$ [4 marks]

Worked Solution

(a) Vertex at $2x+2=0⟹x=-1$, $y=0-3=-3$, so vertex $(-1,-3)$. Y-intercept at $x=0$: $|2|-3=-1$, so $(0,-1)$. X-intercepts: set $y=0$, $|2x+2|=3⟹x=0.5$ or $x=-2.5$, so $(0.5,0)$ and $(-2.5,0)$. The graph is a V-shape with gradient 2 for $x>-1$, gradient -2 for $x<-1$. (b) RHS must be $\ge -3$, so $2x+1\ge -3⟹x\ge -2$. Case 1: $x\ge -1$, $|2x+2|=2x+2$: $2x+2-3=2x+1⟹-1=1$, no solution. Case 2: $-2\le x<-1$, $|2x+2|=-2x-2$: $-2x-2-3=2x+1⟹-4x=6⟹x=-1.5$, which is in the valid range. Solution: $x=-1.5$.

Question 2

(a) Show that $(2x+1)$ is a factor of $P(x)=4x^3+8x^2-x-2$ [2 marks] (b) Decompose $\frac{5x+3}{4x^3+8x^2-x-2}$ into partial fractions [5 marks]

Worked Solution

(a) Use factor theorem: $P\left(-\frac{1}{2}\right)=4\left(-\frac{1}{8}\right)+8\left(\frac{1}{4}\right)-\left(-\frac{1}{2}\right)-2=-0.5+2+0.5-2=0$, so $(2x+1)$ is a factor. (b) Factor $P(x)$: divide by $(2x+1)$ to get $2x^2+3x-2=(2x-1)(x+2)$. So fraction is $\frac{5x+3}{(2x+1)(2x-1)(x+2)}$. Let equal $\frac{A}{2x+1}+\frac{B}{2x-1}+\frac{C}{x+2}$. Substitute $x=-0.5$: $A=-\frac{1}{3}$. Substitute $x=0.5$: $B=\frac{11}{15}$. Substitute $x=-2$: $C=-\frac{7}{15}$. Final decomposition: $-\frac{1}{3(2x+1)}+\frac{11}{15(2x-1)}-\frac{7}{15(x+2)}$.

Question 3

(a) Expand $(1+3x{)}^{-1/3}$ in ascending powers of $x$ up to the $x^2$ term. [4 marks] (b) State the range of $x$ for which the expansion is valid. [1 mark] (c) Use your expansion to estimate $\frac{1}{(1.03{)}^{1/3}}$ to 4 decimal places. [2 marks]

Worked Solution

(a) Apply binomial formula with $n=-\frac{1}{3}$, $x$ term = $3x$: $$1+\left(-\frac{1}{3}\right)(3x)+\frac{\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)}{2!}(3x{)}^2+...=1-x+2x^2+...$$ (b) Valid for $|3x|<1⟹|x|<\frac{1}{3}$. (c) Let $3x=0.03⟹x=0.01$. Substitute into expansion: $1-0.01+2\ast (0.01{)}^2=1-0.01+0.0002=0.9902$.

9. Quick Reference Cheatsheet

10. What's Next

The algebraic techniques covered in this guide are foundational to almost every other topic in A-Level Mathematics Paper 3. Partial fractions are used to integrate rational functions in the calculus topic, binomial expansion is used to approximate values and solve differential equations, and modulus function rules are extended to complex numbers and vectors later in the syllabus. Mastering these techniques will save you significant time on longer multi-part exam questions that combine multiple topics, as you will not need to waste time recalling basic operations mid-problem.

If you struggle with any of the concepts, worked examples, or practice questions in this guide, you can ask Ollie, our AI tutor, for personalized explanations, extra practice problems, or step-by-step walkthroughs tailored to your learning pace. You can also find more study guides and past paper practice for A-Level Mathematics Paper 3 on the OwlsPrep homepage to keep building your exam readiness and identify gaps in your knowledge.

Aligned with the Cambridge International AS & A Level Mathematics 9709 syllabus. OwlsAi is not affiliated with Cambridge Assessment International Education.