P2.1 Proof and algebraic division
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Pure Mathematics 2 (AS), specification sections 1.1, 1.2, 1.3 and 2.1: proof by deduction and exhaustion, disproof by counterexample, and dividing a polynomial by a linear expression.
What this page covers
7 points
- The structure of a mathematical proof: what you assume, the logical steps, and the conclusion
- Proof by deduction for algebraic results, including completing the square and simple inequalities
- Proof by exhaustion: checking every member of a finite set, or splitting into cases that cover every possibility
- Disproof by counterexample
- Dividing a polynomial (up to a cubic) by a linear expression \((ax + b)\) or \((ax - b)\)
- The remainder theorem and the factor theorem
- Factorising a cubic fully and finding unknown coefficients
In one minute
6 points
- A proof starts from facts you are allowed to assume, moves through steps that each follow logically, and ends with a conclusion that restates what was to be proved.
- Deduction: write the general case in algebra (an even number is \(2n\), an odd number is \(2n + 1\)) and manipulate it into the required form.
- Exhaustion: split every possibility into a finite number of cases (even/odd, remainders on division by 3, …) and prove the result in each case.
- Counterexample: one specific case where a statement is false is enough to disprove it. Checking many cases where it is true never proves it.
- Division identity: \(f(x) \equiv (ax - b)\,q(x) + r\), where the remainder \(r\) is a constant.
- Remainder theorem: \(r = f\!\left(\frac{b}{a}\right)\). Factor theorem: \((ax - b)\) is a factor \(\Leftrightarrow f\!\left(\frac{b}{a}\right) = 0\).
Contents
13 sections
- What does a mathematical proof need?
- How do you prove a result by deduction?
- How do you prove a result by exhaustion?
- How do you disprove a statement?
- How do you divide a polynomial by \((ax + b)\)?
- What do the remainder and factor theorems say?
- How do you find unknown coefficients?
- Exam tips for P2.1
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
What does a mathematical proof need?
This page joins two skills that look different but share one habit: every line must follow from the line before. A proof is a chain of justified steps that ends in a stated conclusion; algebraic division is a chain of exact subtractions that ends in a quotient and a remainder. The factor and remainder theorems then turn a long division into a single substitution.
Every proof, however short, has the same three parts. If any part is missing, the reader cannot be sure the argument covers everything.
| Part of the proof | What it contains | Example |
|---|---|---|
| Assumptions (the starting point) | definitions and facts you may use; what the variables stand for | "Let \(n\) be an integer. Then \(2n + 1\) is any odd integer." |
| Logical steps | algebra or reasoning in which each line follows from the previous ones | expanding, factorising, using that a square is never negative |
| Conclusion | a sentence that states what has been shown, linked back to the original claim | "\(3(2n + 1)\) is a multiple of 3, so the sum is always a multiple of 3." |
Language and notation you will use
- Integer forms: any integer \(n\); even \(2n\); odd \(2n + 1\) (or \(2n - 1\)); consecutive integers \(n, n + 1, n + 2\); a multiple of 5 is \(5k\); a number leaving remainder 2 on division by 3 is \(3k + 2\).
- Different letters for different numbers. Two odd numbers need \(2m + 1\) and \(2n + 1\). Using the same letter only covers two equal odd numbers.
- \(\Rightarrow\) "implies": if the left is true, the right is true. \(\Leftrightarrow\) "if and only if": each implies the other. \(\therefore\) "therefore".
- Identity \(\equiv\): true for every value of the variable, e.g. \((x + 1)^2 \equiv x^2 + 2x + 1\).
- Rational number: can be written as \(\frac{p}{q}\) with \(p, q\) integers and \(q \ne 0\). A prime has exactly two factors, 1 and itself (so 1 is not prime).
Checking a few numbers and writing "so it is always true". Examples can suggest a result, but they only prove it when they cover every possible case, which is exactly what proof by exhaustion does.
How do you prove a result by deduction?
The key move is the first one: turning words into algebra that represents every case at once. "Three consecutive odd numbers" is not 1, 3, 5; it is \(2n - 1\), \(2n + 1\), \(2n + 3\) for an unknown integer \(n\). Whatever you prove about these expressions is then true for every choice of \(n\).
A reliable method
- Translate the words into algebra: name a general integer, an odd number, a function.
- Manipulate: expand, collect, factorise, complete the square.
- Reach a form that shows the property: a factor of 3 outside a bracket shows a multiple of 3; \(2(\dots) + 1\) shows odd; \((\dots)^2\) plus a positive number shows positive.
- Write the conclusion in words.
Prove that the sum of any three consecutive odd numbers is a multiple of 3.
Step 1. Let the numbers be \(2n - 1\), \(2n + 1\) and \(2n + 3\), where \(n\) is an integer.
Step 2. Add and factorise: \((2n - 1) + (2n + 1) + (2n + 3) = 6n + 3 = 3(2n + 1)\).
Step 3. \(2n + 1\) is an integer, so the sum is 3 × an integer, a multiple of 3.
Consecutive odd numbers differ by 2, not 1; that is why the terms are \(2n - 1\), \(2n + 1\), \(2n + 3\).
Proving an expression is always positive
Completing the square is the standard tool, because it rewrites a quadratic as "a square plus a constant", and a square has a known sign.
- Complete the square: \(x^2 + bx + c = \left(x + \frac{b}{2}\right)^2 + c - \frac{b^2}{4}\).
- A real number squared is never negative: \((\dots)^2 \ge 0\) for every real \(x\).
- So if the constant left over is positive, the whole expression is positive for every \(x\).
Prove that \(x^2 - 6x + 11 > 0\) for all real values of \(x\).
Step 1. Complete the square: \(x^2 - 6x + 11 = (x - 3)^2 - 9 + 11 = (x - 3)^2 + 2\).
Step 2. \((x - 3)^2 \ge 0\) for all real \(x\), so \((x - 3)^2 + 2 \ge 2\).
Step 3. \(2 > 0\), so \(x^2 - 6x + 11 > 0\) for all real \(x\).
The least value of the expression is 2, when \(x = 3\).
Proving an inequality
- Start from something known to be true, usually "a square is \(\ge 0\)", and work towards the inequality.
- If you found the idea by working backwards from the result, rewrite the proof forwards, or make sure every step is reversible (\(\Leftrightarrow\)), and state any conditions (such as \(a, b > 0\)) that you used.
- Multiplying or dividing an inequality by a negative number reverses it, so check the signs of anything you divide by.
Prove that for all positive real numbers \(a\) and \(b\), \(\dfrac{a + b}{2} \ge \sqrt{ab}\), and state when equality holds.
Step 1. Start from a true statement. As \(a, b > 0\), \(\sqrt{a}\) and \(\sqrt{b}\) are real, so \((\sqrt{a} - \sqrt{b})^2 \ge 0\).
Step 2. Expand: \(a - 2\sqrt{a}\sqrt{b} + b \ge 0\), so \(a + b \ge 2\sqrt{ab}\).
Step 3. Divide by 2 (positive, so the inequality is unchanged): \(\dfrac{a + b}{2} \ge \sqrt{ab}\) for all positive \(a, b\).
Step 4. Equality \(\Leftrightarrow (\sqrt{a} - \sqrt{b})^2 = 0 \Leftrightarrow\) \(a = b\).
How do you prove a result by exhaustion?
- Finite set: if the statement is about a small, listed set of numbers, check every one of them. Every member must be checked.
- Infinite set split into cases: every integer is even or odd (\(2k\) or \(2k + 1\)); every integer is \(3k\), \(3k + 1\) or \(3k + 2\); and so on. A proof for each case covers all integers.
- Say clearly why the cases are exhaustive (nothing is left out) and finish with a conclusion that covers them all.
A finite example: is 97 prime?
- If 97 had a factor other than 1 and itself, it would have a prime factor no bigger than \(\sqrt{97} \approx 9.85\), so only 2, 3, 5 and 7 need checking.
- \(97 = 2 \times 48 + 1\), \(97 = 3 \times 32 + 1\), \(97 = 5 \times 19 + 2\), \(97 = 7 \times 13 + 6\): none divides exactly, so 97 is prime.
(a) Prove that the square of any integer is either a multiple of 4 or one more than a multiple of 4. (b) Hence show that 1 000 003 is not a square number.
(a) Every integer is even or odd, so two cases cover everything.
Case 1, \(n\) even: let \(n = 2k\), \(k\) an integer. Then \(n^2 = 4k^2\), a multiple of 4.
Case 2, \(n\) odd: let \(n = 2k + 1\). Then \(n^2 = 4k^2 + 4k + 1 = 4(k^2 + k) + 1\), one more than a multiple of 4.
In both cases the result holds, so every square is \(4m\) or \(4m + 1\) for some integer \(m\).
(b) \(1\,000\,003 = 4 \times 250\,000 + 3\), which is three more than a multiple of 4. By (a) no square has this form, so 1 000 003 is not a square number.
In an exhaustion proof, name each case before you work on it ("Case 1: \(n\) is even, \(n = 2k\)") and finish with one sentence that brings the cases together. A conclusion for only one case leaves the proof incomplete.
How do you disprove a statement?
- A statement about "all" or "every" is disproved by one counterexample.
- Give the actual values, show the working that makes the statement fail, and say clearly that the statement is therefore false.
- Good places to look: 0, 1, negative numbers, fractions between 0 and 1, and the number that appears in the expression itself.
Disprove each statement by giving a counterexample. (a) \(n^2 + n + 11\) is prime for every positive integer \(n\). (b) \(x^3 > x^2\) for every positive real number \(x\).
(a) Choose \(n\) so that 11 is a common factor: try \(n = 10\). \(10^2 + 10 + 11 = 121 = 11 \times 11\), which is not prime, so the statement is false.
(b) Try a number between 0 and 1, \(x = \frac12\): \(\left(\frac12\right)^3 = \frac18\) and \(\left(\frac12\right)^2 = \frac14\); \(\frac18 < \frac14\), so the statement is false.
Showing that statement (a) holds for \(n = 1, 2, \dots, 9\) proves nothing, and here it is wrong at \(n = 10\).
Stating a counterexample without the evidence. "\(n = 10\)" alone is not enough; you need the value 121, its factorisation \(11 \times 11\), and the sentence "so the statement is false".
How do you divide a polynomial by \((ax + b)\)?
Dividing polynomials works like dividing whole numbers. \(23 \div 5\) is 4 remainder 3 because \(23 = 5 \times 4 + 3\). In the same way, dividing \(f(x)\) by a linear expression gives a quotient and a remainder that satisfy an identity.
- \(f(x)\) is the polynomial being divided (the dividend), \((ax + b)\) is the divisor, \(q(x)\) is the quotient, one degree lower than \(f\), and \(r\) is the remainder, a constant.
- In the formulae list? Not in the formulae booklet: learn it.
If \(r = 0\) the divisor is a factor of \(f(x)\). Dividing a cubic by a linear expression always gives a quadratic quotient, and the remainder is a constant because it must have a lower degree than the divisor.
Long division, step by step
- Write \(f(x)\) in descending powers, with a 0 for any missing power (e.g. \(x^3 + 0x^2 - 7x + 6\)).
- Divide the leading term of what is left by the leading term of the divisor: this gives the next term of the quotient.
- Multiply the whole divisor by that term and subtract. Watch the signs on every term.
- Bring down the next term and repeat until what is left has a lower degree than the divisor: that is the remainder.
Alternative: compare coefficients in \(f(x) \equiv (ax + b)(px^2 + qx + s) + r\). This is often quicker when you only need the quotient.
Divide \(f(x) = 4x^3 + 4x^2 - 5x + 3\) by \((2x + 1)\), giving the quotient and the remainder. Check the remainder by substitution.
Step 1. \(4x^3 \div 2x = 2x^2\); subtract \(2x^2(2x + 1) = 4x^3 + 2x^2\): \((4x^3 + 4x^2) - (4x^3 + 2x^2) = 2x^2\). Bring down \(-5x\): \(2x^2 - 5x\).
Step 2. \(2x^2 \div 2x = x\); subtract \(x(2x + 1) = 2x^2 + x\): \((2x^2 - 5x) - (2x^2 + x) = -6x\). Bring down \(+3\): \(-6x + 3\).
Step 3. \(-6x \div 2x = -3\); subtract \(-3(2x + 1) = -6x - 3\): \((-6x + 3) - (-6x - 3) = 6\).
Step 4. Quotient \(2x^2 + x - 3\), remainder 6: \(f(x) \equiv (2x + 1)(2x^2 + x - 3) + 6\).
Step 5. Check: put \(2x + 1 = 0\), i.e. \(x = -\frac12\), so the first term vanishes. \(f\!\left(-\frac12\right) = 4\left(-\frac18\right) + 4\left(\frac14\right) + \frac52 + 3 = -\frac12 + 1 + \frac52 + 3 = 6\). ✓
What do the remainder and factor theorems say?
Step 5 of Worked example 6 is the whole idea. In the division identity, choose the value of \(x\) that makes the divisor zero. The product term disappears and only the remainder is left. You can find a remainder without dividing at all.
- Why: in \(f(x) \equiv (ax - b)\,q(x) + r\), put \(x = \frac{b}{a}\); the bracket becomes 0, leaving \(f\!\left(\frac{b}{a}\right) = r\).
- Divisor \((x - 3)\): use \(f(3)\). Divisor \((x + 3)\): use \(f(-3)\). Divisor \((2x + 1)\): use \(f\!\left(-\frac12\right)\).
- In the formulae list? Not in the formulae booklet: learn it.
- It is the remainder theorem with remainder 0, and it works both ways: to show a factor, and to find one.
- In the formulae list? Not in the formulae booklet: learn it.
For a cubic with integer coefficients, any integer root divides the constant term, so try the factors of the constant first (\(\pm 1, \pm 2, \dots\)).
Factorising a cubic fully
- Find one root \(p\) with \(f(p) = 0\), given in the question or found by trial.
- Divide by \((x - p)\), by long division or comparing coefficients, to get a quadratic.
- Factorise the quadratic. If it will not factorise, check its discriminant: \(b^2 - 4ac < 0\) means no more real roots.
- "Fully" means a product of linear factors where possible: leaving \((x - 2)(3x^2 + 13x + 4)\) is not finished.
\(f(x) = 3x^3 + 7x^2 - 22x - 8\). (a) Use the factor theorem to show that \((x - 2)\) is a factor of \(f(x)\). (b) Factorise \(f(x)\) completely. (c) Hence solve \(f(x) = 0\).
(a) \(f(2) = 3(8) + 7(4) - 22(2) - 8 = 24 + 28 - 44 - 8 = 0\). Since \(f(2) = 0\), \((x - 2)\) is a factor.
(b) Compare coefficients in \((x - 2)(3x^2 + px + 4)\); the 3 and the 4 make the \(x^3\) term and the constant \(-8\) right. \(x^2\): \(p - 6 = 7\), so \(p = 13\). Check \(x\): \(4 - 2p = -22\). ✓
Factorise the quadratic: \(3x^2 + 13x + 4 = (3x + 1)(x + 4)\), so \(f(x) = (x - 2)(3x + 1)(x + 4)\).
(c) Each factor equal to zero: \(x = 2\), \(x = -\frac13\), \(x = -4\).
Getting the sign wrong in the factor theorem. \(f(3) = 0\) means \((x - 3)\) is a factor, not \((x + 3)\); and for \((3x + 1)\) you substitute \(x = -\frac13\), not \(x = -1\) or \(x = -3\).
How do you find unknown coefficients?
When a polynomial contains unknown constants, each fact you are told about it becomes an equation. The factor theorem gives an equation equal to zero; the remainder theorem gives an equation equal to the stated remainder.
- "\((x + 2)\) is a factor" \(\Rightarrow f(-2) = 0\); "the remainder on dividing by \((x - 1)\) is \(-6\)" \(\Rightarrow f(1) = -6\).
- One unknown needs one condition; two unknowns need two conditions, then solve simultaneously.
- Simplify each equation (divide through by any common factor) before solving, and check your values in both conditions.
\(f(x) = ax^3 + bx^2 - 11x + 6\), where \(a\) and \(b\) are constants. \((x + 2)\) is a factor of \(f(x)\), and when \(f(x)\) is divided by \((x - 1)\) the remainder is \(-6\).
(a) Find the values of \(a\) and \(b\). (b) Factorise \(f(x)\) completely. (c) Find the remainder when \(f(x)\) is divided by \((2x + 1)\).
(a) Factor theorem, \(f(-2) = 0\): \(-8a + 4b + 22 + 6 = 0\), so \(2a - b = 7\).
Remainder theorem, \(f(1) = -6\): \(a + b - 11 + 6 = -6\), so \(a + b = -1\).
Add the two equations: \(3a = 6\), so \(a = 2\), \(b = -3\).
(b) Divide \(2x^3 - 3x^2 - 11x + 6\) by \((x + 2)\): \(2x^3 - 3x^2 - 11x + 6 = (x + 2)(2x^2 - 7x + 3)\), so \(f(x) = (x + 2)(2x - 1)(x - 3)\).
(c) Remainder theorem with \(x = -\frac12\): \(f\!\left(-\frac12\right) = 2\left(-\frac18\right) - 3\left(\frac14\right) - 11\left(-\frac12\right) + 6 = -\frac14 - \frac34 + \frac{11}{2} + 6 =\) \(\frac{21}{2}\).
Check (a): \(f(1) = 2 - 3 - 11 + 6 = -6\) ✓ and \(f(-2) = -16 - 12 + 22 + 6 = 0\) ✓.
Write each condition as "\(f(-2) = 0\)" or "\(f(1) = -6\)" before substituting. It shows which theorem you are using, and it stops you equating the remainder to 0 by habit.
Exam tips for P2.1
Start every algebraic proof by saying what the letters are: "Let \(n\) be an integer." Then the forms \(2n\), \(2n + 1\) or \(3k + 2\) have a meaning, and the conclusion can refer back to them.
A proof is not finished when the algebra stops. Finish with a statement such as "so the sum is always a multiple of 3" or "so the statement is false", using the words of the question.
When a result is given ("show that \((x - 2)\) is a factor"), the working is the whole answer: write the substitution, each value, the total 0, and the conclusion.
After a long division, check the remainder with the remainder theorem, or multiply back: (divisor) × (quotient) + remainder must give the original polynomial term by term.
"Factorise completely" is only finished when the quadratic factor has been factorised too (or shown not to factorise). "Hence" in the next part means use those factors; do not start again.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| A result that works for lots of examples is proved. | A later case may fail: \(n^2 + n + 11\) is prime for \(n = 1\) to 9 but not for 10. | Examples prove a result only when they exhaust every possible case. |
| Two odd numbers can be written \(2n + 1\) and \(2n + 1\). | That describes the same number twice. | Use different letters: \(2m + 1\) and \(2n + 1\). |
| One counterexample is not enough to disprove a statement. | A statement about "all" cases is false if a single case fails. | One clearly worked counterexample is a complete disproof. |
| If \(f(3) = 0\) then \((x + 3)\) is a factor. | Substituting \(x = 3\) makes \((x - 3)\) zero, not \((x + 3)\). | \(f(p) = 0 \Leftrightarrow (x - p)\) is a factor. |
| The remainder when dividing by \((2x - 1)\) is \(f(1)\) or \(f(2)\). | The divisor is zero at \(x = \frac12\). | Remainder \(= f\!\left(\frac12\right)\). |
| Dividing a cubic by a linear expression can leave a remainder in \(x\). | The remainder always has a lower degree than the divisor. | Dividing by a linear divisor leaves a constant remainder. |
How this topic connects
P1 algebra and graphs
Completing the square, the discriminant and sketching cubics from their factors all feed straight into this page: a fully factorised cubic gives its \(x\)-intercepts at once.
P2.3 Sequences and series
The sum formulae for arithmetic and geometric series are proved by deduction: write the sum forwards and backwards, or multiply by \(r\) and subtract.
Later pure units (P3, P4), mechanics and physics
Polynomial division reappears when simplifying algebraic fractions (P3) and splitting them into partial fractions (P4), and proof by contradiction (for example, that \(\sqrt{2}\) is irrational) completes the set of proof methods in P4. Cubic and quadratic models in mechanics are solved with the factor theorem, and "show that" derivations in M1, M2 and IAL Physics need the same complete chain of reasoning.
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(f(x) \equiv (ax - b)\,q(x) + r\) | division identity: quotient \(q(x)\), constant remainder \(r\) | No — learn it |
| \(r = f\!\left(\frac{b}{a}\right)\) | remainder theorem: remainder on division by \((ax - b)\) | No — learn it |
| \(f\!\left(\frac{b}{a}\right) = 0 \Leftrightarrow (ax - b)\) is a factor | factor theorem: showing or finding factors | No — learn it |
| \(2n\), \(2n + 1\), \(n, n + 1, n + 2\) | even, odd and consecutive integers in a proof | No — learn it |
| \(x^2 + bx + c = \left(x + \frac{b}{2}\right)^2 + c - \frac{b^2}{4}\) | completing the square to prove an expression is positive | No — derive it |
| \((\sqrt{a} - \sqrt{b})^2 \ge 0 \Rightarrow \frac{a + b}{2} \ge \sqrt{ab}\) | a standard starting point for inequality proofs (\(a, b > 0\)) | No — derive it |
Check your understanding
1. Prove that the sum of any two consecutive integers is odd.
Let the integers be \(n\) and \(n + 1\), where \(n\) is an integer. Their sum is \(2n + 1\), which is one more than a multiple of 2, so it is odd for every integer \(n\).
2. Disprove: "if \(n\) is an odd positive integer, then \(n^2 + 4\) is prime".
Try odd values: \(1 \to 5\), \(3 \to 13\), \(5 \to 29\), \(7 \to 53\) are prime, but \(n = 9\) gives \(81 + 4 = 85 = 5 \times 17\), which is not prime. So the statement is false.
3. Find the remainder when \(x^3 - 2x^2 + 5\) is divided by \((x + 2)\).
Remainder theorem: \(f(-2) = -8 - 8 + 5\). The remainder is \(-11\).
4. \((2x - 3)\) is a factor of \(2x^3 - x^2 + kx - 6\). Find \(k\) and the quadratic factor.
\(f\!\left(\frac32\right) = \frac{27}{4} - \frac94 + \frac32 k - 6 = \frac32 k - \frac32 = 0\), so \(k = 1\). Then \(2x^3 - x^2 + x - 6 = (2x - 3)(x^2 + x + 2)\); the quadratic has discriminant \(1 - 8 < 0\), so it has no real roots. \(k = 1\); quadratic factor \(x^2 + x + 2\).
5. A student checks that \(n^2 + n + 1\) is odd for \(n = 1\) to 50 and says this proves it for all integers. Explain what is missing and give a proof.
Fifty values do not cover every integer, so this is not a proof. Exhaustion with two cases does: if \(n = 2k\), then \(n^2 + n + 1 = 4k^2 + 2k + 1 = 2(2k^2 + k) + 1\); if \(n = 2k + 1\), then it equals \(4k^2 + 6k + 3 = 2(2k^2 + 3k + 1) + 1\). Both are odd, and every integer is even or odd, so \(n^2 + n + 1\) is always odd.
6. Find the quotient and remainder when \(x^3 + 0x^2 - 7x + 6\) is divided by \((x - 3)\). What does the result tell you?
\(x^3 \div x = x^2\); subtract \(x^3 - 3x^2\) to leave \(3x^2 - 7x\). \(3x^2 \div x = 3x\); subtract \(3x^2 - 9x\) to leave \(2x + 6\). \(2x \div x = 2\); subtract \(2x - 6\) to leave 12. Quotient \(x^2 + 3x + 2\), remainder 12, so \((x - 3)\) is not a factor. Check: \(f(3) = 27 - 21 + 6 = 12\). ✓
Specification checklist
- (P2 1.1) I can set out a proof with clear assumptions, logical steps and a stated conclusion, using \(\Rightarrow\), \(\Leftrightarrow\) and \(\therefore\) correctly.
- (P2 1.1) I can prove simple algebraic results by deduction, including results about even, odd and consecutive integers, and inequalities proved by completing the square.
- (P2 1.2) I can prove a result by exhaustion, either by checking every member of a finite set or by splitting all cases (e.g. even/odd, remainders on division by 3).
- (P2 1.3) I can disprove a statement by finding and fully showing a counterexample.
- (P2 2.1) I can divide a polynomial of degree up to 3 by \((ax + b)\) or \((ax - b)\), giving the quotient and remainder.
- (P2 2.1) I can use the remainder theorem and the factor theorem, factorise a cubic completely and find unknown coefficients from given factors or remainders.