About this chapter
Algebraic methods opens Pure 2 with two very different skills. The factor and remainder theorems are reliable marks once the working is laid out the way examiners expect. Proof is the opposite: Pearson's examiners singled it out as a weak area in both June 2023 and June 2024, usually because a step is not justified or a conclusion is missing.
The chapter covers polynomial division, the factor and remainder theorems, finding unknown coefficients, factorising cubics completely, and proof by deduction, by exhaustion and by counter-example. The later questions combine these: a cubic written partly in brackets, a curve that touches the x-axis, and an inequality for positive numbers that fails once negatives are allowed.
The ten questions
- Q1 (5 marks): two known factors of a cubic give two equations for a and b, then three linear factors.
- Q2 (4 marks): the quotient and remainder when 2x³ − 3x² − 11x + 7 is divided by (x−3).
- Q3 (5 marks): use the factor theorem with (2x+1) to show a = −11, then factorise completely.
- Q4 (4 marks): a remainder of 10 fixes k, then the remainder on dividing by (2x+1).
- Q5 (5 marks): disprove a claim about primes by counter-example, and prove a divisibility result by exhaustion.
- Q6 (8 marks): a cubic written as (x+1)(2x² − x + k) − 9: a remainder by inspection, k from a factor, and why there is only one real root.
- Q7 (8 marks): factorise a cubic, then solve f(x) = x + 3 exactly using the factorised form.
- Q8 (8 marks): prove x³ + y³ ≥ xy(x+y) for positive x and y, find a counter-example for all reals, then deduce a second inequality.
- Q9 (9 marks): factorise a cubic, sketch it, and read off where f(x) ≥ 0.
- Q10 (10 marks): a cubic touching the x-axis at (2, 0) gives p and q, then the exact values of k giving a remainder of 12.
Key skills tested
Solving polynomial equations. Once f(x) is factorised, set each factor equal to zero. Take out a common factor rather than dividing by it, or a root is lost.
Dividing polynomials. Divide by (x − p) by long division or by inspection, so that f(x) = (x−p)q(x) + r.
The factor theorem. f(a) = 0 exactly when (x − a) is a factor. For (2x−3), show that f(3/2) = 0 and say that this makes (2x−3) a factor.
The remainder theorem. The remainder when f(x) is divided by (ax − b) is f(b/a).
Unknown coefficients. Each factor or remainder gives one equation, so two unknowns need two facts, solved simultaneously.
Factorising completely. Divide out the linear factor, then factorise the quadratic. If its discriminant is negative, it does not factorise.
Proof by deduction. Start from something true, such as (x−y)² ≥ 0, and give a reason for every step, stating the sign of anything you divide by.
Proof by exhaustion. Check every possible case, then write a conclusion.
Disproof by counter-example. One example where the conditions hold but the statement is false, with the working shown.
Worked example
Question 2 from this chapter. Find the quotient and the remainder when 2x³ − 3x² − 11x + 7 is divided by (x−3).
Divide term by term. 2x³ ÷ x = 2x², and 2x²(x−3) = 2x³ − 6x², leaving 3x² − 11x. Next, 3x² ÷ x = 3x, and 3x(x−3) = 3x² − 9x, leaving −2x + 7. Finally, −2x ÷ x = −2, and −2(x−3) = −2x + 6, leaving 1.
The quotient is 2x² + 3x − 2 and the remainder is 1.
Check with the remainder theorem: f(3) = 54 − 27 − 33 + 7 = 1, which matches.
Seen on real papers
Every question in our booklets is original. These are the recent papers where each question type has appeared.
- June 2024, Q4: a remainder read straight from a bracketed cubic, the factor theorem to show a constant, then the number of real roots with a reason, as in Q6.
- June 2023, Q2: a constant from the factor theorem, then showing a cubic has only one real root, as in Q3 and Q6.
- June 2024, Q5: proving an inequality for positive numbers, then a counter-example once the condition is removed, as in Q8.
- June 2023, Q8: a counter-example using prime numbers and a proof by exhaustion, as in Q5.
Where marks are lost
- No equals zero. In June 2024 some students substituted into f(x) but never wrote that the result equals zero, which cost the second mark of a "show that".
- Long division when the factor theorem is asked for. In June 2023 students who divided instead of using the factor theorem lost both marks.
- Half a factorisation. Not writing f(x) as the full product of its factors lost the final mark in June 2024.
- A number of roots with no reason. Stating "one real root" is not enough: show the discriminant of the quadratic factor is negative and refer back to the linear factor.
- Dividing without a reason. In the June 2024 proof almost everyone divided by xy without saying that x and y are positive, and lost the final mark.
Common questions
What must a factor theorem answer say?
Show the substitution, show that it equals zero, and state that the bracket is therefore a factor.
How do I choose a counter-example?
It must satisfy every condition in the statement and still make the statement false. In June 2024 most students broke the wrong part of the statement instead.
Is 1 a prime number?
No. Several students in June 2023 used 1 as a prime in a counter-example and lost the marks.
Where this chapter leads
- Pure 2 Chapter 7: Differentiation: factorised cubics give stationary points and sketches.
- Pure 3 Chapter 1: Algebraic Methods: algebraic fractions and division of polynomials.
- Pure 4 Chapter 1: Proof: proof by contradiction.