Use remainder values and conditions to find unknown coefficients
Evaluate at a linear divisor's root, including non-monic divisors, and turn factor and remainder information into simultaneous equations.
Exam scope and assessment notes
Pure Mathematics 2 (Paper 2, AS only) and Pure Mathematics 3 (Paper 3, A Level) · Shared Algebra foundations, Mathematics 9709, 2026–2027. Paper 1 knowledge is assumed. These modulus and polynomial requirements occur in both sections 2.1 and 3.1. Paper 3 additionally requires partial fractions and rational-power binomial expansions; follow its separate lessons for those methods. The AS route using Papers 1 + 2 cannot be carried forward to A Level. Both sections use linear modulus graphs; nonlinear y = |f(x)| and y = f(|x|) graphs are excluded. Polynomial dividends have degree at most 4 and divisors are linear or quadratic. Written practice is self-checked.
Before you begin
- Evaluate polynomials at signed and fractional inputs.
- Use a quotient–remainder identity and solve simultaneous linear equations.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
At which input does the divisor 2x − 1 vanish?
What you will learn
- Use f(c) as the remainder for division by a linear factor vanishing at c.
- Determine unknown coefficients from specified factors or remainders.
Choose the root of the divisor, not a copied coefficient
For division by ax + b with a ≠ 0, the remainder is a constant r. In f(x) = (ax + b)Q(x) + r, substitute c = −b/a so the divisor term vanishes: r = f(c). The remainder is f(c), with no extra division by a. If f(c) = 0, ax + b is a factor. This is the remainder and factor theorem for a linear divisor; a quadratic divisor can have a linear remainder, so one evaluation does not generally determine it.
For f(x) = 2x³ − x² + 4x − 3 divided by 2x − 1, use c = 1/2. The value is 2(1/8) − 1/4 + 2 − 3 = −1, so the remainder is −1. Direct division gives Q = x² + 2, and (2x − 1)(x² + 2) − 1 reconstructs f. Using x = 1 or dividing f(1/2) by 2 would give an incorrect result.
Each condition supplies an equation
If x − 1 is a factor of f(x) = x³ + px² + qx + 6, then f(1) = 0 gives p + q = −7. If division by x + 2 has remainder 12, then f(−2) = 12 gives 4p − 2q − 2 = 12, or 2p − q = 7. Solving gives p = 0 and q = −7. Check both original evaluations once the coefficients have been found.
Different linear conditions can determine several unknown coefficients, provided the resulting equations supply enough independent information. A stated remainder need not be zero; only a factor condition gives a zero remainder. Retain fractions and signs when the divisor root is negative or non-integral, and distinguish a remainder value from the quotient polynomial.
Pause and explain
Division of f by 2x − 1 gives constant remainder r. Which is correct?
Worked example
Find the remainder when 2x³ − x² + 4x − 3 is divided by 2x − 1.
Show the worked solution
- The divisor vanishes at c = 1/2.
- Evaluate f(1/2) = 2(1/8) − 1/4 + 4(1/2) − 3 = −1.
- Check f = (2x − 1)(x² + 2) − 1; the constant remainder is −1.
Answer Remainder −1, using divisor root 1/2.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find the remainder when x³ − 2x² + 5x − 4 is divided by x − 2.
Give me a hint
Use the divisor root 2 in the original polynomial.
Compare my reasoning
- For a linear divisor x − 2, the remainder is f(2).
- Evaluate f(2) = 8 − 8 + 10 − 4 = 6.
- The remainder is 6, so x − 2 is not a factor of this polynomial.
Remainder 6; x − 2 is not a factor.
Look for these in your work
- Choose the zero of the actual divisor.
- Distinguish a non-zero remainder from a factor condition.
Find the remainder when 4x³ + 6x² − x + 2 is divided by 2x + 3, and verify with a quotient identity.
Give me a hint
Use x = −3/2 and keep the evaluated fractional powers exact.
Compare my reasoning
- The divisor root is −3/2, so the remainder is f(−3/2).
- The cubic and quadratic terms cancel; the remaining value is 3/2 + 2 = 7/2.
- Indeed f(x) = (2x + 3)(2x² − 1/2) + 7/2, confirming the same remainder.
Remainder 7/2; quotient check uses 2x² − 1/2.
Look for these in your work
- Use the non-monic divisor's negative fractional root.
- Verify the evaluated remainder in a rebuilt identity.
Let f(x) = x³ + px² + qx + 6. Given that x − 1 is a factor and the remainder on division by x + 2 is 12, find p and q and check both conditions.
Give me a hint
Translate the conditions into f(1) = 0 and f(−2) = 12 before solving.
Compare my reasoning
- f(1) = 0 gives p + q = −7; f(−2) = 12 gives 2p − q = 7.
- Adding the equations gives 3p = 0, hence p = 0 and q = −7.
- For f(x) = x³ − 7x + 6, f(1) = 0 and f(−2) = −8 + 14 + 6 = 12, so both conditions hold.
p = 0 and q = −7; both original remainder conditions hold.
Look for these in your work
- Translate factor and non-zero remainder information correctly.
- Check the recovered polynomial against both conditions.
Common mistakes
- For ax + b, substitute −b/a rather than b or −b alone.
- Do not divide the evaluated remainder by the leading coefficient.
- A non-zero remainder condition is not a factor condition.
What can you explain now?
For f(x) = 4x³ + 6x² − x + 2 divided by 2x + 3, which input gives the remainder and what is it?
Compare with the explanation
Use x = −3/2; remainder 7/2.
At −3/2 the cubic and quadratic contributions −27/2 and 27/2 cancel, leaving 3/2 + 2 = 7/2.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Derive the remainder theorem from the rebuilt identity at a divisor root. Show how a non-monic divisor changes the input but adds no extra scale to the remainder.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.