Use polynomial remainder and factor theorems
Evaluate a polynomial at a candidate root, interpret the remainder and reduce a cubic to a quadratic factor.
Section 3 · Factors of polynomials
Cambridge IGCSE Additional Mathematics 0606, 2028–2030 syllabus, Version 1. Paper 1 is non-calculator; Paper 2 requires a scientific calculator. These original lessons provide self-checked topic practice, not a Cambridge grade or a timed paper. Full-paper assessment and human specialist review are still outstanding.
Before you begin
- Cambridge assumes knowledge of IGCSE Mathematics or an equivalent course. Check the domain before using an inverse or logarithm, and retain exact values where the question asks for them.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the remainder when P(x) is divided by x − a?
What you will learn
- Apply the remainder theorem to P(a)
- Use P(a) = 0 to establish a factor x − a
- Find the remaining factor by division or a valid algebraic method
A polynomial value gives the remainder
When P(x) is divided by x − a, the remainder is P(a). Therefore x − a is a factor exactly when P(a) = 0.
For a cubic, finding one linear factor leaves a quadratic factor. Obtain it by algebraic long division or careful coefficient comparison, then check by multiplying the factors back together.
Keep the sign in the factor theorem consistent
A root a gives factor x − a. A root −2 gives factor x + 2. Confusing the root with the factor's constant sign is a common source of errors.
The factor theorem identifies a factor; it does not guarantee that the remaining quadratic has real roots.
Pause and explain
What does P(3) = 0 prove?
Worked example
Factor P(x) = x³ − 2x² − 5x + 6, given that x = 1 is a root.
Show the worked solution
- P(1) = 1 − 2 − 5 + 6 = 0, so x − 1 is a factor.
- Divide P(x) by x − 1 to obtain x² − x − 6.
- Factor the quadratic: x² − x − 6 = (x − 3)(x + 2).
Answer P(x) = (x − 1)(x − 3)(x + 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.
For P(x) = x³ + x² − 4x − 4, verify that x = 2 is a root and state its factor.
Give me a hint
Substitute the candidate before naming the factor.
Compare my reasoning
- P(2) = 8 + 4 − 8 − 4 = 0.
- The remainder on division by x − 2 is zero.
- Therefore x − 2 is a factor.
P(2) = 0, so x − 2 is a factor.
Look for these in your work
- Substitution arithmetic correct
- Remainder interpreted
- Factor sign correct
Divide x³ + 2x² − x − 2 by x + 2 and identify the remaining factor.
Give me a hint
The divisor x + 2 corresponds to a = −2.
Compare my reasoning
- Use synthetic or polynomial division with root −2.
- The quotient is x² − 1.
- Check (x + 2)(x² − 1) expands to the original polynomial.
Quotient x² − 1; factorisation (x + 2)(x − 1)(x + 1).
Look for these in your work
- Correct divisor root
- Quotient verified
- Full factorisation stated
A computer algebra result claims that x − 4 is a factor of P(x) = x³ − 5x² + 2x + 8. Verify the claim without performing full division.
Give me a hint
The remainder theorem reduces the check to one value.
Compare my reasoning
- Evaluate P(4) = 64 − 80 + 8 + 8.
- The result is 0.
- A zero remainder verifies divisibility by x − 4.
The claim is correct because P(4) = 0.
Look for these in your work
- Substitution done
- Value checked
- Theorem applied
Common mistakes
- Using x + a when the root is a
- Accepting a proposed root without evaluating P(a)
- Forgetting to check the quotient by multiplication
What can you explain now?
If P(−2) = 0, which linear factor must divide P(x)?
Compare with the explanation
x + 2
The factor is x − a with a = −2.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
State the remainder theorem and factor theorem in one sentence each.
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.