IGCSE Additional Mathematics 0606 · Topic 5

IGCSE Additional Mathematics: Factors of Polynomials Practice Questions

The remainder theorem states that dividing f(x) by (x minus a) leaves a remainder of f(a). The factor theorem is the special case where f(a) equals zero, meaning (x minus a) is a factor.

Cambridge IGCSE Additional Mathematics (0606) · Topic 5: Factors of Polynomials

Topic 5 of Cambridge IGCSE Additional Mathematics 0606 is highly mechanical once the two theorems are learned. The step most often skipped is stating that the remainder is zero before concluding a factor. The questions below cover remainders, factors, unknown coefficients and full cubic solutions.

What you need to know for Factors of Polynomials

IGCSE Additional Mathematics Factors of Polynomials questions and answers

4 exam-style questions written to the 0606 syllabus. Try each one on paper first, then open the worked answer to check your method against the marks.

Question 1[3 marks]

Find the remainder when f(x) = x3 minus 2x2 + 3x minus 5 is divided by (x minus 2).

Show the worked answer
Answer: 1
  1. By the remainder theorem, the remainder equals f(2).
  2. f(2) = 23 minus 2 x 22 + 3 x 2 minus 5.
  3. = 8 minus 8 + 6 minus 5 = 1. The remainder is 1, so (x minus 2) is not a factor.
How the marks are awarded. 1 mark for using the remainder theorem with x = 2. 1 mark for correct substitution. 1 mark for a remainder of 1.
Where students lose the mark. Substituting x = minus 2. The divisor (x minus 2) is zero when x = 2, so that is the value to use.
Question 2[5 marks]

Show that (x minus 1) is a factor of f(x) = x3 + 2x2 minus 5x + 2, and factorise f(x) as far as possible.

Show the worked answer
Answer: f(x) = (x minus 1)(x2 + 3x minus 2)
  1. Evaluate f(1) = 1 + 2 minus 5 + 2 = 0.
  2. Since f(1) = 0, by the factor theorem (x minus 1) is a factor.
  3. Divide f(x) by (x minus 1), by long division or by comparing coefficients, giving x2 + 3x minus 2.
  4. So f(x) = (x minus 1)(x2 + 3x minus 2).
  5. The quadratic does not factorise over the integers, since its discriminant is 9 + 8 = 17, which is not a perfect square. This is the fully factorised form.
How the marks are awarded. 1 mark for evaluating f(1). 1 mark for stating f(1) = 0 so (x minus 1) is a factor. 1 mark for a correct division method. 1 mark for the quadratic factor. 1 mark for stating the quadratic does not factorise further.
Where students lose the mark. Stopping after showing f(1) = 0. The question also asks for the factorisation, which requires the division.
Question 3[4 marks]

Given that (x + 2) is a factor of f(x) = x3 + ax2 + x + 6, find the value of a.

Show the worked answer
Answer: a = 1
  1. If (x + 2) is a factor then f(minus 2) = 0.
  2. f(minus 2) = (minus 2)3 + a(minus 2)2 + (minus 2) + 6.
  3. = minus 8 + 4a minus 2 + 6 = 4a minus 4.
  4. Set equal to zero: 4a minus 4 = 0, so a = 1.
How the marks are awarded. 1 mark for using f(minus 2) = 0. 1 mark for correct substitution with brackets. 1 mark for the expression 4a minus 4. 1 mark for a = 1.
Where students lose the mark. Substituting x = 2. The factor (x + 2) is zero when x = minus 2, so the sign reverses.
Question 4[5 marks]

Solve the equation x3 minus 6x2 + 11x minus 6 = 0.

Show the worked answer
Answer: x = 1, x = 2 or x = 3
  1. Try factors of the constant term 6. f(1) = 1 minus 6 + 11 minus 6 = 0, so (x minus 1) is a factor.
  2. Divide to obtain the quadratic factor: x2 minus 5x + 6.
  3. So the equation becomes (x minus 1)(x2 minus 5x + 6) = 0.
  4. Factorise the quadratic: (x minus 2)(x minus 3).
  5. The full factorisation is (x minus 1)(x minus 2)(x minus 3) = 0, giving x = 1, 2 or 3.
How the marks are awarded. 1 mark for finding a root by trial. 1 mark for identifying the linear factor. 1 mark for a correct quadratic factor. 1 mark for factorising the quadratic. 1 mark for all three solutions.
Where students lose the mark. Trying random values rather than the factors of the constant term. Any integer root must divide the constant, so only 1, 2, 3 and 6 and their negatives need testing.

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Factors of Polynomials FAQs

What is the remainder theorem?

When a polynomial f(x) is divided by (x minus a), the remainder equals f(a). For a divisor of the form (bx minus a), substitute x equals a divided by b. Setting the divisor equal to zero always tells you which value to use.

What is the factor theorem?

If f(a) equals zero then (x minus a) is a factor of f(x), and if (x minus a) is a factor then f(a) equals zero. It is the special case of the remainder theorem where the remainder is zero, and it is the standard way to start factorising a cubic.

How do I factorise a cubic?

Test the factors of the constant term, both positive and negative, until one gives zero. That identifies a linear factor. Divide the cubic by it, using long division or comparison of coefficients, then factorise the resulting quadratic if possible.

How do I find an unknown coefficient?

Substitute the value that makes the divisor zero into the polynomial, set the result equal to the given remainder, or to zero if it is stated to be a factor, then solve the resulting equation for the unknown.

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Related IGCSE Additional Mathematics topics

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Written to the published Cambridge IGCSE Additional Mathematics (0606) syllabus. Check your school entry code and syllabus year, because Core and Extended candidates are assessed on different content. Last reviewed 2026-08-12.