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Polynomials: notes and questions
Revision for the factor and remainder theorems, algebraic division and solving cubic equations. Both boards.
Cambridge 0606: section 3 Edexcel 4PM1: section 3A–C 14 questions Key points
Remainder theorem: when f(x) is divided by (x − a), the remainder is f(a). For (ax − b), it is f(b/a). Factor theorem: if f(a) = 0, then (x − a) is a factor of f(x). To solve a cubic, find one root by trying factors of the constant term, divide to get a quadratic factor, then factorise or use the formula. Two unknown coefficients need two equations: one from each piece of information about a factor or remainder. Check your factorisation by expanding it back, or by substituting each root. Watch out: Dividing by (2x − 1) uses x = ½, not x = 1 or x = 2. Set the divisor equal to zero to find the value to substitute.
Worked example · 4 marks
\(f(x) = x^3 - 4x^2 + x + 6\)
(a) Show that \((x - 2)\) is a factor of \(f(x)\).
(b) Hence factorise \(f(x)\) completely.
M1 (a) f(2) evaluatedA1 (a) f(2) = 0 so (x − 2) is a factorM1 (b) quadratic factor x² − 2x − 3A1 (b) (x − 2)(x − 3)(x + 1)f(2) = 8 − 16 + 2 + 6 = 0, so by the factor theorem (x − 2) is a factor. Divide: x³ − 4x² + x + 6 = (x − 2)(x² − 2x − 3). x² − 2x − 3 = (x − 3)(x + 1), so f(x) = (x − 2)(x − 3)(x + 1). Answer: (b) \((x - 2)(x - 3)(x + 1)\)
Polynomials questions
Original exam-style questions, written for this site and each re-solved independently. Tags show the courses whose content a question tests and whether it can be done without a calculator (Cambridge 0606 Paper 1 is non-calculator).
Question 1 · 2 marks 0606 4PM1 no calculator
Use the remainder theorem to find the remainder when \(3x^3 - 2x^2 + x + 7\) is divided by \((x + 2)\).
Show the mark scheme and worked solution
M1 substitute x = −2A1 −27By the remainder theorem, the remainder is f(−2). f(−2) = 3(−8) − 2(4) + (−2) + 7 = −24 − 8 − 2 + 7 = −27. Answer: \(-27\)
Question 2 · 4 marks 0606 4PM1 no calculator
\(p(x) = x^3 + ax^2 + bx - 6\), where \(a\) and \(b\) are constants.
\((x - 1)\) is a factor of \(p(x)\), and when \(p(x)\) is divided by \((x + 1)\) the remainder is \(-8\).
Find the value of \(a\) and the value of \(b\).
Show the mark scheme and worked solution
M1 p(1) = 0: a + b = 5M1 p(−1) = −8: a − b = −1A1 a = 2A1 b = 3p(1) = 1 + a + b − 6 = 0, so a + b = 5. p(−1) = −1 + a − b − 6 = −8, so a − b = −1. Add: 2a = 4, so a = 2 and b = 3. Answer: \(a = 2\), \(b = 3\)
Question 3 · 2 marks 0606 4PM1 no calculator
Given that \((x - 2)\) is a factor of \(6x^3 - 11x^2 - 3x + 2\), factorise the expression completely.
Show the mark scheme and worked solution
M1 quadratic factor 6x² + x − 1A1 (x − 2)(2x + 1)(3x − 1)Divide by (x − 2): 6x³ − 11x² − 3x + 2 = (x − 2)(6x² + x − 1). 6x² + x − 1 = (2x + 1)(3x − 1). So the expression is (x − 2)(2x + 1)(3x − 1). Answer: \((x - 2)(2x + 1)(3x - 1)\)
Question 4 · 3 marks 4PM1 no calculator
Divide \(x^3 + 4x^2 - 3x + 2\) by \((x + 3)\), stating the quotient and the remainder.
Show the mark scheme and worked solution
M1 algebraic division startedA1 quotient x² + x − 6A1 remainder 20x³ ÷ x = x²; x²(x + 3) = x³ + 3x², leaving x² − 3x + 2. x² ÷ x = x; x(x + 3) = x² + 3x, leaving −6x + 2. −6x ÷ x = −6; −6(x + 3) = −6x − 18, leaving 20. Quotient x² + x − 6, remainder 20 (check: f(−3) = −27 + 36 + 9 + 2 = 20). Answer: quotient \(x^2 + x - 6\), remainder \(20\)
9 more polynomials questions
Each with a mark scheme and a worked solution. Included with IGCSE plans, free trials and school licences.
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