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O-Level Polynomials and partial fractions
What the O-Level syllabus expects for Polynomials and partial fractions, and how to practise it.
What the syllabus expects
- Multiplying polynomials together and dividing one by another
- Applying the remainder theorem and factor theorem, which covers factorising polynomials and solving cubic equations
- Applying the identities a^3 + b^3 = (a + b)(a^2 - ab + b^2) and a^3 - b^3 = (a - b)(a^2 + ab + b^2)
- Splitting into partial fractions where the denominator is at most: (ax + b)(cx + d), or (ax + b)(cx + d)^2, or (ax + b)(x^2 + c^2)
Scope: The denominators are limited to no more than these three forms
How it's examined
Questions on this topic most often ask you to compare, find, solve, express. About 7% of the past-paper style questions in Rae's bank for this subject sit in this topic.
Worked examples
Example 1 (5 marks)
Hence resolve (4x³ + 5x² + x − 1)/(x²(x + 1)) into partial fractions.
Show the worked answer
Numerator degree 3 = denominator degree 3, so first divide out. Denominator x²(x+1) = x³ + x². 4x³ + 5x² + x - 1 = 4(x³ + x²) + (x² + x - 1), so expression = 4 + (x² + x - 1)/(x²(x+1)). Let (x² + x - 1)/(x²(x+1)) = A/x + B/x² + C/(x+1). Then x² + x - 1 = A x(x+1) + B(x+1) + C x². x = 0: -1 = B → B = -1. x = -1: 1 - 1 - 1 = C → C = -1. Compare x² coefficients: 1 = A + C → A = 2. Hence expression = 4 + 2/x - 1/x² - 1/(x+1).
Example 2 (5 marks)
The polynomial f(x) = x^4 - px^3 + 7x^2 + x - q is divisible by the quadratic x^2 - 2x - 3. Prove that p = 5 and q = 12.
Show the worked answer
f(x) = x⁴ − px³ + 7x² + x − q is divisible by x² − 2x − 3 = (x − 3)(x + 1), so f(3) = 0 and f(−1) = 0. f(3) = 81 − 27p + 63 + 3 − q = 0 ⟹ 147 − 27p − q = 0 ⟹ 27p + q = 147 ... (1) f(−1) = 1 + p + 7 − 1 − q = 0 ⟹ 7 + p − q = 0 ⟹ q = p + 7 ... (2) Substitute (2) into (1): 27p + (p + 7) = 147 ⟹ 28p = 140 ⟹ p = 5. Then q = 5 + 7 = 12. Hence p = 5 and q = 12 (proved).
Example 3 (3 marks)
For the same f(x) = x^4 - px^3 + 7x^2 + x - q with quadratic factor x^2 - 2x - 3 and p = 5, q = 12, solve the equation f(x) = 0.
Show the worked answer
With p=5, q=12: f(x) = x⁴ - 5x³ + 7x² + x - 12. Since x² - 2x - 3 is a factor, write f(x) = (x² - 2x - 3)(x² + bx + c). Comparing constant terms: -3c = -12 => c = 4. Comparing x³: b - 2 = -5 => b = -3. (Check x²: c - 2b - 3 = 4 + 6 - 3 = 7; x¹: -2c - 3b = -8 + 9 = 1.) So f(x) = (x² - 2x - 3)(x² - 3x + 4) = (x - 3)(x + 1)(x² - 3x + 4). x² - 3x + 4 has discriminant 9 - 16 = -7 < 0, so no real roots. Hence the real solutions are x = 3 and x = -1.
More worked questions on this topic
- Resolve (3x² + 5x − 2)/[(x+1)(x²+4)] into partial fractions. (5 marks)
- Explain why f(x) = 3x³ + 2x² + 16 = 0 has only one real root. [5] (5 marks)
- Solve 3x³ − 11x² + 3x + 2 = 0, giving any non-integer roots in surd form. (5 marks)
- Hence solve 4x³ + ax + b = 0, writing the non-integer roots as (c + √d)/2. (4 marks)
More O-Level Additional Maths (A-Maths) topics
Quadratic functions · Equations and inequalities · Surds · Binomial expansions · Exponential and logarithmic functions · Trigonometric functions, identities and equations · all of O-Level Additional Maths (A-Maths)