Algebra: Partial Fractions, Permutations, Combinations and Binomial Theorem
What the exam tests
This domain mixes four algebraic tools: resolving proper rational expressions into partial fractions, counting arrangements and selections, and expanding binomials. Exam questions are short but exacting, and one wrong factorial or sign can lose the whole mark.
Core rules
- Partial fractions require degree(numerator) < degree(denominator). If not, divide first.
- Distinct linear factors: A/(x−a) + B/(x−b). A repeated factor (x−a)² needs A/(x−a) + B/(x−a)². An irreducible quadratic needs (Ax+B)/(ax²+bx+c).
- nPr = n!/(n−r)!; nCr = n!/[r!(n−r)!]; nPr = r! · nCr; nCr = nC(n−r).
- Binomial theorem: (a+b)ⁿ = Σ nCr a^(n−r)b^r. General term T_(r+1) = nCr a^(n−r)b^r.
Worked example
Resolve (x+3)/[(x−1)(x+2)]. Write (x+3)/[(x−1)(x+2)] = A/(x−1) + B/(x+2). Then x+3 = A(x+2) + B(x−1). Put x = 1: 4 = 3A, so A = 4/3. Put x = −2: 1 = −3B, so B = −1/3. Hence the expression equals (4/3)/(x−1) − (1/3)/(x+2).
Binomial check: the coefficient of x⁵ in (2x+3)⁸ uses r = 5: 8C5 · 2⁵ · 3³ = 56 × 32 × 27 = 48,384.
Traps
- Forgetting to divide when the numerator's degree is not lower.
- Giving only one constant for a repeated factor.
- Treating nCr and nPr as interchangeable; order matters only in permutations.
- Starting the binomial expansion at r = 1 instead of r = 0.