Algebra
The Algebra domain of CAPE Pure Mathematics Unit 1 tests whether you can manipulate algebraic structure fluently: polynomials and their factors, surds and indices, logarithms, inequalities including modulus, and finite and infinite series. Questions are usually multi-step, so one sign slip early costs the whole item.
Polynomials: remainder and factor theorems
When a polynomial f(x) is divided by (x − a) the remainder is f(a); dividing by (ax − b) gives remainder f(b/a). The factor theorem is the special case f(a) = 0. For a cubic, find one root by testing small integers, divide to obtain a quadratic, then factorise or use the discriminant.
Surds, indices and logarithms
Rationalise denominators using the conjugate, e.g. 1/(√5 − 2) = (√5 + 2)/((√5)² − 2²) = √5 + 2. Log laws: log(ab) = log a + log b, log(a/b) = log a − log b, log aⁿ = n log a, and change of base log_a b = log b / log a. To solve aˣ = b, take logs of both sides.
Inequalities and modulus
For b > 0, |x − a| < b means a − b < x < a + b, while |x − a| > b splits into two rays. Rational inequalities are solved with a sign table built from the zeros of numerator and denominator, remembering the denominator cannot be zero.
Series
AP: Tₙ = a + (n − 1)d, Sₙ = (n/2)(2a + (n − 1)d). GP: Tₙ = arⁿ⁻¹, Sₙ = a(1 − rⁿ)/(1 − r), and S∞ = a/(1 − r) only when |r| < 1.
Worked example
Factorise f(x) = 2x³ − 5x² − 4x + 3 completely. Test x = 3: f(3) = 54 − 45 − 12 + 3 = 0, so (x − 3) is a factor. Synthetic division on coefficients 2, −5, −4, 3 with root 3 gives 2, 1, −1, 0, so the quotient is 2x² + x − 1. Factorising: 2x² + x − 1 = (2x − 1)(x + 1). Hence f(x) = (x − 3)(2x − 1)(x + 1).
Traps
- Sign errors when substituting a negative value into f(a).
- Accepting solutions that make a log argument non-positive or a denominator zero.
- Using S∞ when |r| ≥ 1.
- Squaring a modulus inequality without checking for extraneous roots.