Junkao Mathematics - independent shared foundations study materials
Junkao Mathematics - independent shared foundations study resources

Junkao Mathematics - independent shared foundations: Algebra and functions

Junkao Mathematics - independent shared foundations · Work through your exam, one topic at a time.

All guide sections

Algebra and functions

Algebra and functions tests manipulation of polynomials and rational expressions, and the behaviour of functions as objects: domain, range, inverse, composition and graph transformations. Expect factor and remainder theorem questions, partial fractions, modulus equations and inequalities, and sketching transformed curves.

Polynomials and the factor theorem

The remainder when f(x) is divided by (x − a) is f(a). If f(a) = 0 then (x − a) is a factor. For a divisor (ax − b) the remainder is f(b/a). After extracting one factor, divide to obtain a quadratic, then factorise or use the discriminant.

Functions, inverses and graphs

Domain is the set of permitted inputs; range is the set of outputs. An inverse f⁻¹ exists only where f is one-to-one, and its domain equals the range of f. The graph of f⁻¹ is the reflection of y = f(x) in the line y = x. For transformations: y = f(x + a) shifts left by a, y = af(x) stretches vertically by a, for a ≠ 0, y = f(ax) scales horizontal distances by 1/|a|, with a reflection if a < 0, and y = |f(x)| reflects any negative portion above the axis.

Partial fractions

A proper rational function with distinct linear factors splits as A/(x − a) + B/(x − b). A repeated factor needs A/(x − a) + B/(x − a)². An irreducible quadratic factor needs a linear numerator.

Worked example

Given f(x) = 2x³ − 5x² − 4x + 3, show that (x − 3) is a factor and factorise fully.

Evaluate f(3) = 2(27) − 5(9) − 4(3) + 3 = 54 − 45 − 12 + 3 = 0, so (x − 3) is a factor. Dividing gives 2x² + x − 1, which factorises as (2x − 1)(x + 1). Hence f(x) = (x − 3)(2x − 1)(x + 1), with roots 3, 1/2 and −1.

Traps that cost marks

  • Sign errors when dividing or when substituting a negative value into f(a).
  • Writing the inverse of a function without restricting its domain to make it one-to-one.
  • Misreading y = f(x + a) as a shift right.
  • Forgetting to check whether a rational function is proper before splitting into partial fractions.
  • Losing the second solution when solving modulus equations such as |x − 2| = 5.