Fluency topics: expanding, factorising (including \(ax^2 + bx + c\)), rearranging formulae where the subject appears twice, simplifying algebraic fractions, identities versus equations, and \(n\)th terms of linear and quadratic sequences. Kinematics graphs (area under speed-time = distance) are also here.
Contents: Manipulation (M4.1-M4.8)Sequences and graphs in context (M4.18, M4.19)Must-knowTrapsReal questions
Manipulation (M4.1-M4.8)
- Rearranging when the subject appears twice: collect the subject terms on one side and factorise: \(y(x - 1) = 2x + 3 \Rightarrow x(y - 2) = y + 3\).
- Algebraic fractions: factorise numerator and denominator fully, cancel common factors; add with a common denominator.
- Identity (\(\equiv\)): true for all \(x\), e.g. \((x + 3)^2 - (x - 3)^2 \equiv 12x\). Equation: true for particular \(x\).
- Factorise \(6x^2 - 7x - 20\): find two numbers with product \(-120\) and sum \(-7\) (\(-15, 8\)), split the middle term.
- Substitution into formulae: \(v^2 = u^2 + 2as\) with care over signs and squares.
Sequences and graphs in context (M4.18, M4.19)
- Linear sequence: \(n\)th term \(= dn + (a - d)\). Quadratic sequence: second difference \(2a\); fit \(an^2 + bn + c\) from the first terms.
- Term-to-term rules can be constant or cyclic: \(x \to x^2 - 2\) from \(2\) stays at \(2\).
- Speed-time graph: area = distance; gradient = acceleration. Distance-time graph: gradient = speed; average speed = total distance \(\div\) total time (not displacement).
Must-know (no formula booklet)
- \(n\)th term of a quadratic sequence: second difference \(= 2a\)
- area under \(v\)-\(t\) = distance
- identity: true for all \(x\)
Classic traps
- Cancelling: only common factors cancel, never terms (\(\frac{x^2 - 9}{x - 3} = x + 3\) but \(\frac{x + 9}{x + 3}\) does not simplify).
- \((n + 1)^2 - 1 = n^2 + 2n\): two option forms can be the same expression.
- 'For how long is the ball above 15 m' is the length of the interval between the two times.
Worked example
Q. \(n\)th term of \(3, 8, 15, 24, 35\)?
A. Differences \(5, 7, 9, 11\); second difference \(2\), so \(n^2 + bn + c\); \(n = 1, 2\) give \(b = 2\), \(c = 0\): \(n^2 + 2n\).