Mini-Lesson

Algebra and functions

This mini-lesson covers the CCEA algebra spine: indices and surds, quadratics (completing the square, the discriminant), polynomials (factor and remainder theorems), partial fractions, the modulus function, composite and inverse functions and graph transformations.

Where this sits: Unit AS 1: Pure Mathematics (indices, surds, quadratics, simultaneous equations, inequalities, polynomials, graphs) extended in Unit A2 1: Pure Mathematics (functions, modulus, partial fractions).

Work through each screen, answer the questions as you go (some are reasoning, some are calculations) and collect ⭐ stars. Press Start when you are ready.

Foundations

Indices and surds

aᵐ × aⁿ = aᵐ⁺ⁿ · aᵐ ÷ aⁿ = aᵐ⁻ⁿ · (aᵐ)ⁿ = aᵐⁿa⁰ = 1 · a⁻ⁿ = 1/aⁿ · a^(1/n) = ⁿ√a · a^(m/n) = (ⁿ√a)ᵐ

Surds. Simplify by pulling out square factors: √50 = √25 × √2 = 5√2. Rationalise a denominator by multiplying by the conjugate:

Worked example

6/(3 − √3) = 6(3 + √3) / ((3 − √3)(3 + √3)) = 6(3 + √3) / (9 − 3) = 6(3 + √3)/6 = 3 + √3

Common slip: √(a + b) ≠ √a + √b. Check: √(9 + 16) = 5, but √9 + √16 = 7.

Quadratics

Completing the square

Writing y = ax² + bx + c as a(x + p)² + q instantly gives the vertex (−p, q), the line of symmetry x = −p and the maximum/minimum value q.

Worked example — 2x² − 8x + 3

Take out the 2 from the x terms: 2(x² − 4x) + 3

Complete the square inside: x² − 4x = (x − 2)² − 4

So 2[(x − 2)² − 4] + 3 = 2(x − 2)² − 8 + 3 = 2(x − 2)² − 5

Minimum value −5 at x = 2, since 2(x − 2)² ≥ 0 with equality at x = 2.

Calculate

Complete the square

1Write y = 2x² − 8x + 3 in the form a(x + p)² + q. State the minimum value of y.

Hint: 2(x − 2)² − 5. The square is never negative, so the least value occurs when x = 2, giving y = −5.

Quadratics

The discriminant

b² − 4ac> 0 → two distinct real roots · = 0 → one repeated root · < 0 → no real roots

Worked example — equal roots

kx² + (k + 3)x + 4 = 0 has equal roots, so b² − 4ac = 0:

(k + 3)² − 16k = 0 ⇒ k² + 6k + 9 − 16k = 0 ⇒ k² − 10k + 9 = 0

(k − 1)(k − 9) = 0 ⇒ k = 1 or k = 9

Watch the trap: if a question says "the equation has two distinct roots" you need b² − 4ac > 0 and a ≠ 0 — otherwise it is not a quadratic at all.

Calculate

Discriminant condition

2The equation kx² + (k + 3)x + 4 = 0 has equal roots. Find the largest possible value of k.

k

Hint: b² − 4ac = 0 gives (k + 3)² − 16k = k² − 10k + 9 = 0, so (k − 1)(k − 9) = 0 and k = 1 or 9. The largest is 9.

Quick check

Read the discriminant

?How many real roots does 3x² − 4x + 5 = 0 have?

Polynomials

Factor and remainder theorems

f(a) = 0 ⇔ (x − a) is a factor of f(x)Remainder theorem: dividing f(x) by (x − a) leaves remainder f(a)

Worked example — f(x) = 2x³ − 5x² − 4x + 3

f(3) = 2(27) − 5(9) − 12 + 3 = 54 − 45 − 12 + 3 = 0, so (x − 3) is a factor.

Divide: 2x³ − 5x² − 4x + 3 = (x − 3)(2x² + x − 1) = (x − 3)(2x − 1)(x + 1)

Roots: x = 3, x = ½, x = −1.

For a factor (ax − b) test f(b/a). Here f(½) = 2(⅛) − 5(¼) − 2 + 3 = 0.25 − 1.25 + 1 = 0 ✓.

Calculate

Remainder theorem

3f(x) = 2x³ − 5x² − 4x + 3. Find the remainder when f(x) is divided by (x − 2).

Hint: Remainder = f(2) = 2(8) − 5(4) − 4(2) + 3 = 16 − 20 − 8 + 3.

A2 skill

Partial fractions

Splitting an algebraic fraction into simpler pieces is essential for integration and for binomial expansions later.

Worked example

(7x − 4)/((x − 2)(2x + 1)) ≡ A/(x − 2) + B/(2x + 1)

Multiply through: 7x − 4 ≡ A(2x + 1) + B(x − 2)

Let x = 2: 10 = 5A ⇒ A = 2

Let x = −½: −7.5 = −2.5B ⇒ B = 3

Check: 2(2x + 1) + 3(x − 2) = 4x + 2 + 3x − 6 = 7x − 4 ✓

Repeated factor (x − a)²: you need A/(x − a) + B/(x − a)². Improper fraction (top degree ≥ bottom degree): divide first.

Calculate

Find the constant

4Express (7x − 4)/((x − 2)(2x + 1)) as A/(x − 2) + B/(2x + 1). Find the value of B.

B

Hint: 7x − 4 ≡ A(2x + 1) + B(x − 2). Substituting x = −½ kills A: −3.5 − 4 = B(−2.5), so −7.5 = −2.5B.

Functions

Composite and inverse functions

Composite fg(x) = f(g(x)) — apply g first. Order matters: fg ≠ gf in general.
Inverse f⁻¹ undoes f. It exists only if f is one-to-one; swap x and y and make y the subject.
The graph of y = f⁻¹(x) is the reflection of y = f(x) in the line y = x. Domain of f⁻¹ = range of f.

Worked example

f(x) = 3x − 2, g(x) = x² + 1.

fg(2) = f(g(2)) = f(5) = 3(5) − 2 = 13. But gf(2) = g(4) = 17 — different!

f⁻¹(x): y = 3x − 2 ⇒ x = (y + 2)/3 ⇒ f⁻¹(x) = (x + 2)/3

Calculate

Composite function

5f(x) = 3x − 2 and g(x) = x² + 1. Evaluate fg(2).

Hint: Apply g first: g(2) = 2² + 1 = 5. Then f(5) = 3(5) − 2 = 13.

Functions

The modulus function

|x| is the distance of x from 0, so it is never negative. To solve an equation with a modulus, split it into two cases — then check both answers in the original.

Worked example — |2x − 1| = 5

Case 1: 2x − 1 = 5 ⇒ x = 3

Case 2: 2x − 1 = −5 ⇒ x = −2

Both check out: |2(3) − 1| = 5 ✓ and |2(−2) − 1| = |−5| = 5 ✓

Graph: y = |f(x)| reflects the negative part of y = f(x) above the x-axis. y = f(|x|) keeps x ≥ 0 and reflects it in the y-axis. These are different graphs.

Sort it

Sort by discriminant

Tap a quadratic, then tap the number of real roots its discriminant gives.

✌️ Two distinct roots

☝️ One repeated root

🚫 No real roots

Quick check

Transformations

?The graph of y = f(x) is transformed to y = f(x − 3) + 2. Describe the transformation.

Quick check

Inverse functions

?f(x) = x² for x ∈ ℝ has no inverse. Why not?

Match it

Match the algebra

Tap an expression on the left, then its simplified equivalent on the right.

Statement

Answer

Quick check

Inequality care

?Solve x² > 4x. Which is the full solution set?

Examiner traps

Where algebra marks disappear

Dividing an inequality by x. x may be negative or zero. Rearrange to make one side 0 and factorise instead.
Discriminant without a ≠ 0. "Two distinct roots" needs both b² − 4ac > 0 and a ≠ 0.
fg vs gf. fg(x) means do g first. They are almost never equal.
Forgetting the domain of f⁻¹. Domain of f⁻¹ = range of f.
|x| equations. Solve both cases, then check both solutions in the original — one can be spurious.
Transformations inside the bracket act in reverse: f(x − 3) moves the curve right, not left.

Recap

The big ideas to know

Indices/surds: a^(m/n) = (ⁿ√a)ᵐ; rationalise with the conjugate

Quadratics: a(x + p)² + q gives the vertex; b² − 4ac decides the roots

Polynomials: f(a) = 0 ⇔ (x − a) is a factor; remainder = f(a)

Partial fractions: cover-up/substitution to find A and B — needed for A2 integration

Functions: fg means g first; f⁻¹ reflects in y = x; |x| splits into two cases

Transformations: f(x − a) → right a; f(x) + b → up b; af(x) → stretch ×a vertically

These techniques feed straight into coordinate geometry, calculus and the binomial expansion. Press Finish to see your score.

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