Topic 2 is the engine room of 9MA0. This mini-lesson covers surds and indices, the discriminant, the factor and remainder theorems, partial fractions, the modulus function, composite and inverse functions with their domains and ranges, and the four graph transformations.
Work through each screen, answer the questions as you go (some are wordy, most are calculations) and collect ⭐ stars. Watch for the Exam trap notes — that is where the marks get thrown away. Press Start when you are ready.
For ax² + bx + c = 0 the discriminant is Δ = b² − 4ac. It tells you how the parabola meets the x-axis.
Find k such that x² + kx + 9 = 0 has equal roots, k > 0.
Equal roots ⇒ b² − 4ac = 0 ⇒ k² − 4(1)(9) = 0 ⇒ k² = 36 ⇒ k = ±6, so k = 6.
Check: x² + 6x + 9 = (x + 3)², repeated root x = −3. ✓
Exam trap: “the curve meets the axis” means Δ ≥ 0; “the line is a tangent” means Δ = 0; “the curve is always positive” means a > 0 and Δ < 0. Read the wording, then set up the correct inequality.
Δ = k² − 4(1)(9) = k² − 36. Equal roots ⇒ Δ = 0 ⇒ k² = 36 ⇒ k = ±6.
k > 0 so k = 6, giving (x + 3)² = 0. ✓
For a polynomial f(x):
For a divisor (bx − a), the root is x = a/b, so the remainder is f(a/b).
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.
Exam trap: the remainder on dividing by (x − 2) is f(2), not f(−2). Sign errors here are the single most common lost mark in this topic.
f(2) = 2(2)³ − 5(2)² − 4(2) + 3 = 16 − 20 − 8 + 3 = −9.
So the remainder is −9 (and (x − 2) is therefore not a factor).
Split a proper algebraic fraction into simpler pieces. 9MA0 requires denominators with up to three distinct linear factors, or a repeated linear factor.
9x − 5 ≡ A(x + 3) + B(x − 1).
Let x = 1: 9 − 5 = 4 = A(4) ⇒ A = 1.
Let x = −3: −27 − 5 = −32 = B(−4) ⇒ B = 8.
So (9x − 5)/[(x − 1)(x + 3)] ≡ 1/(x − 1) + 8/(x + 3).
Check at x = 5: LHS = 40/(4 × 8) = 1.25; RHS = 1/4 + 8/8 = 0.25 + 1 = 1.25 ✓
Why it matters: partial fractions are the gateway to integrating rational functions (Topic 8) and to binomial expansions of them (Topic 4). If the fraction is improper (degree of top ≥ degree of bottom), divide first.
Substituting x = −3 kills the A term: 9(−3) − 5 = B(−3 − 1)
−32 = −4B ⇒ B = 8. (And x = 1 gives A = 1.)
|x| is the distance from zero: |x| = x if x ≥ 0, and |x| = −x if x < 0. The graph of y = |f(x)| reflects anything below the x-axis up above it.
Positive branch: 2x − 5 = x + 1 ⇒ x = 6. Check: |7| = 7 and 6 + 1 = 7 ✓
Negative branch: 2x − 5 = −(x + 1) ⇒ 3x = 4 ⇒ x = 4/3. Check: |8/3 − 5| = 7/3 and 4/3 + 1 = 7/3 ✓
Exam trap: always check each root back in the original equation — the negative branch can throw up solutions that do not actually work. And |a| = |b| ⇒ a = ±b, so squaring both sides is legitimate but doubles the work.
2x − 5 = x + 1 ⇒ x = 6 (check: |12 − 5| = 7 = 6 + 1 ✓)
2x − 5 = −(x + 1) ⇒ 3x = 4 ⇒ x = 4/3 (check: |8/3 − 5| = 7/3 = 4/3 + 1 ✓)
Larger solution: x = 6.
A function maps each element of the domain to exactly one element of the range.
fg(2) = f(g(2)) = f(2² + 1) = f(5) = 3(5) − 4 = 11.
gf(2) = g(f(2)): here f(2) = 3(2) − 4 = 2, so gf(2) = g(2) = 2² + 1 = 5. Different answer — order matters.
f⁻¹: let y = 3x − 4 ⇒ x = (y + 4)/3, so f⁻¹(x) = (x + 4)/3. Check: f⁻¹(11) = (11 + 4)/3 = 5, which is exactly the g(2) that f turned into 11. ✓
Exam trap: f(x) = x² on ℝ has no inverse — it is many-to-one (both 3 and −3 map to 9). Restrict the domain to x ≥ 0 and it becomes one-to-one, so f⁻¹(x) = √x exists.
g(2) = 2² + 1 = 5.
f(5) = 3(5) − 4 = 11.
(Watch the order: gf(2) = g(2) = 5, which is different.)
Tap a transformation of y = f(x), then the family it belongs to.
Changes inside the bracket affect x and do the opposite of what you expect; changes outside affect y and behave normally.
Exam trap: f(2x) squashes the graph towards the y-axis (factor ½), it does not stretch it. And when combining, apply the inside changes in the order that undoes the bracket.
Tap a card on the left, then the matching result.
These are assumed knowledge but examined constantly inside longer questions.
Exam trap: √(a + b) ≠ √a + √b. Test it: √(9 + 16) = 5, but 3 + 4 = 7.
Discriminant: b² − 4ac: > 0 two roots · = 0 equal roots (tangent) · < 0 none
Factor / remainder: f(a) = 0 ⇔ (x − a) is a factor; remainder on ÷(x − a) is f(a)
Partial fractions: substitute the roots; a repeated factor needs A/(x−a) AND B/(x−a)²
Modulus: |f(x)| reflects the negative part upwards; solve both branches and check
Functions: fg means g first; f⁻¹ exists only for one-to-one f; domain and range swap
Transformations: inside the bracket = opposite, outside = as expected
Topic 2 underpins the whole qualification — every calculus question is an algebra question in disguise. Press Finish to see your score.
You have worked through Algebra and functions for Edexcel A-level Mathematics (9MA0). 🎉
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Next: test yourself in the Verify stage.