Lesson 3: Polynomials and Rational Functions

Quadratics generalise to polynomials — functions built from powers of x with constant coefficients — and pair naturally with rational functions, ratios of polynomials. In the AA guide this material (Topic 2) is examined through behaviour: roots and factors, end behaviour, asymptotes, and the shape of a graph. HL candidates meet the factor and remainder theorems head-on; SL candidates need the same graph sense for cubic and reciprocal-type curves. The unifying principle is that a polynomial's algebra (its factors) and its geometry (its intercepts and turning points) are two readings of one object.

Polynomial fundamentals

A polynomial of degree n has the form aₙxⁿ + … + a₁x + a₀, aₙ ≠ 0. Facts IB assumes: a degree-n polynomial has at most n real roots and at most n − 1 turning points; its end behaviour is governed by the leading term (odd degree with positive leading coefficient falls left and rises right; even degree rises at both ends if aₙ > 0). The remainder theorem says that dividing p(x) by (x − a) leaves remainder p(a); the factor theorem follows immediately: (x − a) is a factor exactly when p(a) = 0. These two theorems turn "find the roots" into "evaluate cleverly, then divide".

Worked Example 1 — remainder and factor theorems (past-paper style, HL). Let p(x) = x³ + kx² − 5x + 6. When p(x) is divided by (x − 2), the remainder is −4. Find k, and hence factorise p(x) completely.

Solution. By the remainder theorem, the remainder equals p(2) = 8 + 4k − 10 + 6 = 4k + 4. Setting 4k + 4 = −4 gives k = −2, so p(x) = x³ − 2x² − 5x + 6. Hunt for a factor using the integer candidates ±1, ±2, ±3, ±6 (factors of the constant term): p(1) = 1 − 2 − 5 + 6 = 0, so (x − 1) is a factor. Divide: x³ − 2x² − 5x + 6 = (x − 1)(x² − x − 6) = (x − 1)(x − 3)(x + 2). The roots are 1, 3 and −2. Internalise the rhythm: evaluate the candidates, divide once one works, factor the quadratic that remains. If every candidate fails on a Paper 1 question, suspect a sign slip — examiners design these to split.

Rational functions and asymptotes

For f(x) = p(x)/q(x), the vertical asymptotes stand where q(x) = 0 (and p(x) ≠ 0 there), and the horizontal asymptote comes from end behaviour: if deg p < deg q it is y = 0; if deg p = deg q it is the ratio of leading coefficients; if deg p > deg q there is none (possibly an oblique asymptote — HL). The AA course features f(x) = (ax + b)/(cx + d) most heavily: one vertical asymptote, one horizontal asymptote, and a graph that is a translated reciprocal curve. Intercepts come from p(x) = 0 (x-intercepts) and x = 0 (y-intercept), exactly as for polynomials.

Worked Example 2 — a full rational-function analysis. Let f(x) = (2x + 4)/(x − 1). State the asymptotes, the intercepts, and the domain and range.

Solution. Vertical asymptote: denominator zero at x = 1, so x = 1; domain x ≠ 1. Horizontal asymptote: degrees equal, ratio of leading coefficients 2/1, so y = 2; range y ≠ 2. x-intercept: numerator zero, x = −2. y-intercept: f(0) = −4. The graph has two branches hugging the asymptotes; on Paper 1 you sketch this from the analysis alone, on Paper 2 the GDC confirms — but note the GDC may draw a false near-vertical line at x = 1: that artefact is the calculator joining the branches, not part of the graph.

Worked Example 3 — solving a rational equation. Solve (x + 5)/(x − 2) = 3.

Solution. Multiply both sides by (x − 2), valid for x ≠ 2: x + 5 = 3(x − 2) = 3x − 6, so 11 = 2x and x = 5.5. Check legality: 5.5 ≠ 2, accepted. The habit that protects you: state the excluded value *before* clearing denominators, because clearing can manufacture a "solution" that was never legal — equations like (x² − 4)/(x − 2) = 4 tempt x = 2, which must be rejected.

Zeros, multiplicity and touching (HL flavour)

If (x − a)² divides p(x), the graph touches the x-axis at a and turns back; an odd multiplicity crosses, an even multiplicity touches. This is the fastest way to sketch a factorised cubic or quartic on Paper 1: plot the intercepts, mark each as cross or touch, and let the end behaviour join the dots.

Real-world modelling: cost and concentration

Rational functions model averages: if a journey of 120 km is driven at speed v, the time is 120/v — a reciprocal relationship with the vertical axis as asymptote (speeds near zero are meaningless, another contextual domain). Drug concentration after a dose often follows c(t) = 12t/(t² + 4): it rises, peaks, and decays towards the horizontal asymptote c = 0 — the model in the exam experience below. Average-cost functions in economics, A(x) = (fixed costs)/x + unit cost, have horizontal asymptotes equal to the unit cost, a genuinely interpretable limit.

Common mistakes

• Cancelling terms, not factors: (x + 2)/(x + 3) cannot be simplified; only common *factors* cancel.
• Horizontal asymptote from "the biggest term" alone when degrees are equal — the ratio of leading coefficients is required.
• Forgetting the excluded values when solving rational equations, or reporting an asymptote as part of the domain's complement incorrectly (domain excludes the vertical asymptote's x; range excludes the horizontal asymptote's y — candidates swap these).
• Remainder theorem sign slip: the remainder on division by (x + a) is p(−a), not p(a).
• GDC artefact line at a vertical asymptote drawn into a sketch.

IB Exam Experience

Question 1 [2 marks]. When x³ + 2x² + kx − 6 is divided by (x − 1), the remainder is 0. Find k. *Marking notes:* M1 for applying the remainder theorem, p(1) = 0; A1 for 1 + 2 + k − 6 = 0 giving k = 3.

Question 2 [3 marks]. Let f(x) = (4 − 2x)/(x + 3). Write down (a) the vertical asymptote, (b) the horizontal asymptote, (c) the x-intercept. *Marking notes:* A1 each: x = −3; y = −2 (ratio of leading coefficients −2/1 — the most-missed mark in this family); x = 2.

Question 3 [4 marks]. A cubic p(x) has roots −1, 2 and 4, and p(0) = 8. Write p(x) in the form a(x − r)(x − s)(x − t) and find a. *Marking notes:* M1 for p(x) = a(x + 1)(x − 2)(x − 4); M1 for p(0) = a(1)(−2)(−4) = 8a; A1 for a = 1; A1 for a correct expansion check p(x) = x³ − 5x² + 2x + 8 if requested. Sign discipline in the factors is the examiner's focus.

Extended problem (Paper 2 style) [9 marks]. The concentration of a drug in a patient's bloodstream is modelled by c(t) = 12t/(t² + 4) mg/L, t hours after the dose. (a) Find the concentration after 1 hour. [1] (b) Find the maximum concentration and when it occurs. [4] (c) The drug is effective while c(t) ≥ 2. Find the length of the effective window. [4] *Marking notes:* (a) A1: c(1) = 12/5 = 2.4 mg/L. (b) M1 for using the GDC maximum or calculus; A1 for t = 2 h; A1 for c(2) = 3 mg/L; A1 for units stated. (Paper 1 route: by AM-GM or by the derivative in Lesson 10 — c(t) ≤ 3 with equality at t = 2.) (c) M1 for solving 12t/(t² + 4) = 2, i.e. t² − 6t + 4 = 0 (GDC graph intersection acceptable); A1 for t = 3 ± √5 ≈ 0.764 and 5.236; M1 for the window length as the difference; A1: ≈ 4.47 hours. Examiner's note: candidates who report the two times but not their difference leave the final accuracy mark behind.

Exploration (IA) idea: measure how the average speed of a real commute depends on departure time, model segments with reciprocal/rational functions, and discuss asymptotes as physical limits — or model dilution in repeated water-tank exchanges (a classic rational-function IA with genuine iteration).

Going deeper: partial fractions (HL) and sign analysis

HL candidates meet partial fractions: splitting (ax + b)/((x − r)(x − s)) into A/(x − r) + B/(x − s) — the reverse of adding fractions, and the key that unlocks integrals in Lesson 23. At both levels, a sign chart for p(x)/q(x) solves rational inequalities: mark the zeros of numerator and denominator on a number line; the sign alternates at each simple zero.

Extra worked example. (a) Express (5x − 4)/((x − 1)(x + 2)) in partial fractions. (b) Hence solve (5x − 4)/((x − 1)(x + 2)) > 0.

Solution. (a) Write (5x − 4) = A(x + 2) + B(x − 1). Set x = 1: 1 = 3A → A = 1/3. Set x = −2: −14 = −3B → B = 14/3. So the split is (1/3)/(x − 1) + (14/3)/(x + 2). Verify at x = 0: the original gives −4/((−1)(2)) = 2; the split gives (1/3)(−1) + (14/3)(1/2) = −1/3 + 7/3 = 2 ✓. (b) The critical values are the numerator zero x = 4/5 and the denominator zeros x = −2 and x = 1 — all three split the line. Test one point per region: at −3: −19/4 (negative); at 0: 2 (positive); at 0.9: negative; at 2: 6/4 (positive). So the sign pattern is −, +, −, + across (−∞, −2), (−2, 4/5), (4/5, 1), (1, ∞), and the solution of > 0 is −2 < x < 4/5 or x > 1. Include the numerator's zeros in the chart — omitting x = 4/5 is the designed trap.

Extra exam practice [3 marks]. Solve (x − 3)/(x + 1) ≤ 0. *Marking notes:* M1 for critical values −1 and 3 with a sign chart or test points; A1 for the solution interval −1 < x ≤ 3; A1 for the endpoint discipline (x = −1 excluded — denominator; x = 3 included — numerator zero satisfies ≤).