Polynomial and Rational Functions

IB Mathematics: Analysis and Approaches HL· 6 min read

1. Polynomials: Definitions and Core Theorems★★☆☆☆⏱ 15 min

📘 Definition

Polynomial Function

P(x) = a_n x^n + a_{n-1}x^{n-1} + ... + a_1 x + a_0

A function where $n$ is a non-negative integer, $a_{i}$ are real constants, and $a_{n} \neq = 0$ . The leading term is $a_{n} x^{n}$ , and $n$ is the degree of the polynomial ( $de g (P)$ ).

Example:

$P (x) = 2 x^{3} - 5 x^{2} + 3 x - 7$ is a degree 3 polynomial.

📘 Definition

Remainder and Factor Theorems

When $P (x)$ is divided by $(x - c)$ , the remainder equals $P (c)$ . If $P (c) = 0$ , then $(x - c)$ is a factor of $P (x)$ (this is the Factor Theorem).

📐 Worked Example

Find the remainder when $P (x) = 3 x^{4} - 2 x^{3} + x^{2} - 5$ is divided by $(x - 2)$ .

1
By the Remainder Theorem, the remainder equals $P (2)$ .
2
Substitute $x = 2$ into $P (x)$ :
3
$P(2) = 3(2)^4 - 2(2)^3 + (2)^2 - 5$
4
$= 3(16) - 2(8) + 4 - 5 = 48 - 16 + 4 - 5 = 31$
5
The remainder is 31.

2. Factoring and Root Finding★★★☆☆⏱ 20 min

By the Fundamental Theorem of Algebra, a degree $n$ polynomial has exactly $n$ roots (counting multiplicities) over the complex numbers. For real polynomials, complex roots occur in conjugate pairs.

📘 Definition

Multiplicity of a Root

If $(x - c)^{k}$ is a factor of $P (x)$ but $(x - c)^{k + 1}$ is not, then $c$ is a root of multiplicity $k$ .

📐 Worked Example

Factorize $P (x) = x^{3} - x^{2} - 8 x + 12$ completely, given that $x = 2$ is a root of multiplicity 2.

1
Since $x = 2$ is multiplicity 2, $(x - 2)^{2}$ is a factor. $P (x)$ is degree 3, so the remaining factor is linear: $(a x + b)$ .
2
$P(x) = (x^2 - 4x + 4)(ax + b) = ax^3 + (b - 4a)x^2 + (4a - 4b)x + 4b$
3
Equate coefficients with the original polynomial:
4
Leading coefficient: $a = 1$ , constant term: $4 b = 12 \to b = 3$ . Check other coefficients: $b - 4 a = 3 - 4 = - 1$ (matches $x^{2}$ term), $4 a - 4 b = 4 - 12 = - 8$ (matches $x$ term).
5
Final factorization:
6
$P(x) = (x - 2)^2(x + 3)$

✓ Quick check

What is the sum of the roots of $2 x^{3} - 6 x^{2} + 4 x - 10 = 0$ ?
- -3
- 3
- -6
- 6
Reveal answer
1 —
By Vieta's formula, sum of roots for $a x^{3} + b x^{2} + c x + d = 0$ is $- b / a = - (- 6) /2 = 3$ .

3. Rational Functions: Asymptotes and Holes★★★☆☆⏱ 20 min

📘 Definition

Rational Function

R(x) = \frac{P(x)}{Q(x)}

A function written as the ratio of two polynomials $P (x)$ (numerator) and $Q (x)$ (denominator, not the zero polynomial). The domain excludes all roots of $Q (x)$ .

Vertical asymptotes: Occur at roots of the denominator that are not roots of the numerator. Common roots create holes, not asymptotes.
Horizontal asymptotes: If $de g (P) < de g (Q)$ , asymptote at $y = 0$ . If $de g (P) = de g (Q)$ , asymptote at $y = \frac{leading coefficient of P}{leading coefficient of Q}$ . No horizontal asymptote if $de g (P) > de g (Q)$ .
Oblique (slant) asymptotes: Occur when $de g (P) = de g (Q) + 1$ , found via polynomial long division.

📐 Worked Example

Find all asymptotes of $R (x) = \frac{2 x ^{2} - 2 x - 12}{x ^{2} - 4 x + 3}$ .

1
Factor numerator and denominator:
2
$R(x) = \frac{2(x-3)(x+2)}{(x-1)(x-3)}$
3
The common factor $(x - 3)$ means there is a hole at $x = 3$ , not a vertical asymptote. Simplify to $R (x) = \frac{2 ( x + 2 )}{x - 1}, x \neq = 3$ .
4
Vertical asymptote at the root of the simplified denominator: $x = 1$ .
5
Degrees of numerator and denominator are equal, so horizontal asymptote is $y = 2/1 = 2$ .
6
Final result: Vertical asymptote $x = 1$ , horizontal asymptote $y = 2$ , hole at $(3, 5)$ .

4. Sketching Graphs★★★★☆⏱ 25 min

Graph sketching questions require you to mark all key features: intercepts, roots (with multiplicity behavior), turning points, asymptotes, and end behavior. For polynomials, end behavior is determined by the leading term:

Even degree: Both ends go the same direction (up if leading coefficient positive, down if negative)
Odd degree: Ends go opposite directions (right end up if leading coefficient positive, right end down if negative)
Root behavior: Odd multiplicity = graph crosses x-axis; Even multiplicity = graph touches x-axis and turns

📐 Worked Example

Sketch the graph of $P (x) = x (x + 1) (x - 2)^{2}$ .

1
Degree is $1 + 1 + 2 = 4$ , leading coefficient positive, so both ends of the graph point upwards.
2
Roots: $x = - 1$ (multiplicity 1, crosses), $x = 0$ (multiplicity 1, crosses), $x = 2$ (multiplicity 2, touches).
3
Y-intercept at $P (0) = 0$ , so the graph passes through the origin.
4
Check sign of $P (x)$ in each interval: $x < - 1$ (positive), $- 1 < x < 0$ (negative), $0 < x < 2$ (negative), $x > 2$ (positive). Connect the points, showing crossing at $- 1$ and $0$ , touching at $2$ , with both ends pointing up.

Exam tip:

Sketching questions are usually 4-6 marks in IB exams, so allocate your time accordingly.

5. Common Pitfalls

Wrong move:

Calling a common root of numerator/denominator a vertical asymptote instead of a hole

Why:

IB examiners explicitly test the distinction between holes and asymptotes

Correct move:

Always factor both polynomials first, cancel common factors before identifying asymptotes, and note the position of any holes

Wrong move:

Using the constant term instead of the leading coefficient to find horizontal asymptotes when degrees are equal

Why:

Confusion between intercept formulas and asymptote rules

Correct move:

Remember: horizontal asymptotes depend on leading terms, not constant terms

Wrong move:

Forgetting that multiplicity changes how the graph behaves at a root

Why:

Focusing only on the location of the root, not its multiplicity

Correct move:

Memorize: odd multiplicity = cross the x-axis, even multiplicity = touch the x-axis

Wrong move:

Claiming an asymptote does not exist because the graph crosses it at a finite x-value

Why:

Misunderstanding that asymptotes describe end behavior, not no intersection

Correct move:

Only vertical asymptotes cannot be crossed; horizontal and oblique asymptotes can be crossed at finite x-values

Wrong move:

Assuming an odd-degree polynomial has all real roots

Why:

Misremembering the Fundamental Theorem of Algebra, which only guarantees at least one real root for odd degree

Correct move:

Check possible roots with the Rational Root Theorem or Descartes' Rule of Signs before confirming the number of real roots

6. Quick Reference Cheatsheet

Feature	Polynomial Rule	Rational Function Rule
Degree	$n =$ highest power of $x$	No standard degree
Vertical Asymptotes	None	Roots of simplified denominator
Horizontal Asymptote	None	$de g (P) < de g (Q) \to y = 0$ ; $de g (P) = de g (Q) \to y = \frac{a _{n}}{b _{m}}$ ; else none
Oblique Asymptote	None	Exists if $de g (P) = de g (Q) + 1$ from division
Root Behavior	Odd multiplicity: cross x-axis; Even: touch	Same for roots of the numerator
End Behavior	Determined by leading term	Determined by leading terms of numerator/denominator

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 1
Find roots of a cubic polynomial
2022 · 2
Sketch graph of rational function
2021 · 1
Remainder theorem calculation

Going deeper

syllabusIB AA HL Syllabus Topic 2

What's Next

Polynomial and rational functions are foundational for almost all other topics in IB AA HL, from differential and integral calculus to complex numbers and statistical modeling. Mastering root finding, factoring, and graph behavior is critical for solving polynomial equations, integrating rational functions via partial fractions, and analyzing the end behavior of a wide range of functions beyond this sub-topic. IB examiners frequently test combinations of these properties with calculus, so a solid understanding here will save you time and points on exam day.