Polynomials: roots, factor theorem, remainder theorem

IB Mathematics: Applications and Interpretation HL· 25 min read

1. Roots of Polynomials★★☆☆☆⏱ 15 min

📘 Definition

Root (Zero) of a Polynomial

A root or zero of a polynomial $P (x)$ is a value $x = a$ such that $P (a) = 0$ . Graphically, roots correspond to x-intercepts of the curve $y = P (x)$ .

Example:

$x = 2$ is a root of $P (x) = x^{2} - 3 x + 2$ , since $P (2) = 4 - 6 + 2 = 0$

By the Fundamental Theorem of Algebra, a polynomial of degree $n$ with real coefficients has exactly $n$ roots when counting repeated roots and complex roots. Complex roots always come in conjugate pairs for polynomials with real coefficients, and you will mostly work with real roots in IB AI HL.

📐 Worked Example

Find all real roots of $P (x) = x^{3} - 4 x^{2} + x + 6$

1
Test small integer values per the rational root theorem (possible roots are factors of the constant term over factors of the leading coefficient):
2
Test $x = - 1$ :
$P(-1) = (-1)^3 - 4(-1)^2 + (-1) + 6 = -1 - 4 - 1 + 6 = 0$
3
Confirm $x = - 1$ is a root, then factor out $(x + 1)$ from the polynomial:
$x^3 - 4x^2 + x + 6 = (x+1)(x^2 - 5x + 6)$
4
Factorize the quadratic, then set each factor equal to zero:
$(x+1)(x-2)(x-3) = 0 \implies x=-1, x=2, x=3$
5
The three distinct real roots are $- 1$ , $2$ , and $3$ .

2. Remainder Theorem★★☆☆☆⏱ 15 min

📘 Definition

Remainder Theorem

When a polynomial $P (x)$ is divided by a linear divisor $(x - a)$ , the remainder is equal to $P (a)$ . This eliminates the need for full polynomial long division to find the remainder.

Example:

When dividing $P (x) = x^{2} - 2 x + 3$ by $(x - 1)$ , the remainder is $P (1) = 1 - 2 + 3 = 2$ , which matches the result of long division.

The remainder theorem can be generalized for any linear divisor: when dividing by $(k x - a)$ , the remainder is equal to $P (\frac{a}{k})$ .

📐 Worked Example

Find the remainder when $P (x) = 2 x^{3} - 5 x^{2} + 7 x - 9$ is divided by $(2 x - 3)$

1
For divisor $(k x - a)$ , we have $k = 2$ and $a = 3$ , so we evaluate $P (\frac{3}{2})$ :
2
$P\left(\frac{3}{2}\right) = 2\left(\frac{3}{2}\right)^3 - 5\left(\frac{3}{2}\right)^2 + 7\left(\frac{3}{2}\right) - 9$
3
$= \frac{27}{4} - \frac{45}{4} + \frac{42}{4} - \frac{36}{4} = \frac{-12}{4} = -3$
4
By the remainder theorem, the remainder is $- 3$ .

Exam tip:

If an exam question asks only for the remainder, always use the remainder theorem instead of long division to save 2-3 minutes per question.

3. Factor Theorem★★★☆☆⏱ 20 min

📘 Definition

Factor Theorem

The factor theorem is a special case of the remainder theorem: $(x - a)$ is a factor of polynomial $P (x)$ if and only if $P (a) = 0$ . This means if $a$ is a root of $P (x)$ , then $(x - a)$ divides $P (x)$ with no remainder.

The factor theorem is the primary tool for factorizing higher-degree polynomials (up to degree 4) in IB AI HL. Once you find one root by testing, you can factor out the linear term and reduce the problem to solving a lower-degree polynomial, which is far simpler.

📐 Worked Example

Given that $x = 1$ is a root of $P (x) = 2 x^{3} + x^{2} - 5 x + 2$ , fully factorize $P (x)$ and find all roots.

1
By the factor theorem, since $P (1) = 2 (1)^{3} + (1)^{2} - 5 (1) + 2 = 0$ , $(x - 1)$ is a factor of $P (x)$ .
2
Use polynomial division or equating coefficients to write $P (x)$ as $(x - 1)$ times a quadratic:
$2x^3 + x^2 -5x + 2 = (x-1)(2x^2 + 3x - 2)$
3
Factorize the quadratic:
$2x^2 + 3x - 2 = (2x - 1)(x + 2)$
4
The full factorization of $P (x)$ is:
$P(x) = (x-1)(2x-1)(x+2)$
5
Set each factor equal to zero to get all roots:
$x = 1, x = \frac{1}{2}, x = -2$

✓ Quick check

Test your understanding of the factor theorem:

If $(x + 2)$ is a factor of $P (x)$ , which of the following is true?
- $P (2) = 0$
- $P (- 2) = 0$
- Remainder = 2
- $P (0) = - 2$
Reveal answer
1 —
Remember: $(x - a)$ is a factor if $P (a) = 0$ . Rewrite $(x + 2)$ as $(x - (- 2))$ , so $a = - 2$ , so $P (- 2) = 0$ .

4. Common Pitfalls

Wrong move:

When testing for roots, test $x = a$ for the factor $(x + a)$ instead of $x = - a$

Why:

Signs are easily mixed up when extracting the root from a linear factor

Correct move:

Rewrite the factor as $(x - (- a))$ to clearly see that the root to test is $x = - a$

Wrong move:

Assume all nth degree polynomials have n distinct real roots

Why:

The Fundamental Theorem of Algebra counts repeated and complex roots, not just distinct real roots

Correct move:

Always check for repeated roots by looking at factor exponents, and only report real roots unless complex roots are explicitly requested

Wrong move:

Calculate remainder as $P (a)$ when dividing by $(a x + b)$ , instead of $P (- b / a)$

Why:

The theorem is often memorized only for the $(x - a)$ form, leading to errors for non-monic divisors

Correct move:

For any linear divisor $k x - a$ , the remainder is always $P (a / k)$ regardless of the value of $k$

Wrong move:

Stop after finding the first root, when the question asks for all roots

Why:

Higher-degree polynomials have multiple roots, so finding one does not complete the problem

Correct move:

After finding one root, factor out the linear term and continue factorizing the remaining polynomial to find all roots

5. Quick Reference Cheatsheet

Term	Rule	Formula
Root of $P (x)$	Value of $x$ where $P (x) = 0$	$P (a) = 0$
Remainder: divide by $(x - a)$	Remainder equals $P (a)$	$Remainder = P (a)$
Remainder: divide by $(k x - a)$	Remainder equals $P (a / k)$	$Remainder = P (\frac{a}{k})$
Factor Theorem	$(x - a)$ is a factor iff $P (a) = 0$	$(x - a) ∣ P (x) ⟺ P (a) = 0$

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.

2021 · 1
Find all roots of a cubic polynomial
2023 · 2
Factorize quartic given one root

What's Next

Now that you understand how to use the remainder and factor theorems to find roots of polynomials, you can apply these skills to more advanced topics in polynomial functions, including sketching polynomial graphs, solving polynomial inequalities, and modelling real-world phenomena with higher-degree polynomials. These tools are foundational for all work with polynomial functions, and you will use them frequently in calculus and statistical modelling topics later in the IB AI HL course. Mastery of these theorems significantly speeds up your work on polynomial questions in both Paper 1 and Paper 2, so consistent practice testing roots quickly will pay off in the final exam.