Introduction to Polynomial Functions

A polynomial function is one of the most fundamental structures in mathematics — built entirely from addition, multiplication, and non-negative integer exponents.

Polynomial Function

A function of the form $f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}$ where $a_{n}, a_{n - 1}, \dots, a_{0}$ are real constants (called coefficients), $a_{n} \neq = 0$ , and $n$ is a non-negative integer.

Degree of a Polynomial

The highest power of the variable in the polynomial. For example, $f (x) = 3 x^{4} - 2 x + 7$ has degree 4.

Some key vocabulary you'll need throughout this subtopic:

Leading coefficient: the coefficient $a_{n}$ of the highest-degree term
Constant term: $a_{0}$ , the value of $f (0)$ (the $y$ -intercept)
Roots / Zeros: values of $x$ for which $f (x) = 0$ (the $x$ -intercepts)

Note

A polynomial of degree $n$ has at most $n$ real roots, though some roots may be repeated or complex.

The Factor Theorem

Factor Theorem

A polynomial $f (x)$ has a factor $(x - a)$ if and only if $f (a) = 0$ . Equivalently, $a$ is a root of $f (x)$ if and only if $(x - a)$ divides $f (x)$ exactly (with zero remainder).

The "if and only if" means it works in both directions:

If you suspect $a$ is a root, evaluate $f (a)$ . If $f (a) = 0$ , then $(x - a)$ is a factor.
If you know $(x - a)$ is a factor, then $f (a) = 0$ is guaranteed.

This gives you a powerful method for factorising polynomials: test candidate integer values until you find one that gives $f (a) = 0$ , then perform polynomial division.

Exam Tip

Rational Root Theorem (useful for finding candidates): For a polynomial with integer coefficients, any rational root $\frac{p}{q}$ (in lowest terms) satisfies: $p$ divides the constant term $a_{0}$ and $q$ divides the leading coefficient $a_{n}$ . For monic polynomials ( $a_{n} = 1$ ), only integer factors of $a_{0}$ need to be tested.

Example

Find a factor of $f (x) = x^{3} - 4 x^{2} + x + 6$ .

Test integer factors of 6: $\pm 1, \pm 2, \pm 3, \pm 6$

$f (1) = 1 - 4 + 1 + 6 = 4 \neq = 0$

$f (2) = 8 - 16 + 2 + 6 = 0$ ✓

Since $f (2) = 0$ , by the Factor Theorem, $(x - 2)$ is a factor of $f (x)$ .

Continuing: $f (3) = 27 - 36 + 3 + 6 = 0$ ✓, so $(x - 3)$ is also a factor.

$f (- 1) = - 1 - 4 - 1 + 6 = 0$ ✓, so $(x + 1)$ is a factor.

Therefore $f (x) = (x - 2) (x - 3) (x + 1)$ .

Common Mistake

The Factor Theorem says $(x - a)$ is a factor when $f (a) = 0$ . Students sometimes write the factor as $(x + a)$ — remember the sign flip: if $f (2) = 0$ , the factor is $(x - 2)$ , not $(x + 2)$ .

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AHL 2.12—Factor and remainder theorems, sum and product of roots

Introduction to Polynomial Functions

The Factor Theorem

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