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Chapter 2: Polynomial Functions

Precalculus · Updated February 2026 · 25 min read

Polynomial functions are among the most fundamental objects in all of mathematics. They appear everywhere: in physics when modeling projectile motion, in economics when analyzing cost and revenue, and throughout higher mathematics as the building blocks of more complex functions. In this chapter we develop a thorough understanding of polynomial functions—their structure, their zeros, their graphs, and how to solve polynomial inequalities. Mastering this material is essential preparation for calculus, where polynomials serve as the local approximations to virtually every function you will encounter.

2.1 Polynomial Basics

We begin with precise language. A polynomial function is a function that can be written as a sum of terms, each consisting of a constant multiplied by a non-negative integer power of $x$. Despite this simple definition, polynomial functions exhibit rich and varied behavior that rewards careful study.

Definition: Polynomial Function

A polynomial function of degree $n$ is a function of the form

$$f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$$

where $a_n, a_{n-1}, \ldots, a_1, a_0$ are real numbers called coefficients, $n$ is a non-negative integer called the degree, and $a_n \neq 0$. The term $a_n x^n$ is called the leading term, $a_n$ is the leading coefficient, and $a_0$ is the constant term.

When the terms are arranged from highest to lowest power of $x$, we say the polynomial is in standard form. For example, $f(x) = 3x^4 - 2x^3 + 7x - 5$ is a degree-4 polynomial (also called a quartic) in standard form with leading coefficient $3$ and constant term $-5$.

Common Polynomial Types

Note that expressions like $f(x) = x^{-1}$ or $f(x) = \sqrt{x} = x^{1/2}$ are not polynomials because the exponents are not non-negative integers.

End Behavior

The end behavior of a polynomial describes what happens to $f(x)$ as $x \to +\infty$ and as $x \to -\infty$. For large values of $|x|$, the leading term dominates all other terms, so end behavior is determined entirely by the degree $n$ and the leading coefficient $a_n$.

End Behavior of Polynomials

Let $f(x) = a_n x^n + \cdots$ be a polynomial of degree $n$ with leading coefficient $a_n$.

Even degree ($n$ even): Both ends point the same direction. If $a_n > 0$, both ends rise ($f(x) \to +\infty$ as $x \to \pm\infty$). If $a_n < 0$, both ends fall.

Odd degree ($n$ odd): The ends point in opposite directions. If $a_n > 0$, the left end falls and the right end rises. If $a_n < 0$, the left end rises and the right end falls.

A useful mnemonic: for odd-degree polynomials with positive leading coefficient, the graph goes from lower-left to upper-right, like the graph of $y = x$. For even-degree polynomials with positive leading coefficient, the graph opens upward like a bowl, similar to $y = x^2$.

Example 2.1: Identifying Polynomial Features

Describe the degree, leading coefficient, constant term, and end behavior of $f(x) = -2x^5 + 4x^3 - x + 6$.

Solution. The polynomial is already in standard form. Reading off the components:

2.2 Zeros and Factoring

A zero (or root) of a polynomial $f(x)$ is a value $c$ such that $f(c) = 0$. Geometrically, zeros are the $x$-intercepts of the graph. Finding zeros is one of the central problems in algebra, and several powerful theorems help us locate them.

Factor Theorem

$c$ is a zero of the polynomial $f(x)$ if and only if $(x - c)$ is a factor of $f(x)$.

The Factor Theorem connects two problems: finding zeros and factoring. If you know that $f(3) = 0$, then you know immediately that $(x - 3)$ divides $f(x)$ evenly. Conversely, if you can factor out $(x - 3)$, then $x = 3$ is a zero.

Remainder Theorem

If a polynomial $f(x)$ is divided by $(x - c)$, then the remainder is $f(c)$.

The Remainder Theorem is more general: it tells you that the remainder upon dividing $f(x)$ by $(x - c)$ is simply $f(c)$. The Factor Theorem is a special case where the remainder is zero.

The Rational Root Theorem

When we need to find zeros of a polynomial with integer coefficients, the Rational Root Theorem dramatically narrows the search.

Rational Root Theorem

If $f(x) = a_n x^n + \cdots + a_1 x + a_0$ has integer coefficients and $\frac{p}{q}$ is a rational zero (in lowest terms), then:

$p$ is a factor of the constant term $a_0$, and $q$ is a factor of the leading coefficient $a_n$.

This theorem does not tell us which rational numbers are zeros—it tells us which rational numbers could possibly be zeros. We must test each candidate, but the list of candidates is finite.

Example 2.2: Using the Rational Root Theorem

Find all rational zeros of $f(x) = 2x^3 + 3x^2 - 8x + 3$.

Solution. The constant term is $a_0 = 3$ with factors $\pm 1, \pm 3$. The leading coefficient is $a_n = 2$ with factors $\pm 1, \pm 2$. The possible rational zeros are:

$$\frac{p}{q} = \pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2}$$

We test these systematically. Starting with $x = 1$:

$$f(1) = 2(1) + 3(1) - 8(1) + 3 = 0 \checkmark$$

Since $f(1) = 0$, we know $(x - 1)$ is a factor. Dividing by $(x - 1)$ using synthetic division (detailed below), we get $f(x) = (x - 1)(2x^2 + 5x - 3)$. Factoring the quadratic:

$$2x^2 + 5x - 3 = (2x - 1)(x + 3)$$

Therefore $f(x) = (x - 1)(2x - 1)(x + 3)$, and the zeros are $x = 1$, $x = \frac{1}{2}$, and $x = -3$.

Synthetic Division

Synthetic division is a streamlined method for dividing a polynomial by a linear factor $(x - c)$. It is faster than long division and particularly convenient when combined with the Rational Root Theorem.

To divide $f(x) = 2x^3 + 3x^2 - 8x + 3$ by $(x - 1)$, we write the coefficients and the value $c = 1$:

Example 2.3: Synthetic Division Step by Step

Divide $f(x) = 2x^3 + 3x^2 - 8x + 3$ by $(x - 1)$.

Setup: Write $c = 1$ on the left and the coefficients $2, 3, -8, 3$ in a row.

Step 1: Bring down the first coefficient: $2$.

Step 2: Multiply $2 \times 1 = 2$. Add to the next coefficient: $3 + 2 = 5$.

Step 3: Multiply $5 \times 1 = 5$. Add to the next coefficient: $-8 + 5 = -3$.

Step 4: Multiply $-3 \times 1 = -3$. Add to the last coefficient: $3 + (-3) = 0$.

The bottom row reads $2, 5, -3, 0$. The last number is the remainder ($0$, confirming $x = 1$ is a zero), and the other numbers are coefficients of the quotient: $2x^2 + 5x - 3$.

Multiplicity of Zeros

When a factor $(x - c)$ appears more than once in the factorization of a polynomial, we say the zero $c$ has multiplicity greater than one. The multiplicity determines how the graph behaves at that zero.

Definition: Multiplicity

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

Behavior at Zeros by Multiplicity

Odd multiplicity (1, 3, 5, ...): The graph crosses the $x$-axis at the zero. At multiplicity 1, it crosses cleanly; at higher odd multiplicities, it flattens near the axis before crossing.

Even multiplicity (2, 4, 6, ...): The graph touches the $x$-axis at the zero and turns around (bounces) without crossing.

Example 2.4: Multiplicity and Graph Behavior

Describe the behavior of $g(x) = (x + 1)^3(x - 2)^2$ at each zero.

Solution.

The degree of $g$ is $3 + 2 = 5$ (odd), and the leading coefficient is positive (expand: the leading term is $x^5$), so the end behavior goes from lower-left to upper-right.

2.3 Graphing Polynomials

We now combine everything from the previous sections into a systematic approach for graphing polynomial functions by hand. Even with graphing technology readily available, understanding how to sketch a polynomial by hand deepens your understanding of its algebraic structure.

Steps for Graphing a Polynomial

Step 1: Determine the end behavior from the degree and leading coefficient.

Step 2: Find all real zeros by factoring (using the Rational Root Theorem and synthetic division as needed). Note each zero's multiplicity.

Step 3: Find the $y$-intercept by computing $f(0)$.

Step 4: Plot the zeros and $y$-intercept. Determine sign changes and the behavior at each zero (cross or touch). Connect the points smoothly, respecting end behavior.

Example 2.5: Graphing a Cubic Polynomial

Sketch the graph of $f(x) = x^3 - 4x^2 + x + 6$.

Step 1 (End behavior): Degree 3, leading coefficient $+1$. The graph goes from lower-left to upper-right.

Step 2 (Zeros): Testing $x = -1$: $f(-1) = -1 - 4 - 1 + 6 = 0$. So $(x + 1)$ is a factor. Synthetic division gives $f(x) = (x + 1)(x^2 - 5x + 6) = (x + 1)(x - 2)(x - 3)$. The zeros are $x = -1, 2, 3$, each with multiplicity 1 (all cross).

Step 3 ($y$-intercept): $f(0) = 6$. The point $(0, 6)$ is on the graph.

Step 4 (Sketch): Plot $(-1, 0)$, $(2, 0)$, $(3, 0)$, and $(0, 6)$. The graph rises from lower-left, crosses at $x = -1$, rises to a local maximum near $(0, 6)$, falls through $(2, 0)$, reaches a local minimum between $x = 2$ and $x = 3$, crosses at $(3, 0)$, and rises toward upper-right.

Interactive Polynomial Graph

Use the Desmos graph below to explore how changing the leading coefficient $a$ and the roots $r_1, r_2, r_3$ affects the shape of the cubic polynomial $y = a(x - r_1)(x - r_2)(x - r_3)$. Drag the sliders to move the roots and change the stretch factor.

Interactive: Adjust sliders for leading coefficient $a$ and roots $r_1, r_2, r_3$ to explore cubic polynomial behavior.

Example 2.6: Graphing a Quartic with Repeated Root

Sketch the graph of $h(x) = -x^4 + 4x^3 - 4x^2$.

Step 1 (End behavior): Degree 4 (even), leading coefficient $-1$ (negative). Both ends fall: $h(x) \to -\infty$ as $x \to \pm\infty$.

Step 2 (Zeros): Factor out $-x^2$:

$$h(x) = -x^2(x^2 - 4x + 4) = -x^2(x - 2)^2$$

The zeros are $x = 0$ (multiplicity 2, touches) and $x = 2$ (multiplicity 2, touches).

Step 3 ($y$-intercept): $h(0) = 0$.

Step 4 (Sketch): Both zeros are "bounce" points. The graph falls from upper-left, touches and bounces at $x = 0$, rises to a local maximum between $0$ and $2$, touches and bounces at $x = 2$, then falls toward the lower-right. To find the local max, evaluate the midpoint: $h(1) = -1 + 4 - 4 = -1$. So the graph peaks at $(1, -1)$, which is still below the $x$-axis. This means the entire graph lies on or below the $x$-axis.

2.4 Polynomial Inequalities

A polynomial inequality is an inequality of the form $f(x) > 0$, $f(x) \geq 0$, $f(x) < 0$, or $f(x) \leq 0$. The key idea for solving these is sign analysis: a polynomial can only change sign at its zeros, so we find the zeros, divide the number line into intervals, and test the sign on each interval.

Sign Analysis Method

Step 1: Move all terms to one side so the inequality has the form $f(x) > 0$ (or $\geq, <, \leq$).

Step 2: Factor $f(x)$ completely and find all real zeros.

Step 3: Plot the zeros on a number line, dividing it into intervals.

Step 4: Test one value in each interval to determine the sign of $f(x)$ there.

Step 5: Select the intervals that satisfy the inequality. Include endpoints for $\geq$ or $\leq$; exclude them for $>$ or $<$.

The reason this works is the Intermediate Value Theorem: a continuous function (and all polynomials are continuous) can only change sign by passing through zero. So the sign of $f(x)$ is constant on each interval between consecutive zeros.

Example 2.7: Solving a Quadratic Inequality

Solve $x^2 - 5x + 6 \leq 0$.

Step 1: The inequality is already in the form $f(x) \leq 0$ with $f(x) = x^2 - 5x + 6$.

Step 2: Factor: $f(x) = (x - 2)(x - 3)$. Zeros at $x = 2$ and $x = 3$.

Step 3: Three intervals: $(-\infty, 2)$, $(2, 3)$, $(3, \infty)$.

Step 4: Test values:

Step 5: We need $f(x) \leq 0$. The sign is negative on $(2, 3)$, and we include the endpoints because of $\leq$.

Answer: $[2, 3]$

Example 2.8: Solving a Higher-Degree Inequality

Solve $x^3 - x^2 - 6x > 0$.

Step 1: Already in the form $f(x) > 0$.

Step 2: Factor out $x$: $f(x) = x(x^2 - x - 6) = x(x - 3)(x + 2)$. Zeros at $x = -2, 0, 3$.

Step 3: Four intervals: $(-\infty, -2)$, $(-2, 0)$, $(0, 3)$, $(3, \infty)$.

Step 4: Test values:

Step 5: We need $f(x) > 0$. The sign is positive on $(-2, 0)$ and $(3, \infty)$. Endpoints are excluded because the inequality is strict.

Answer: $(-2, 0) \cup (3, \infty)$

A useful shortcut when all factors are linear and each has multiplicity 1: the sign alternates between consecutive zeros, starting from the sign you determine for the leftmost interval. When zeros have even multiplicity, the sign does not change at that zero (the graph bounces).

2.5 Practice Problems

Test your understanding of polynomial functions with the following problems. Each has a detailed solution you can reveal after attempting the problem on your own.

Problem 1

State the degree, leading coefficient, and end behavior of $f(x) = -3x^6 + 2x^4 - x + 7$.

Show Solution

Degree: $6$ (even).

Leading coefficient: $-3$ (negative).

End behavior: Even degree with negative leading coefficient means both ends fall. As $x \to +\infty$, $f(x) \to -\infty$, and as $x \to -\infty$, $f(x) \to -\infty$.

Problem 2

Use the Rational Root Theorem to list all possible rational zeros of $f(x) = 3x^3 - 7x^2 + 4$.

Show Solution

Constant term $a_0 = 4$, factors: $\pm 1, \pm 2, \pm 4$.

Leading coefficient $a_n = 3$, factors: $\pm 1, \pm 3$.

Possible rational zeros: $\pm 1, \pm 2, \pm 4, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{4}{3}$.

Problem 3

Find all zeros of $f(x) = x^3 - 6x^2 + 11x - 6$.

Show Solution

Possible rational zeros: $\pm 1, \pm 2, \pm 3, \pm 6$.

Test $x = 1$: $f(1) = 1 - 6 + 11 - 6 = 0$. So $(x - 1)$ is a factor.

Synthetic division: $x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6) = (x - 1)(x - 2)(x - 3)$.

The zeros are $x = 1, 2, 3$.

Problem 4

When $f(x) = 2x^3 + kx^2 - 5x + 3$ is divided by $(x - 1)$, the remainder is $4$. Find $k$.

Show Solution

By the Remainder Theorem, the remainder when dividing by $(x - 1)$ is $f(1)$.

$$f(1) = 2(1)^3 + k(1)^2 - 5(1) + 3 = 2 + k - 5 + 3 = k$$

Setting $f(1) = 4$: $k = 4$.

Problem 5

Describe the behavior of $g(x) = (x - 1)^2(x + 3)^3$ at each zero. What is the degree and end behavior?

Show Solution

At $x = 1$ (multiplicity 2, even): the graph touches the $x$-axis and bounces back.

At $x = -3$ (multiplicity 3, odd): the graph crosses the $x$-axis with a flattening effect.

Degree: $2 + 3 = 5$ (odd). Leading coefficient: expanding, the leading term is $x^5$ (positive).

End behavior: falls on the left ($x \to -\infty, f(x) \to -\infty$) and rises on the right ($x \to +\infty, f(x) \to +\infty$).

Problem 6

Use synthetic division to divide $f(x) = 3x^4 - 2x^3 + x - 5$ by $(x + 1)$. State the quotient and remainder.

Show Solution

Dividing by $(x + 1) = (x - (-1))$, so $c = -1$. Coefficients: $3, -2, 0, 1, -5$ (note the $0$ for the missing $x^2$ term).

Synthetic division:

Bring down $3$. Then $3 \times (-1) = -3$; $-2 + (-3) = -5$.

$-5 \times (-1) = 5$; $0 + 5 = 5$.

$5 \times (-1) = -5$; $1 + (-5) = -4$.

$-4 \times (-1) = 4$; $-5 + 4 = -1$.

Quotient: $3x^3 - 5x^2 + 5x - 4$. Remainder: $-1$.

Check: $f(-1) = 3(1) - 2(-1) + (-1) - 5 = 3 + 2 - 1 - 5 = -1$. Confirmed by the Remainder Theorem.

Problem 7

Solve $x^2 - 4x - 5 > 0$.

Show Solution

Factor: $(x - 5)(x + 1) > 0$. Zeros at $x = -1$ and $x = 5$.

Test intervals:

  • $x = -2$: $(-7)(-1) = 7 > 0$
  • $x = 0$: $(-5)(1) = -5 < 0$
  • $x = 6$: $(1)(7) = 7 > 0$

The polynomial is positive on $(-\infty, -1) \cup (5, \infty)$.

Answer: $(-\infty, -1) \cup (5, \infty)$

Problem 8

Solve $x^3 + 2x^2 - 4x - 8 \leq 0$.

Show Solution

Factor by grouping: $x^2(x + 2) - 4(x + 2) = (x + 2)(x^2 - 4) = (x + 2)(x - 2)(x + 2) = (x + 2)^2(x - 2)$.

Zeros: $x = -2$ (multiplicity 2) and $x = 2$ (multiplicity 1).

Test intervals:

  • $x = -3$: $(1)(-5) = -5 < 0$
  • $x = 0$: $(4)(-2) = -8 < 0$
  • $x = 3$: $(25)(1) = 25 > 0$

Note: The sign does not change at $x = -2$ because the multiplicity is even.

We need $f(x) \leq 0$: negative on $(-\infty, -2)$ and $(-2, 2)$, and zero at $x = -2$ and $x = 2$.

Answer: $(-\infty, 2]$

Problem 9

Find a polynomial of degree 4 with leading coefficient $1$ that has zeros at $x = -1$ (multiplicity 2) and $x = 3$ (multiplicity 2).

Show Solution

Using the Factor Theorem, the polynomial is:

$$f(x) = (x + 1)^2(x - 3)^2$$

Expanding: $(x + 1)^2 = x^2 + 2x + 1$ and $(x - 3)^2 = x^2 - 6x + 9$.

$f(x) = (x^2 + 2x + 1)(x^2 - 6x + 9)$

$= x^4 - 6x^3 + 9x^2 + 2x^3 - 12x^2 + 18x + x^2 - 6x + 9$

$= x^4 - 4x^3 - 2x^2 + 12x + 9$

The graph touches the $x$-axis at both $x = -1$ and $x = 3$ (both even multiplicity).

Problem 10

Sketch a rough graph of $p(x) = -(x + 2)(x - 1)^2(x - 4)$, identifying zeros, end behavior, and $y$-intercept.

Show Solution

Zeros: $x = -2$ (multiplicity 1, crosses), $x = 1$ (multiplicity 2, touches/bounces), $x = 4$ (multiplicity 1, crosses).

Degree: $1 + 2 + 1 = 4$ (even). Leading coefficient: $-1$ (negative, from the leading $-$ sign). Both ends fall.

$y$-intercept: $p(0) = -(0 + 2)(0 - 1)^2(0 - 4) = -(2)(1)(-4) = 8$. The point $(0, 8)$ is on the graph.

Sketch: The graph falls from the upper-left, crosses the $x$-axis at $x = -2$, rises through $(0, 8)$, bounces at $x = 1$, rises to a local max, then crosses downward through $x = 4$, and falls toward the lower-right.

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