Chapter 3: Polynomial Functions
In This Chapter
Polynomial functions form the backbone of Algebra 2 and appear everywhere in mathematics, science, and engineering. In this chapter, we build a complete toolkit for analyzing polynomials: classifying them by degree, predicting their graphs through end behavior and multiplicity, dividing them efficiently, applying powerful theorems to locate zeros, and solving polynomial inequalities. By the end, you will be able to take any polynomial, find all of its zeros (real and complex), sketch its graph, and solve related inequalities.
3.1 Polynomial Basics
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 $a_n, a_{n-1}, \ldots, a_0$ are real numbers and $a_n \neq 0$. The number $a_n$ is the leading coefficient, $a_n x^n$ is the leading term, and $a_0$ is the constant term. This expanded form is called standard form.
Degree and Classification
The degree of a polynomial is the largest exponent that appears on the variable. The degree tells us the maximum number of zeros, the maximum number of turning points ($n - 1$), and the general shape of the graph. We classify polynomials by degree with special names:
- Degree 0 — Constant (e.g., $f(x) = 5$)
- Degree 1 — Linear (e.g., $f(x) = 3x + 2$)
- Degree 2 — Quadratic (e.g., $f(x) = x^2 - 4x + 3$)
- Degree 3 — Cubic (e.g., $f(x) = 2x^3 - x + 7$)
- Degree 4 — Quartic (e.g., $f(x) = x^4 - 6x^2 + 1$)
- Degree 5 — Quintic (e.g., $f(x) = x^5 - x$)
We also classify polynomials by the number of terms: a monomial has one term, a binomial has two, and a trinomial has three. A polynomial with four or more terms is simply called a polynomial.
Example 3.1
Identify the degree, leading coefficient, and constant term of $f(x) = -3x^4 + 7x^2 - x + 5$.
Solution: The highest power of $x$ is $4$, so the degree is $4$ (quartic). The leading coefficient is $-3$. The constant term is $5$.
Writing in Standard Form
A polynomial is in standard form when its terms are written in descending order of degree. If you are given a polynomial with terms out of order, rearrange them. For instance, $4 + 3x^3 - x$ in standard form is $3x^3 - x + 4$.
Study Tip: When working with polynomials, always write them in standard form first. This makes it easy to identify the degree, leading coefficient, and to set up division problems correctly.
3.2 End Behavior & Graphing
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. This gives us a simple rule.
Leading Term Test for End Behavior
The end behavior of $f(x) = a_n x^n + \cdots$ depends only on the degree $n$ and the sign of $a_n$:
Even degree, $a_n > 0$: Both ends rise. $f(x) \to +\infty$ as $x \to \pm\infty$.
Even degree, $a_n < 0$: Both ends fall. $f(x) \to -\infty$ as $x \to \pm\infty$.
Odd degree, $a_n > 0$: Falls left, rises right. $f(x) \to -\infty$ as $x \to -\infty$; $f(x) \to +\infty$ as $x \to +\infty$.
Odd degree, $a_n < 0$: Rises left, falls right. $f(x) \to +\infty$ as $x \to -\infty$; $f(x) \to -\infty$ as $x \to +\infty$.
Example 3.2
Describe the end behavior of $g(x) = -2x^5 + 4x^3 - x$.
Solution: The leading term is $-2x^5$. The degree is $5$ (odd) and the leading coefficient is $-2$ (negative). By the leading term test: $g(x) \to +\infty$ as $x \to -\infty$ and $g(x) \to -\infty$ as $x \to +\infty$. The graph rises to the left and falls to the right.
Turning Points
A turning point is a point where the graph changes from increasing to decreasing or vice versa. A polynomial of degree $n$ has at most $n - 1$ turning points. For example, a cubic ($n = 3$) can have at most $2$ turning points, while a quartic ($n = 4$) can have at most $3$.
Zeros and Multiplicity
The $x$-intercepts (or real zeros) of a polynomial are the values of $x$ where $f(x) = 0$. If a polynomial in factored form has a factor $(x - c)^k$, we say $c$ is a zero of multiplicity $k$. The multiplicity determines how the graph behaves at the intercept:
- Odd multiplicity (1, 3, 5, ...): The graph crosses the $x$-axis at $x = c$.
- Even multiplicity (2, 4, 6, ...): The graph touches (bounces off) the $x$-axis at $x = c$ and turns around.
Example 3.3
Find the zeros and describe the graph behavior at each zero for $f(x) = 2(x + 1)^2(x - 3)$.
Solution: Set $f(x) = 0$: the zeros are $x = -1$ (multiplicity $2$) and $x = 3$ (multiplicity $1$).
- At $x = -1$: even multiplicity, so the graph bounces off the $x$-axis.
- At $x = 3$: odd multiplicity, so the graph crosses the $x$-axis.
The degree is $2 + 1 = 3$ (odd) with a positive leading coefficient, so the graph falls left and rises right. The $y$-intercept is $f(0) = 2(1)^2(-3) = -6$.
3.3 Polynomial Long Division & Synthetic Division
Just as we can divide integers, we can divide polynomials. If $f(x)$ and $d(x)$ are polynomials with $\deg(d) \leq \deg(f)$, then there exist unique polynomials $q(x)$ (the quotient) and $r(x)$ (the remainder) such that:
$$f(x) = d(x) \cdot q(x) + r(x), \quad \text{where } \deg(r) < \deg(d)$$
This is called the Division Algorithm for polynomials.
Polynomial Long Division
Polynomial long division follows the same pattern as long division with numbers: divide, multiply, subtract, bring down, repeat.
Example 3.4: Long Division
Divide $f(x) = 2x^3 + 3x^2 - 5x + 1$ by $d(x) = x + 2$.
Step 1 Divide the leading term of the dividend by the leading term of the divisor: $\dfrac{2x^3}{x} = 2x^2$.
Step 2 Multiply: $2x^2(x + 2) = 2x^3 + 4x^2$. Subtract from the dividend:
$$(2x^3 + 3x^2) - (2x^3 + 4x^2) = -x^2$$
Bring down $-5x$ to get $-x^2 - 5x$.
Step 3 Divide: $\dfrac{-x^2}{x} = -x$. Multiply: $-x(x + 2) = -x^2 - 2x$. Subtract:
$$(-x^2 - 5x) - (-x^2 - 2x) = -3x$$
Bring down $+1$ to get $-3x + 1$.
Step 4 Divide: $\dfrac{-3x}{x} = -3$. Multiply: $-3(x + 2) = -3x - 6$. Subtract:
$$(-3x + 1) - (-3x - 6) = 7$$
Result: $\dfrac{2x^3 + 3x^2 - 5x + 1}{x + 2} = 2x^2 - x - 3 + \dfrac{7}{x + 2}$
Synthetic Division
Synthetic division is a shortcut for dividing a polynomial by a linear divisor of the form $(x - c)$. It uses only the coefficients and is much faster than long division.
Synthetic Division Algorithm
To divide $f(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_0$ by $(x - c)$:
1. Write $c$ on the left and list all coefficients of $f(x)$ in order (include zeros for missing terms).
2. Bring down the first coefficient.
3. Multiply by $c$, write the result under the next coefficient, and add.
4. Repeat step 3 until all coefficients are processed.
5. The last number is the remainder; the other numbers are coefficients of the quotient (one degree less).
Example 3.5: Synthetic Division
Divide $f(x) = 2x^3 + 3x^2 - 5x + 1$ by $(x + 2)$, i.e., $(x - (-2))$, so $c = -2$.
Reading the bottom row: the quotient is $2x^2 - x - 3$ and the remainder is $7$.
This confirms our long division result: $\dfrac{2x^3 + 3x^2 - 5x + 1}{x + 2} = 2x^2 - x - 3 + \dfrac{7}{x + 2}$.
When to Use Which: Use synthetic division when the divisor is linear $(x - c)$. Use long division when the divisor has degree 2 or higher.
3.4 Factor & Remainder Theorems
Two powerful theorems connect polynomial division to polynomial evaluation.
Remainder Theorem
If a polynomial $f(x)$ is divided by $(x - c)$, the remainder equals $f(c)$.
This means you can find the remainder of any polynomial division by $(x - c)$ simply by plugging $c$ into $f(x)$, without performing the division at all.
Example 3.6
Find the remainder when $f(x) = x^4 - 3x^2 + 2x - 5$ is divided by $(x - 2)$.
Solution: By the Remainder Theorem, the remainder is $f(2)$:
$$f(2) = (2)^4 - 3(2)^2 + 2(2) - 5 = 16 - 12 + 4 - 5 = 3$$
The remainder is $3$.
Factor Theorem
$(x - c)$ is a factor of $f(x)$ if and only if $f(c) = 0$.
The Factor Theorem is the special case of the Remainder Theorem where the remainder is zero. It tells us: if you plug a number $c$ into a polynomial and get $0$, then $(x - c)$ divides the polynomial evenly.
Example 3.7
Show that $(x - 1)$ is a factor of $f(x) = x^3 - 6x^2 + 11x - 6$, then factor completely.
Solution: Check $f(1) = 1 - 6 + 11 - 6 = 0$. Since $f(1) = 0$, the Factor Theorem guarantees $(x - 1)$ is a factor.
Divide by $(x - 1)$ using synthetic division:
The quotient is $x^2 - 5x + 6 = (x - 2)(x - 3)$.
Therefore $f(x) = (x - 1)(x - 2)(x - 3)$.
Rational Root Theorem
When we need to find rational zeros of a polynomial, the Rational Root Theorem narrows the search dramatically.
Rational Root Theorem
If $\dfrac{p}{q}$ is a rational zero of $f(x) = a_n x^n + \cdots + a_0$ (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 of these candidates are actual zeros; it gives us a finite list to test. We test each candidate using synthetic division or direct substitution.
Example 3.8
List all possible rational zeros of $f(x) = 2x^3 - 3x^2 - 8x + 12$.
Solution: Factors of the constant term $12$: $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$.
Factors of the leading coefficient $2$: $\pm 1, \pm 2$.
Possible rational zeros $= \dfrac{p}{q}$: $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm \frac{1}{2}, \pm \frac{3}{2}$.
3.5 Finding All Zeros
The goal of this section is to combine all the tools from Sections 3.3 and 3.4 into a systematic strategy for finding every zero of a polynomial.
Fundamental Theorem of Algebra
Every polynomial of degree $n \geq 1$ has exactly $n$ zeros (counting multiplicity) in the complex number system.
This means a degree-$3$ polynomial has exactly $3$ zeros, a degree-$5$ polynomial has exactly $5$ zeros, and so on. Some (or all) of these zeros may be complex numbers.
Complex Conjugate Pairs Theorem
If $f(x)$ has real coefficients and $a + bi$ (with $b \neq 0$) is a zero, then its conjugate $a - bi$ is also a zero.
Complex roots of polynomials with real coefficients always come in conjugate pairs. This means a polynomial of odd degree must have at least one real zero.
Strategy for Finding All Zeros
- List possible rational zeros using the Rational Root Theorem.
- Test candidates using synthetic division. When you find a zero $c$, you get a quotient polynomial of lower degree.
- Repeat on the reduced polynomial until you reach a quadratic (or simpler).
- Solve the remaining quadratic by factoring or the quadratic formula.
- Check that the total number of zeros (counting multiplicity) equals the degree.
Example 3.9: Finding All Zeros
Find all zeros of $f(x) = 2x^4 - x^3 - 7x^2 + 4x - 4$.
Step 1 Possible rational zeros: $\pm 1, \pm 2, \pm 4, \pm \frac{1}{2}$.
Step 2 Test $x = 2$: $f(2) = 2(16) - 8 - 28 + 8 - 4 = 0$. So $x = 2$ is a zero.
Synthetic division by $(x - 2)$:
Quotient: $2x^3 + 3x^2 - x + 2$.
Step 3 Test $x = -2$ on the quotient: $2(-8) + 3(4) - (-2) + 2 = -16 + 12 + 2 + 2 = 0$. So $x = -2$ is a zero.
Synthetic division of $2x^3 + 3x^2 - x + 2$ by $(x + 2)$:
Quotient: $2x^2 - x + 1$.
Step 4 Solve $2x^2 - x + 1 = 0$ with the quadratic formula:
$$x = \frac{1 \pm \sqrt{1 - 8}}{4} = \frac{1 \pm \sqrt{-7}}{4} = \frac{1 \pm i\sqrt{7}}{4}$$
All four zeros: $x = 2, \; x = -2, \; x = \dfrac{1 + i\sqrt{7}}{4}, \; x = \dfrac{1 - i\sqrt{7}}{4}$.
Note that the complex zeros come as a conjugate pair, as expected.
Efficiency Tip: When testing rational root candidates, start with the simplest values ($\pm 1, \pm 2$) before testing fractions. Also, use Descartes' Rule of Signs to narrow the possibilities: count sign changes in $f(x)$ for positive zeros and in $f(-x)$ for negative zeros.
3.6 Polynomial Inequalities
A polynomial inequality asks where a polynomial is positive, negative, or zero. The method relies on the fact that a polynomial can only change sign at its real zeros.
Sign Analysis Method
To solve a polynomial inequality like $f(x) > 0$, $f(x) \leq 0$, etc.:
1. Move all terms to one side so you have $f(x) \; [\text{inequality sign}] \; 0$.
2. Factor $f(x)$ completely and find all real zeros.
3. Plot the zeros on a number line, dividing it into intervals.
4. Pick a test point in each interval and evaluate the sign of $f(x)$.
5. Determine which intervals satisfy the inequality. Include endpoints for $\leq$ or $\geq$.
Example 3.10
Solve $x^3 - 4x^2 + x + 6 \leq 0$.
Step 1 Factor. Test $x = -1$: $(-1) - 4 - 1 + 6 = 0$. So $(x + 1)$ is a factor.
Synthetic division gives $x^2 - 5x + 6 = (x - 2)(x - 3)$.
So $f(x) = (x + 1)(x - 2)(x - 3)$. The zeros are $x = -1, 2, 3$.
Step 2 Test intervals:
| Interval | Test point | $(x+1)$ | $(x-2)$ | $(x-3)$ | $f(x)$ |
|---|---|---|---|---|---|
| $(-\infty, -1)$ | $x = -2$ | $-$ | $-$ | $-$ | $-$ |
| $(-1, 2)$ | $x = 0$ | $+$ | $-$ | $-$ | $+$ |
| $(2, 3)$ | $x = 2.5$ | $+$ | $+$ | $-$ | $-$ |
| $(3, \infty)$ | $x = 4$ | $+$ | $+$ | $+$ | $+$ |
Step 3 We need $f(x) \leq 0$ (negative or zero). This occurs on $(-\infty, -1]$ and $[2, 3]$.
Answer: $(-\infty, -1] \cup [2, 3]$
Multiplicity Matters: At a zero with even multiplicity, the polynomial does not change sign (it bounces). At a zero with odd multiplicity, the polynomial changes sign (it crosses). Use this to fill in sign charts without testing every interval.
Example 3.11
Solve $(x - 1)^2(x + 3) > 0$.
Zeros: $x = 1$ (multiplicity 2, even) and $x = -3$ (multiplicity 1, odd).
Test $x = -4$: $(25)(-1) = -25 < 0$. So $f(x) < 0$ on $(-\infty, -3)$.
At $x = -3$, odd multiplicity: sign changes. So $f(x) > 0$ on $(-3, 1)$.
At $x = 1$, even multiplicity: sign does not change. So $f(x) > 0$ on $(1, \infty)$ as well.
Answer: $(-3, 1) \cup (1, \infty)$
(We exclude $x = -3$ and $x = 1$ because the inequality is strict.)
3.7 Practice Problems
Test your understanding with these problems. Try each one before revealing the solution.
Problem 1
Identify the degree, leading coefficient, and constant term of $f(x) = -5x^6 + 3x^4 - x^2 + 9$. Classify by degree.
Show Solution
Degree: $6$ (sixth-degree polynomial). Leading coefficient: $-5$. Constant term: $9$.
There is no standard single-word name commonly used for degree 6, but it may be called a "sextic" or simply a "degree-6 polynomial."
Problem 2
Describe the end behavior of $h(x) = 4x^3 - x + 7$.
Show Solution
The leading term is $4x^3$. Degree $3$ is odd and the leading coefficient $4$ is positive.
End behavior: $h(x) \to -\infty$ as $x \to -\infty$ and $h(x) \to +\infty$ as $x \to +\infty$.
The graph falls to the left and rises to the right.
Problem 3
Find the zeros and their multiplicities for $f(x) = -3(x + 4)^3(x - 1)^2$. At each zero, state whether the graph crosses or bounces.
Show Solution
$x = -4$, multiplicity $3$ (odd) — the graph crosses the $x$-axis.
$x = 1$, multiplicity $2$ (even) — the graph bounces off the $x$-axis.
Total degree: $3 + 2 = 5$. With leading coefficient $-3$ (negative, odd degree), the graph rises left and falls right.
Problem 4
Use polynomial long division to divide $3x^3 - 2x^2 + 5x - 1$ by $x^2 + 1$.
Show Solution
Step 1 $\dfrac{3x^3}{x^2} = 3x$. Multiply: $3x(x^2 + 1) = 3x^3 + 3x$. Subtract: $(3x^3 - 2x^2 + 5x - 1) - (3x^3 + 3x) = -2x^2 + 2x - 1$.
Step 2 $\dfrac{-2x^2}{x^2} = -2$. Multiply: $-2(x^2 + 1) = -2x^2 - 2$. Subtract: $(-2x^2 + 2x - 1) - (-2x^2 - 2) = 2x + 1$.
Result: $\dfrac{3x^3 - 2x^2 + 5x - 1}{x^2 + 1} = 3x - 2 + \dfrac{2x + 1}{x^2 + 1}$
Problem 5
Use synthetic division to divide $x^4 - 3x^3 + 2x - 8$ by $(x - 3)$.
Show Solution
Note: the $x^2$ term is missing, so use $0$ as its coefficient.
Quotient: $x^3 + 0x^2 + 0x + 2 = x^3 + 2$. Remainder: $-2$.
$$\frac{x^4 - 3x^3 + 2x - 8}{x - 3} = x^3 + 2 + \frac{-2}{x - 3}$$
Problem 6
Use the Remainder Theorem to find $f(-3)$ for $f(x) = 2x^4 + 5x^3 - x + 7$ without full evaluation.
Show Solution
By the Remainder Theorem, $f(-3)$ equals the remainder when dividing by $(x + 3)$.
The remainder is $37$, so $f(-3) = 37$.
We can verify: $2(81) + 5(-27) - (-3) + 7 = 162 - 135 + 3 + 7 = 37$.
Problem 7
Find all zeros of $f(x) = x^3 - 4x^2 + x + 6$.
Show Solution
Possible rational zeros: $\pm 1, \pm 2, \pm 3, \pm 6$.
Test $x = -1$: $f(-1) = -1 - 4 - 1 + 6 = 0$. So $x = -1$ is a zero.
Synthetic division:
Quotient: $x^2 - 5x + 6 = (x - 2)(x - 3)$.
All zeros: $x = -1, \; x = 2, \; x = 3$.
Problem 8
Find all zeros (real and complex) of $g(x) = x^4 + x^3 - x^2 + x - 2$.
Show Solution
Possible rational zeros: $\pm 1, \pm 2$.
Test $x = 1$: $1 + 1 - 1 + 1 - 2 = 0$. So $x = 1$ is a zero.
Quotient: $x^3 + 2x^2 + x + 2$. Factor by grouping:
$x^2(x + 2) + 1(x + 2) = (x + 2)(x^2 + 1)$.
Solve $x^2 + 1 = 0$: $x = \pm i$.
All zeros: $x = 1, \; x = -2, \; x = i, \; x = -i$.
The complex zeros $i$ and $-i$ form a conjugate pair.
Problem 9
Solve the inequality $x^3 - x^2 - 6x \geq 0$.
Show Solution
Factor: $x(x^2 - x - 6) = x(x - 3)(x + 2)$.
Zeros: $x = -2, \; 0, \; 3$ (all multiplicity 1).
| Interval | Test point | $x$ | $(x-3)$ | $(x+2)$ | Product |
|---|---|---|---|---|---|
| $(-\infty, -2)$ | $x = -3$ | $-$ | $-$ | $-$ | $-$ |
| $(-2, 0)$ | $x = -1$ | $-$ | $-$ | $+$ | $+$ |
| $(0, 3)$ | $x = 1$ | $+$ | $-$ | $+$ | $-$ |
| $(3, \infty)$ | $x = 4$ | $+$ | $+$ | $+$ | $+$ |
We need $f(x) \geq 0$ (positive or zero): $[-2, 0] \cup [3, \infty)$.
Problem 10
Solve $(x + 1)^2(x - 4) < 0$.
Show Solution
Zeros: $x = -1$ (multiplicity 2, even — no sign change) and $x = 4$ (multiplicity 1, odd — sign change).
Test $x = 0$: $(1)(- 4) = -4 < 0$.
Since $x = -1$ has even multiplicity, the sign does not change there. So $f(x) < 0$ on $(-\infty, -1)$ and $(-1, 4)$.
At $x = 4$ (odd multiplicity), the sign changes: $f(x) > 0$ on $(4, \infty)$.
Answer: $(-\infty, -1) \cup (-1, 4)$
(We exclude $x = -1$ and $x = 4$ because the inequality is strict.)
Problem 11
Find a polynomial of degree 3 with real coefficients that has zeros $x = 2$ and $x = 1 + i$.
Show Solution
Since the coefficients are real and $1 + i$ is a zero, its conjugate $1 - i$ must also be a zero. The three zeros are $2, \; 1 + i, \; 1 - i$.
$$f(x) = (x - 2)\bigl[(x - (1+i))(x - (1-i))\bigr]$$
Expand the complex factor pair:
$(x - 1 - i)(x - 1 + i) = (x-1)^2 - (i)^2 = (x-1)^2 + 1 = x^2 - 2x + 2$
Therefore:
$$f(x) = (x - 2)(x^2 - 2x + 2) = x^3 - 4x^2 + 6x - 4$$
Problem 12
A polynomial $f(x)$ of degree 4 has leading coefficient $1$, a zero at $x = -1$ (multiplicity 2), passes through the origin, and $f(2) = 18$. Find $f(x)$.
Show Solution
Since $f$ has degree 4 with leading coefficient 1 and $x = -1$ is a zero of multiplicity 2, we can write:
$$f(x) = (x + 1)^2(x - r)(x - s)$$
Since $f$ passes through the origin, $f(0) = 0$, so $x = 0$ is a zero:
$$f(x) = (x + 1)^2 \cdot x \cdot (x - s)$$
This is degree 4, which matches. Use $f(2) = 18$:
$f(2) = (3)^2(2)(2 - s) = 18(2 - s) = 18$
$2 - s = 1$, so $s = 1$.
$$f(x) = x(x - 1)(x + 1)^2 = x(x-1)(x^2 + 2x + 1)$$
Expanding: $x(x-1) = x^2 - x$, then $(x^2 - x)(x^2 + 2x + 1) = x^4 + x^3 - x^2 - x$.
Answer: $f(x) = x^4 + x^3 - x^2 - x$.