AP · Polynomial functions and complex zeros · 14 min read · Updated 2026-05-10

Polynomial functions and complex zeros — AP Precalculus Study Guide

For: AP Precalculus candidates sitting AP Precalculus.

Covers: The Fundamental Theorem of Algebra, Complex Conjugate Root Theorem, factoring polynomials with complex zeros, multiplicity of complex roots, degree-zero count relationship, and discriminant analysis for polynomials with real coefficients.

You should already know: Basic arithmetic of complex numbers in rectangular form. Factoring of polynomials with real rational/irrational zeros. Definition of polynomial degree and multiplicity of zeros.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Precalculus style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Polynomial functions and complex zeros?

This topic extends the study of polynomial solutions beyond real zeros to include the full set of complex zeros for polynomials with real coefficients, which is the standard case on the AP Precalculus exam. Per the AP Precalculus Course and Exam Description (CED), this topic accounts for approximately 5-7% of total exam score, and questions appear in both multiple-choice (MCQ) and free-response (FRQ) sections. A core result of this topic is that any degree $n$ polynomial has exactly $n$ zeros (counting multiplicity) when we expand our solution set to the complex plane, which unifies all polynomial factoring behavior. Exam questions regularly ask you to find all zeros of a polynomial given one complex zero, factor high-degree polynomials completely over the reals or complex numbers, or distinguish between the number of real vs. complex zeros for a given polynomial. Unlike real zeros, complex non-real zeros do not correspond to x-intercepts on the real coordinate plane, so questions often test this key distinction.

2. The Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra (FTA) is the core result that governs all polynomial zero-counting. The theorem states that every non-constant single-variable polynomial with complex coefficients has at least one complex zero. A direct corollary that is used almost exclusively on the AP exam is: A polynomial of degree $n \geq 1$ has exactly $n$ complex zeros, when you count each zero according to its multiplicity. Intuitively, the FTA confirms that the complex number system is "complete" for solving polynomials: unlike the real numbers, which cannot solve all polynomial equations, the complex numbers guarantee a solution for every non-constant polynomial. The requirement to count multiplicity is critical: for example, $(x - 4)^{3}$ is a degree 3 polynomial with only one unique zero at $x = 4$ , but it counts as 3 zeros per the FTA because of multiplicity 3. It is important to distinguish between "total zeros (counting multiplicity)" and "unique zeros" or "real zeros" — the AP exam tests this wording distinction regularly.

Worked Example

How many total complex zeros (counting multiplicity) does the polynomial $p (x) = 2 x^{3} (x - 1)^{2} (3 x^{2} - 7 x + 2)$ have? How many unique complex zeros does it have?

First calculate the degree of the polynomial by adding the exponents of all factors: the leading term is $2 x^{3} \cdot x^{2} \cdot 3 x^{2} = 6 x^{3 + 2 + 2} = 6 x^{7}$ , so the degree of $p (x)$ is 7.
By the Fundamental Theorem of Algebra, the total number of complex zeros counting multiplicity equals the degree of the polynomial, so the answer for total zeros is 7.
To find unique zeros, factor $p (x)$ completely: we have $x = 0$ (multiplicity 3), $x = 1$ (multiplicity 2), and the quadratic factors to $(3 x - 1) (x - 2)$ , giving unique zeros at $x = \frac{1}{3}$ and $x = 2$ .
Listing all unique zeros gives $0, 1, \frac{1}{3}, 2$ , so there are 4 unique complex zeros.

Exam tip: Always scan the question prompt for wording about "unique" or "real" zeros. If no qualifier is given, "number of zeros" by convention means total zeros counting multiplicity per the Fundamental Theorem of Algebra on the AP exam.

3. The Complex Conjugate Root Theorem

The Complex Conjugate Root Theorem is the most commonly tested result for this topic on the AP exam, and it applies exclusively to polynomials with real coefficients (which is all polynomials you will encounter on the exam). The theorem states: If $a + bi$ is a non-real complex zero of a polynomial with real coefficients, then its complex conjugate $a - bi$ is also a zero of the polynomial. This works because complex conjugates preserve addition and multiplication: $\overline{z + w} = \overline{z} + \overline{w}$ and $\overline{z w} = \overline{z} \overline{w}$ , so plugging the conjugate of a zero into a polynomial with real coefficients (whose coefficients are their own conjugates) gives the conjugate of zero, which is zero. This theorem is extremely useful because exam questions often give you one complex zero and ask you to find all remaining zeros: the theorem immediately gives you a second zero, so you can form a quadratic factor with real coefficients and divide to find the rest, without doing division with complex coefficients.

Worked Example

Given that $p (x) = x^{4} - 4 x^{3} + 14 x^{2} - 28 x + 45$ has a complex zero $1 + 2 i$ , find all zeros of $p (x)$ .

By the Complex Conjugate Root Theorem, since $p (x)$ has real coefficients and $1 + 2 i$ is a zero, its conjugate $1 - 2 i$ must also be a zero.
Calculate the quadratic factor with these two zeros: $(x - (1 + 2 i)) (x - (1 - 2 i)) = (x - 1)^{2} - (2 i)^{2} = x^{2} - 2 x + 1 + 4 = x^{2} - 2 x + 5$
Divide $p (x)$ by this quadratic to get the remaining quadratic factor: $\frac{x ^{4} - 4 x ^{3} + 14 x ^{2} - 28 x + 45}{x ^{2} - 2 x + 5} = x^{2} - 2 x + 9$ .
Solve for the roots of the remaining quadratic: $x = \frac{2 \pm 4 - 36}{2} = 1 \pm 22 i$ .
All four zeros are $1 + 2 i, 1 - 2 i, 1 + 22 i, 1 - 22 i$ .

Exam tip: You can use the shortcut $x^{2} - 2 a x + (a^{2} + b^{2})$ for the quadratic factor for roots $a \pm bi$ to skip expanding the product every time, which saves time on both MCQ and FRQ.

4. Factoring Over Reals vs. Over Complex Numbers

Once you know all zeros of a polynomial, you can factor it in two different ways depending on whether you are working over the real numbers or the complex numbers. The Linear Factorization Theorem, a corollary of the Fundamental Theorem of Algebra, states that any degree $n$ polynomial can be factored completely over the complex numbers into $n$ linear factors: if the leading coefficient is $a_{n}$ and zeros are $r_{1}, r_{2}, ..., r_{n}$ , then: $p (x) = a_{n} (x - r_{1}) (x - r_{2}) ... (x - r_{n})$ When factoring over the real numbers, however, non-real complex zeros cannot be written as real numbers, so their linear factors have complex coefficients. Instead, we pair up conjugate non-real zeros into irreducible quadratic factors with real coefficients (irreducible meaning they cannot be factored further into linear terms with real coefficients). A key consequence of this pairing is that the number of non-real complex zeros (counting multiplicity) is always even, so the number of real zeros (counting multiplicity) has the same parity as the degree of the polynomial. This means all odd-degree polynomials with real coefficients have at least one real zero.

Worked Example

Factor $p (x) = 2 x^{4} - 2 x^{3} + 4 x^{2} + 14 x - 10$ completely (a) over the real numbers, (b) over the complex numbers, given $1 + 2 i$ is a zero.

By the Complex Conjugate Root Theorem, $1 - 2 i$ is also a zero, giving the quadratic factor $x^{2} - 2 x + 5$ as in the previous example.
Divide $p (x)$ by $x^{2} - 2 x + 5$ to get the remaining factor $2 x^{2} + 2 x - 2 = 2 (x^{2} + x - 3)$ .
The quadratic $x^{2} + x - 3$ has discriminant $Δ = 1 + 12 = 13 > 0$ , so it has two real zeros: $x = \frac{- 1 \pm 13}{2}$ .
(a) Factoring over the reals, we leave the pair of conjugate non-real zeros grouped in their irreducible quadratic: $p (x) = 2 (x^{2} - 2 x + 5) (x - \frac{- 1 + 13}{2}) (x - \frac{- 1 - 13}{2})$ , or equivalently $p (x) = 2 (x^{2} - 2 x + 5) (x^{2} + x - 3)$ .
(b) Factoring over the complex numbers, we split all factors into linear terms: $p (x) = 2 (x - (1 + 2 i)) (x - (1 - 2 i)) (x - \frac{- 1 + 13}{2}) (x - \frac{- 1 - 13}{2})$ .

Exam tip: If a question asks for factoring over the reals, do not split the irreducible quadratic into linear factors with complex coefficients — this will cost you points on FRQs.

5. Common Pitfalls (and how to avoid them)

Wrong move: Claiming a 3rd degree polynomial can have 2 total complex zeros (counting multiplicity). Why: Confuses the rule that non-real complex zeros come in pairs with the FTA's total zero count rule. Students forget pairing only applies to non-real zeros, not all complex zeros. Correct move: First apply FTA to get total complex zeros equal to degree, then note that the number of non-real complex zeros must be even.
Wrong move: Given $2 + 3 i$ is a zero, conclude the other conjugate zero is $- 2 + 3 i$ . Why: Students confuse flipping the sign of the real part instead of the imaginary part when finding conjugates. Correct move: Always find the conjugate by flipping only the sign of the imaginary term: $\overline{a + bi} = a - bi$ .
Wrong move: Fails to count multiplicity when asked for the number of complex zeros. Why: Assumes "number of zeros" always means unique zeros. Correct move: Always check the question prompt for wording about uniqueness; if it does not specify unique zeros, default to counting multiplicity per FTA convention.
Wrong move: Claims an odd-degree polynomial with real coefficients can have zero real zeros. Why: Forgets non-real complex zeros come in pairs, so the number of real zeros must have the same parity as the degree. Correct move: Remember that for any odd-degree polynomial with real coefficients, there must be at least one real zero, because you cannot pair all odd number of zeros into non-real conjugate pairs.
Wrong move: When factoring over the reals, writes the linear factors for non-real zeros, claiming they are valid over the reals. Why: Forgets that non-real zeros have non-real coefficients in their linear factors, which are not allowed when factoring over the reals. Correct move: Always leave pairs of conjugate non-real zeros grouped into their irreducible quadratic with real coefficients when factoring over the reals.

6. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

How many non-real complex zeros does the polynomial $p (x) = (x + 3)^{2} (x^{2} - 5 x + 9) (x^{2} + 2 x + 5) (x - 7)$ have, counting multiplicity? A) 2 B) 3 C) 4 D) 5

Worked Solution: Non-real complex zeros only come from quadratic factors with negative discriminant, since linear factors give real zeros, and non-real zeros come in pairs for polynomials with real coefficients. Calculate the discriminant of each quadratic: for $x^{2} - 5 x + 9$ , $Δ = 25 - 36 = - 11 < 0$ , so it has two non-real zeros. For $x^{2} + 2 x + 5$ , $Δ = 4 - 20 = - 16 < 0$ , so it also has two non-real zeros. All other factors are linear or squared linear, which only give real zeros. Adding the non-real zeros gives $2 + 2 = 4$ . The correct answer is C.

Question 2 (Free Response)

Let $p (x) = x^{3} - 5 x^{2} + 11 x - 15$ , and it is known that $x = 1 + 2 i$ is a zero of $p (x)$ . (a) Find all other zeros of $p (x)$ . (b) Write $p (x)$ as a product of linear factors with complex coefficients. (c) Explain why $p (x)$ must cross the x-axis at least once, justifying your answer using properties of complex zeros.

Worked Solution: (a) By the Complex Conjugate Root Theorem, $1 - 2 i$ is also a zero, since $p (x)$ has real coefficients. A cubic has 3 total zeros per FTA, so we only need one more zero. The quadratic factor from the two complex zeros is $x^{2} - 2 x + 5$ . Dividing $p (x)$ by this quadratic gives a quotient of $x - 3$ , so the third zero is $x = 3$ . All other zeros are $1 - 2 i$ and $3$ . (b) Writing as a product of linear factors using all three zeros gives: $p (x) = (x - (1 + 2 i)) (x - (1 - 2 i)) (x - 3)$ . (c) The degree of $p (x)$ is 3, which is odd. Non-real complex zeros of polynomials with real coefficients always come in pairs, so the number of non-real zeros must be even. Subtracting an even number from an odd degree gives an odd number of real zeros, meaning there is at least 1 real zero, which corresponds to an x-intercept. Thus, $p (x)$ must cross the x-axis at least once.

Question 3 (Application / Real-World Style)

An electrical engineer is modeling the transfer function of a filter circuit, where the characteristic polynomial is $P (s) = s^{4} + 6 s^{3} + 15 s^{2} + 18 s + 10$ , with $s$ being the complex frequency variable. A filter is stable if all zeros of $P (s)$ have negative real parts. One zero of $P (s)$ is known to be $s = - 2 + i$ . Find all zeros of $P (s)$ and determine if the filter is stable.

Worked Solution:

By the Complex Conjugate Root Theorem, the conjugate $s = - 2 - i$ is also a zero of $P (s)$ , since $P (s)$ has real coefficients.
The quadratic factor for these two zeros is $(s - (- 2 + i)) (s - (- 2 - i)) = (s + 2)^{2} + 1 = s^{2} + 4 s + 5$ .
Divide $P (s)$ by this quadratic to get the remaining quadratic factor: $\frac{s ^{4} + 6 s ^{3} + 15 s ^{2} + 18 s + 10}{s ^{2} + 4 s + 5} = s^{2} + 2 s + 2$ .
Solve for the zeros of the remaining quadratic: $s = \frac{- 2 \pm 4 - 8}{2} = - 1 \pm i$ .
All four zeros are $- 2 + i, - 2 - i, - 1 + i, - 1 - i$ , all of which have negative real parts.

Interpretation: All zeros of the characteristic polynomial have negative real parts, so the modeled filter is stable.

7. Quick Reference Cheatsheet

Category	Formula / Rule	Notes
Fundamental Theorem of Algebra	Degree $n$ polynomial has exactly $n$ complex zeros (counting multiplicity)	Applies to all non-constant polynomials, all AP Precalculus problems meet this
Complex Conjugate Root Theorem	If $a + bi$ is a zero of a polynomial with real coefficients, $a - bi$ is also a zero	Only applies to polynomials with real coefficients (all AP problems use this)
Quadratic from conjugate zeros	$(x - (a + bi)) (x - (a - bi)) = x^{2} - 2 a x + (a^{2} + b^{2})$	Always gives a quadratic with real coefficients, no complex division needed
Linear Factorization Theorem	$p (x) = a_{n} (x - r_{1}) (x - r_{2}) ... (x - r_{n})$	Full factorization over the complex numbers, all factors are linear
Factoring over the reals	Product of (linear factors from real zeros) × (irreducible quadratics from conjugate complex zeros)	Do not split irreducible quadratics when factoring over the reals
Parity of real zeros	Number of non-real zeros (counting multiplicity) is even; number of real zeros has same parity as degree	Odd-degree polynomials always have at least one real zero
Discriminant for quadratic	$Δ = b^{2} - 4 a c$ for $a x^{2} + b x + c$ : $Δ < 0$ means two non-real complex zeros	Applies only to quadratics with real coefficients

8. What's Next

This topic lays the foundation for all further work with polynomials and rational functions in AP Precalculus, and it is a prerequisite for advanced work with polynomials in calculus and engineering. Immediately after this topic, you will study the Remainder and Factor Theorems and rational roots of polynomials, which rely on understanding the structure of complex zeros to find all roots of high-degree polynomials. This topic also feeds directly into the study of rational functions, where you will use the count of real and complex zeros to find x-intercepts and analyze end behavior, holes, and asymptotes. Without mastering the properties of complex zeros, you will not be able to correctly count intercepts or simplify rational functions completely.

← Back to topic

Stuck on a specific question?
Snap a photo or paste your problem — Ollie (our AI tutor) walks through it step-by-step with diagrams.
Try Ollie free →

Polynomial functions and complex zeros — AP Precalculus Study Guide

1. What Is Polynomial functions and complex zeros?

2. The Fundamental Theorem of Algebra

Worked Example

3. The Complex Conjugate Root Theorem

Worked Example

4. Factoring Over Reals vs. Over Complex Numbers

Worked Example

5. Common Pitfalls (and how to avoid them)

6. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

Question 2 (Free Response)

Question 3 (Application / Real-World Style)

7. Quick Reference Cheatsheet

8. What's Next

More study guides