# Polynomial functions and complex zeros

> AP Precalculus · CED — Polynomial and Rational Functions
> Source: https://www.owlsprep.com/study/ap-precalculus-u1-polynomial-functions-and-complex-zeros/

Everything a polynomial's zeros tell you: how zeros and linear factors determine each other, what multiplicity means and why a degree-n polynomial has exactly n complex zeros, how real zeros place x-intercepts and bound the intervals where a polynomial inequality holds, why non-real zeros arrive in conjugate pairs, when the graph crosses the x-axis versus resting tangent to it, how successive differences reveal the degree, and how to tell whether a polynomial function is even or odd.

**Prerequisites:** Factoring quadratics and recognising a difference of squares; Arithmetic with complex numbers, including i² = −1; Reading local and global extrema from a polynomial graph (Topic 1.4)

## Learning objectives

- Connect the zeros of a polynomial function to its linear factors, in both directions.
- Use multiplicity to account for all n complex zeros of a degree-n polynomial function.
- Locate x-intercepts from real zeros and use those zeros as endpoints of the intervals where a polynomial inequality holds.
- Pair each non-real zero with its complex conjugate.
- Decide from the multiplicity of a real zero whether the graph crosses the x-axis or is tangent to it.
- Find the degree of a polynomial function from successive differences of outputs over equal input steps.
- Determine whether a polynomial function is even or odd, graphically and analytically.

## Zeros, Linear Factors, and Multiplicity

A **zero** of a polynomial function $p$ is an input that makes the output $0$. The same number is called a **root** of the equation $p(x)=0$ — two names for one idea. Zeros may be real or non-real complex numbers.

**Zero and linear factor** — If $a$ is a complex number and $p(a)=0$, then $a$ is a **zero** of $p$. If $a$ is a **real** number, then $(x-a)$ is a **linear factor** of $p$ **if and only if** $a$ is a zero of $p$.

*Notation:* p(a)=0 \iff (x-a) \mid p(x)

> **tip**
>
> Read that biconditional both ways — it is what makes factoring useful. Spotting a factor $(x-3)$ tells you $3$ is a zero; discovering the zero $3$ tells you $(x-3)$ divides $p$.

A factor can appear more than once, and that repetition is recorded by the **multiplicity**.

**Multiplicity** — If the linear factor $(x-a)$ is repeated $n$ times in the factorisation, the corresponding zero $a$ has **multiplicity** $n$. Counting multiplicities, a polynomial function of **degree $n$ has exactly $n$ complex zeros**.

*Notation:* \deg p = n \;\Longrightarrow\; n \text{ complex zeros, counted with multiplicity}

> **warning**
>
> Watch the word **complex** here. In mathematics *complex* **includes real** — a real number is just a complex number with no imaginary part. So *$n$ complex zeros* means the **total** number of zeros is $n$, real and non-real together; it is not a count of only the ones with an $i$. In $(x+2)(x-1)^2(x^2+4)$ that total splits as three real ($-2,1,1$) plus two non-real ($\pm 2i$), which is $5$ — the degree.

**Worked example:** List all complex zeros of $p(x) = (x+2)(x-1)^2(x^2+4)$, with their multiplicities, and check the count against the degree.

1. Read the zeros off the factors. From $(x+2)$: the zero $-2$, appearing once, so multiplicity $1$.
2. From $(x-1)^2$: the zero $1$, with multiplicity $2$.
3. From $x^2+4$: solve $x^2=-4$, giving the non-real zeros $2i$ and $-2i$, each of multiplicity $1$.
4. Now the count. Multiplying the factors gives degree $1+2+2=5$, and the zeros counted with multiplicity are $-2,\ 1,\ 1,\ 2i,\ -2i$ — exactly $5$, as the degree requires.
5. **Answer:** $-2$ (multiplicity 1), $1$ (multiplicity 2), $2i$ and $-2i$ (each multiplicity 1); five complex zeros for a degree-5 polynomial.

## Real Zeros: x-Intercepts and Polynomial Inequalities

Only the **real** zeros are visible on the graph, and each one puts the curve on the $x$-axis.

**Real zeros on the graph** — If $a$ is a real zero of $p$, then the graph of $y=p(x)$ has an **$x$-intercept** at the point $(a,\ 0)$. Consequently the real zeros are the **endpoints of the intervals** on which $p(x)\ge 0$ or $p(x)\le 0$.

*Notation:* p(a)=0,\ a \in \mathbb{R} \;\Longrightarrow\; (a,\,0) \text{ on the graph}

The reason the second half follows: a polynomial can only change sign by passing through $0$, so between consecutive real zeros the output keeps one sign throughout. That turns solving an inequality into testing one point per interval.

**Worked example:** For $p(x) = (x+2)(x-1)^2(x^2+4)$, solve $p(x) \le 0$.

1. The real zeros are $-2$ and $1$ (the factor $x^2+4$ is never zero for real $x$, since $x^2+4 \ge 4$). They cut the number line into $x<-2$, $-2<x<1$ and $x>1$.
2. Test one value in each interval. At $x=-3$: $(-1)(16)(13) < 0$. At $x=0$: $(2)(1)(4) = 8 > 0$. At $x=2$: $(4)(1)(8) > 0$.
3. So $p$ is negative only on $x<-2$, and it equals $0$ at the zeros themselves, which the $\le$ sign includes.
4. **Answer:** $p(x)\le 0$ for $x \le -2$, together with the single point $x=1$.

> **warning**
>
> Do not forget an isolated point like $x=1$ above. There $p(x)=0$, so it satisfies $p(x)\le 0$ even though $p$ is positive on both sides of it — a direct consequence of the even multiplicity in the next section.

## Non-Real Zeros Come in Conjugate Pairs

Non-real zeros never appear alone in a polynomial with real coefficients.

**Conjugate pairs** — If $a+bi$ is a non-real zero of a polynomial function $p$, then its **conjugate** $a-bi$ is also a zero of $p$.

*Notation:* p(a+bi)=0 \;\Longrightarrow\; p(a-bi)=0

This is why our running example contains $x^2+4$: the pair $2i$ and $-2i$ multiplies back to a real quadratic, $(x-2i)(x+2i)=x^2+4$. Every conjugate pair collapses into one real quadratic factor in the same way.

**Worked example:** A polynomial function $q$ of degree $3$ with real coefficients has zeros $4$ and $1-3i$. Find its third zero and write $q$ as a product of factors with real coefficients.

1. Non-real zeros come in pairs, so the conjugate $1+3i$ must also be a zero. That is three zeros for a degree-3 polynomial — the full set.
2. Combine the conjugate pair into a real quadratic:
3. $$\bigl(x-(1-3i)\bigr)\bigl(x-(1+3i)\bigr) = (x-1)^2 - (3i)^2 = x^2-2x+1+9 = x^2-2x+10$$
4. **Answer:** the third zero is $1+3i$, and $q(x) = a(x-4)(x^2-2x+10)$ for some nonzero real constant $a$.

A polynomial has **odd degree** when its highest power is an odd number — degree $3$, $5$, $7$, and so on. That single fact, together with conjugate pairs, forces something about its real zeros.

> **tip**
>
> A count follows from this: non-real zeros arrive two at a time, so an **odd-degree** polynomial with real coefficients always has at least one real zero — its graph must cross the $x$-axis somewhere.

## Even Multiplicity: Tangent Instead of Crossing

Multiplicity does more than balance the count — it decides what the graph does at that intercept.

**Even multiplicity** — If the real zero $a$ has **even multiplicity**, then the signs of the output values are **the same** for input values near $x=a$ on both sides. For these polynomial functions the graph is **tangent to the $x$-axis** at $x=a$ — it touches and turns back rather than crossing.

*Notation:* \text{even multiplicity at } a \;\Longrightarrow\; \text{graph tangent to the } x\text{-axis at } (a,0)

**Worked example:** For $p(x) = (x+2)(x-1)^2(x^2+4)$, decide at each real zero whether the graph crosses the $x$-axis or is tangent to it.

1. At $x=-2$ the factor is $(x+2)^1$ — multiplicity $1$, which is odd. Check the signs: $p(-2.1) \approx -8.1$ and $p(-1.9) \approx 6.4$, opposite signs, so the graph **crosses** there.
2. At $x=1$ the factor is $(x-1)^2$ — multiplicity $2$, which is even. Check the signs: $p(0.9) \approx 0.14$ and $p(1.1) \approx 0.16$, both positive.
3. Same sign on both sides means the graph comes down to the axis at $(1,0)$ and goes back up without passing through: it is **tangent** to the $x$-axis.
4. **Answer:** crosses at $(-2,0)$; tangent at $(1,0)$.

> **tip**
>
> This is what lets you sketch a polynomial from its factored form alone: each real zero is an $x$-intercept, and its multiplicity tells you whether the curve passes through or bounces off.

## Finding the Degree from Successive Differences

If you are handed a table of values rather than a formula, the degree is still recoverable — by differencing, the same technique that showed a quadratic's rates form a linear pattern in Topic 1.3.

**Successive differences** — Take output values over **equal-interval** input values and difference them repeatedly. The **degree** of the polynomial function is the **least value $n$ for which the successive $n$th differences are constant**.

*Notation:* n\text{th differences constant} \;\Longrightarrow\; \deg p = n

This is not a rule to memorise — you can see **why** it works. Subtracting neighbouring values of a general quadratic $f(x)=ax^2+bx+c$ gives $f(x+1)-f(x) = 2ax+(a+b)$: the $ax^2$ term appears in both and cancels exactly, leaving something one degree lower. The same top-term cancellation happens at every degree, so **each round of differencing drops the degree by one**. A degree-$n$ polynomial therefore descends $n \to n-1 \to \cdots \to 0$, reaching a **constant** (a degree-$0$ row) after exactly $n$ rounds — which is why the first constant row tells you the degree.

> **tip**
>
> This recovers the **degree**, not the polynomial itself. A table of values plus its differences tells you a cubic is a cubic, but finding the actual formula is a separate task.

**Worked example:** A polynomial function $f$ has these values at equally spaced inputs. Find its degree.

| $x$ | $0$ | $1$ | $2$ | $3$ | $4$ |
|---|---|---|---|---|---|
| $f(x)$ | $1$ | $2$ | $9$ | $28$ | $65$ |

1. First differences: $2-1=1$, $9-2=7$, $28-9=19$, $65-28=37$ — that is $1,\ 7,\ 19,\ 37$, not constant.
2. Second differences: $7-1=6$, $19-7=12$, $37-19=18$ — that is $6,\ 12,\ 18$, still not constant.
3. Third differences: $12-6=6$, $18-12=6$ — constant at $6$.
4. The third differences are the first ones to be constant, so the least such $n$ is $3$.
5. **Answer:** $f$ has degree $3$. (The values come from $f(x)=x^3+1$.)

> **warning**
>
> The inputs must be **equally spaced** for this to work. If the $x$-values jump irregularly, difference the outputs and you will get nothing meaningful.

## Even and Odd Polynomial Functions

> **warning**
>
> One word of warning before we start: the **even** in *even function* is **not** the **even** in *even multiplicity* from earlier. Even multiplicity counted how many times a factor repeats; an even *function* is about a symmetry of the whole graph. Same word, unrelated ideas.

The last question this topic asks about a polynomial is whether it has a symmetry. There are two named cases, each with a graphical description and an algebraic test.

The quickest way to feel the difference is to feed in a pair of **opposite inputs** and compare outputs. For $g(x)=x^4-3x^2$: $g(2)=4$ and $g(-2)=4$ — the same. For $h(x)=x^3-3x$: $h(2)=2$ but $h(-2)=-2$ — opposite. Those two behaviours are exactly what the names below describe.

**Even function** — An **even** function is graphically symmetric over the line $x=0$ (the $y$-axis) and analytically satisfies $f(-x)=f(x)$. If $n$ is even, then $p(x)=a_nx^n$ with $n \ge 1$ and $a_n \ne 0$ is an even function.

*Notation:* f(-x)=f(x)

**Odd function** — An **odd** function is graphically symmetric about the point $(0,0)$ (a half-turn about the origin) and analytically satisfies $f(-x)=-f(x)$. If $n$ is odd, then $p(x)=a_nx^n$ with $n \ge 1$ and $a_n \ne 0$ is an odd function.

*Notation:* f(-x)=-f(x)

**Worked example:** Decide whether each function is even, odd, or neither: (a) $f(x)=3x^4-2x^2$, (b) $g(x)=x^3-4x$, (c) $h(x)=x^2+x$.

1. (a) Replace $x$ by $-x$: $f(-x)=3(-x)^4-2(-x)^2 = 3x^4-2x^2 = f(x)$. Every exponent is even, so nothing changes sign — $f$ is **even**.
2. (b) $g(-x)=(-x)^3-4(-x) = -x^3+4x = -(x^3-4x) = -g(x)$. Every exponent is odd, so every term flips sign — $g$ is **odd**.
3. (c) $h(-x)=(-x)^2+(-x) = x^2-x$. This is neither $h(x)$ nor $-h(x)$, because the exponents are mixed — $h$ is **neither**.
4. **Answer:** (a) even, (b) odd, (c) neither.

> **tip**
>
> For a polynomial written as a sum of powers there is a quick read: all exponents even → even function; all exponents odd → odd function; a mix → neither. A nonzero constant term counts as $x^0$, an even power.

**Check your understanding**

Test your understanding with this multiple-choice question:

1. A polynomial function $p$ with real coefficients has degree $4$, a zero of multiplicity $2$ at $x=3$, and a zero at $x=-i$. Which statement must be true?

   - $p$ has four distinct zeros
   - The graph of $p$ is tangent to the $x$-axis at $(3,0)$
   - The graph of $p$ crosses the $x$-axis at $(3,0)$
   - $p$ is an even function

   *Answer:* The graph of $p$ is tangent to the $x$-axis at $(3,0)$

   *Why:* The zero at $x=3$ has even multiplicity, so the outputs keep the same sign on both sides and the graph is tangent to the axis there. The zeros are $3, 3, -i, i$ — only three distinct values, and the conjugate $i$ is forced by $-i$. Nothing here makes $p$ even.

## Common pitfalls

- **Wrong:** Counting only the distinct zeros of a polynomial.
  - Why it fails: The degree counts zeros **with multiplicity**. For $(x+2)(x-1)^2(x^2+4)$ the distinct zeros number four, but the degree is $5$ and the zero $1$ must be counted twice.
  - Correct: List every zero as many times as its factor is repeated, then check the total equals the degree.
- **Wrong:** Assuming the graph crosses the $x$-axis at every real zero.
  - Why it fails: Even multiplicity keeps the output sign the same on both sides, so the graph is tangent to the axis and turns back instead of passing through.
  - Correct: Read the multiplicity of each real zero first: odd → crosses, even → tangent.
- **Wrong:** Giving a non-real zero without its conjugate.
  - Why it fails: For real coefficients, non-real zeros occur in pairs. Reporting $1-3i$ as a zero but not $1+3i$ leaves the polynomial with the wrong number of zeros.
  - Correct: Pair every non-real zero with its conjugate, and combine the pair into a real quadratic when factoring over the reals.
- **Wrong:** Differencing a table whose inputs are not equally spaced.
  - Why it fails: The successive-differences test for the degree assumes equal input steps; with irregular spacing the differences carry no information about the degree.
  - Correct: Check the $x$-values step by a constant amount first, then difference the outputs until a row is constant.
- **Wrong:** Calling any polynomial with an even degree an even function.
  - Why it fails: Degree and parity are different things. $h(x)=x^2+x$ has even degree but $h(-x)=x^2-x \ne h(x)$, so it is neither even nor odd.
  - Correct: Test $f(-x)$ against $f(x)$ and $-f(x)$, or check that the exponents present are all even (even function) or all odd (odd function).

## Cheatsheet

| Concept | Statement | Key Notes |
| --- | --- | --- |
| Zero / root | $p(a)=0$ | Same idea, two names; may be real or non-real |
| Linear factor | For real $a$: $(x-a)$ is a factor $\iff$ $a$ is a zero | Works in both directions |
| Multiplicity | $(x-a)$ repeated $n$ times | Zero $a$ counted $n$ times |
| Zero count | Degree $n$ $\Rightarrow$ exactly $n$ complex zeros | Counted with multiplicity |
| Real zero on the graph | $x$-intercept at $(a,0)$ | Real zeros bound the intervals for $p(x)\ge 0$ / $\le 0$ |
| Conjugate pairs | $a+bi$ a zero $\Rightarrow$ $a-bi$ a zero | So odd degree forces at least one real zero |
| Odd multiplicity | Sign changes across the zero | Graph **crosses** the $x$-axis |
| Even multiplicity | Sign stays the same across the zero | Graph is **tangent** to the $x$-axis |
| Degree from a table | Least $n$ with constant $n$th differences | Inputs must be equally spaced |
| Even function | $f(-x)=f(x)$ | Symmetric over $x=0$; $a_nx^n$ with $n$ even |
| Odd function | $f(-x)=-f(x)$ | Symmetric about $(0,0)$; $a_nx^n$ with $n$ odd |

## What's next

You can now read a polynomial's zeros in full — how many there are, which are visible on the graph, which arrive in pairs, and where the curve crosses versus touches — plus recover the degree from a table and test for symmetry. Topic 1.6, Polynomial Functions and End Behavior, turns to what happens at the far ends of the graph, where the leading term takes over.

- [](https://www.owlsprep.com/study/ap-precalculus-u1-polynomial-functions-and-end-behavior/)

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