Polynomial functions and complex zeros

AP Precalculus· Unit 1 · Polynomial & Rational Functions· 14 min read

1. Zeros, Linear Factors, and Multiplicity★★☆☆☆⏱ 4 min

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.

📘 Definition

Zero and linear factor

p(a)=0 \iff (x-a) \mid p(x)

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 $p$ .

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

📘 Definition

Multiplicity

\deg p = n \;\Longrightarrow\; n \text{ complex zeros, counted with 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.

📐 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 $2 i$ and $- 2 i$ , each of multiplicity $1$ .
4
Now the count. Multiplying the factors gives degree $1 + 2 + 2 = 5$ , and the zeros counted with multiplicity are $- 2, 1, 1, 2 i, - 2 i$ — exactly $5$ , as the degree requires.
5
Answer: $- 2$ (multiplicity 1), $1$ (multiplicity 2), $2 i$ and $- 2 i$ (each multiplicity 1); five complex zeros for a degree-5 polynomial.

2. Real Zeros: x-Intercepts and Polynomial Inequalities★★☆☆☆⏱ 3 min

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

📘 Definition

Real zeros on the graph

p(a)=0,\ a \in \mathbb{R} \;\Longrightarrow\; (a,\,0) \text{ 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 $p (x) \geq 0$ or $p (x) \leq 0$ .

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 $p (x) \leq 0$ .

1
The real zeros are $- 2$ and $1$ (the factor $x^{2} + 4$ is never zero for real $x$ , since $x^{2} + 4 \geq 4$ ). 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 $\leq$ sign includes.
4
Answer: $p (x) \leq 0$ for $x \leq - 2$ , together with the single point $x = 1$ .

3. Non-Real Zeros Come in Conjugate Pairs★★☆☆☆⏱ 3 min

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

📘 Definition

Conjugate pairs

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

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

This is why our running example contains $x^{2} + 4$ : the pair $2 i$ and $- 2 i$ multiplies back to a real quadratic, $(x - 2 i) (x + 2 i) = x^{2} + 4$ . 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 $1 - 3 i$ . 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 $1 + 3 i$ 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 $1 + 3 i$ , and $q (x) = a (x - 4) (x^{2} - 2 x + 10)$ 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.

4. Even Multiplicity: Tangent Instead of Crossing★★★☆☆⏱ 3 min

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

📘 Definition

Even multiplicity

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

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.

📐 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 $(1, 0)$ .

5. Finding the Degree from Successive Differences★★★☆☆⏱ 3 min

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.

📘 Definition

Successive differences

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

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.

This is not a rule to memorise — you can see why it works. Subtracting neighbouring values of a general quadratic $f (x) = a x^{2} + b x + c$ gives $f (x + 1) - f (x) = 2 a x + (a + b)$ : the $a x^{2}$ 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 $n \to n - 1 \to \dots \to 0$ , reaching a constant (a degree- $0$ row) after exactly $n$ rounds — which is why the first constant row tells you the degree.

📐 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 $f (x) = x^{3} + 1$ .)

6. Even and Odd Polynomial Functions★★☆☆☆⏱ 3 min

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 $g (x) = x^{4} - 3 x^{2}$ : $g (2) = 4$ and $g (- 2) = 4$ — the same. For $h (x) = x^{3} - 3 x$ : $h (2) = 2$ but $h (- 2) = - 2$ — opposite. Those two behaviours are exactly what the names below describe.

📘 Definition

Even function

f(-x)=f(x)

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 $p (x) = a_{n} x^{n}$ with $n \geq 1$ and $a_{n} \neq = 0$ is an even function.

📘 Definition

Odd function

f(-x)=-f(x)

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 $p (x) = a_{n} x^{n}$ with $n \geq 1$ and $a_{n} \neq = 0$ is an odd function.

📐 Worked Example

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

1
(a) Replace $x$ by $- x$ : $f (- x) = 3 (- x)^{4} - 2 (- x)^{2} = 3 x^{4} - 2 x^{2} = f (x)$ . Every exponent is even, so nothing changes sign — $f$ is even.
2
(b) $g (- x) = (- x)^{3} - 4 (- x) = - x^{3} + 4 x = - (x^{3} - 4 x) = - g (x)$ . 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.

✓ Quick check

Test your understanding with this multiple-choice question:

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
Reveal answer
1 —
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.

7. Common Pitfalls

Wrong move:

Counting only the distinct zeros of a polynomial.

Why:

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 move:

List every zero as many times as its factor is repeated, then check the total equals the degree.

Wrong move:

Assuming the graph crosses the $x$ -axis at every real zero.

Why:

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 move:

Read the multiplicity of each real zero first: odd → crosses, even → tangent.

Wrong move:

Giving a non-real zero without its conjugate.

Why:

For real coefficients, non-real zeros occur in pairs. Reporting $1 - 3 i$ as a zero but not $1 + 3 i$ leaves the polynomial with the wrong number of zeros.

Correct move:

Pair every non-real zero with its conjugate, and combine the pair into a real quadratic when factoring over the reals.

Wrong move:

Differencing a table whose inputs are not equally spaced.

Why:

The successive-differences test for the degree assumes equal input steps; with irregular spacing the differences carry no information about the degree.

Correct move:

Check the $x$ -values step by a constant amount first, then difference the outputs until a row is constant.

Wrong move:

Calling any polynomial with an even degree an even function.

Why:

Degree and parity are different things. $h (x) = x^{2} + x$ has even degree but $h (- x) = x^{2} - x \neq = h (x)$ , so it is neither even nor odd.

Correct move:

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).

8. Quick Reference 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 $⟺$ $a$ 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 $p (x) \geq 0$ / $\leq 0$
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 $x = 0$ ; $a_{n} x^{n}$ with $n$ even
Odd function	$f (- x) = - f (x)$	Symmetric about $(0, 0)$ ; $a_{n} x^{n}$ 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.