Topic 1.5A

Polynomial Functions and Complex Zeros

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01
If a linear factor $(x - a)$ is repeated $n$ times, the zero at $x = a$ has multiplicity $n$ .
02
The pattern: odd exponent → crosses. even exponent → bounces. higher multiplicity → flatter/wider near that zero.
03
A degree- $n$ polynomial has exactly $n$ zeros, counting multiplicity.

Practice 24 problems Next: Chapter 1.5B Even and Odd Functions

Zeros, x-Intercepts, and Linear Factors — Three Ways of Saying the Same Thing

If $p(a) = 0$ for a polynomial $p$ , that single fact can be described in three connected ways:

CONCEPT

The Vocabulary Triangle

$a$ is called a ZERO of $p$ ( $a$ is an input that makes the output $0$ )

There is an X-INTERCEPT at the point $(a, 0)$ (this is the graphical picture)

If $a$ is a real number, then $(x - a)$ is a LINEAR FACTOR of $p$ (this is the algebraic picture)

These three ideas are just three different lenses on the same fact — knowing one tells you the other two immediately.

Step-by-Step: Finding x-Intercepts by Factoring

Worked example

Find all the x-intercepts of $g(x) = x^3 - 2x^2 - 8x$ .

01
Factor out the GCF first — every term has an $x$ :
$g(x) = x^3 - 2x^2 - 8x = x(x^2 - 2x - 8) = x(x - 4)(x + 2)$
02
Set each factor equal to zero and solve:
$x = 0, \quad x = 4, \quad x = -2$
03
The x-intercepts are $(0, 0)$ , $(4, 0)$ , and $(-2, 0)$ .

Multiplicity

KEY RULE

If a linear factor $(x - a)$ is repeated $n$ times, the zero at $x = a$ has multiplicity $n$ .

Let's build this up slowly, the way you'd figure it out with a tutor sitting next to you. Forget complicated polynomials for a second — let's just look at the SIMPLEST possible examples: $y = x$ , $y = x^2$ , $y = x^3$ , and $y = x^4$ . Each one has a single zero at $x = 0$ , but with a different multiplicity (1, 2, 3, 4). Watch what happens at that zero in each case:

The simplest possible example of each multiplicity — notice the pattern as the exponent goes up.

CONCEPT

Reading the Pattern, One Graph at a Time

$y = x$ (multiplicity 1): this is just a straight line through the origin. Of course it crosses — a line always goes from negative to positive.

$y = x^2$ (multiplicity 2): this is a parabola sitting right on the x-axis. It touches zero, but since $x^2$ can NEVER be negative, the graph can't dip below the axis — it has to bounce back up.

$y = x^3$ (multiplicity 3): this still crosses (odd power), but notice it looks noticeably FLATTER right around $x=0$ compared to $y=x$ — it hugs the axis a little before finally crossing.

$y = x^4$ (multiplicity 4): still bounces (even power), but again, flatter/wider than $y=x^2$ near the zero.

KEY RULE

The pattern: ODD exponent → crosses. EVEN exponent → bounces. HIGHER multiplicity → flatter/wider near that zero.

That same pattern holds even when several zeros with different multiplicities appear together in one polynomial. Here's a clean side-by-side comparison — one zero with multiplicity 1, another with multiplicity 3, both odd (both cross), but look how much flatter the multiplicity-3 crossing is:

Both zeros cross, but the higher-multiplicity zero ( $x=1$ ) flattens out noticeably before continuing through.

And here's a bounce vs. cross comparison in a single polynomial.

Quick check

For $p(x) = (x + 2)^{3}(x - 5)^{2}$ , at which zero does the graph bounce, and at which does it cross?

Why Multiplicity Controls the Sign

Here's the deeper reason bounce vs. cross happens: look at the output values for inputs near $x = a$ .

CONCEPT

The Rule

Odd multiplicity → the sign of the output FLIPS as you pass through $x = a$ (one side positive, other side negative) → the graph crosses.

Even multiplicity → the output has the SAME sign on both sides of $x = a$ → the graph bounces, never actually crossing to the other sign.

Sign Analysis: Solving Polynomial Inequalities

Once you know the zeros (and their multiplicities), you can figure out the sign of a polynomial on EVERY interval — without plugging in more than one test point per region.

Worked Example 1 — All Odd Multiplicities

Worked example

Given $h(x) = x^2 - 2x - 3$ , find all intervals where $h(x) \ge 0$ .

01
Factor:
$h(x) = x^2 - 2x - 3 = (x - 3)(x + 1)$
02
Zeros at $x = -1$ and $x = 3$ (each multiplicity 1, so the sign flips at each).
03
Pick one test point in each of the 3 regions and check the sign:
04
$h(x)$ is positive on $(-\infty, -1)$ and $(3, \infty)$ , and negative on $(-1, 3)$ .
05
Since we want $h(x) \ge 0$ (including where it equals 0, at the zeros themselves): answer = $(-\infty, -1]$ ∪ $[3, \infty)$ .

Worked Example 2 — Mixing Odd and Even Multiplicities

Worked example

Given $p(x) = (x - 2)(x - 5)^4(x + 7)$ , find all intervals where $p(x) \le 0$ .

01
Zeros: $x = 2$ (multiplicity 1, odd), $x = 5$ (multiplicity 4, even), $x = -7$ (multiplicity 1, odd).
02
Key insight: $(x-5)^4$ is raised to an EVEN power, so it can never be negative — it's always $\ge 0$ . That means it never changes the overall sign; only the two odd-multiplicity factors $(x-2)$ and $(x+7)$ control the sign.
03
Build the sign chart using just $(x-2)(x+7)$ , remembering $p(x)=0$ exactly at $x=5$ :
04
$p(x)$ is negative only on $(-7, 2)$ , and touches 0 (without going negative) at $x=5$ .
05
Since we want $p(x) \le 0$ : the interval $(-7, 2)$ qualifies, PLUS the single isolated point $x=5$ (where $p(x)$ is exactly 0). Answer: $[-7, 2]$ ∪ {5}.

COMMON MISTAKE

An even-multiplicity factor never changes sign — don't treat it like the odd ones when building your sign chart.

Don't forget isolated 'touch' points when solving ≤ 0 or ≥ 0 — a bounce point technically satisfies equality even though it doesn't create a whole interval.

Always double check whether the inequality includes equality (≤, ≥) or not (<, >) — that decides whether zeros themselves are included in the answer.

Complex Zeros: the Fundamental Theorem of Algebra

KEY RULE

A degree- $n$ polynomial has EXACTLY $n$ zeros, counting multiplicity.

CONCEPT

"Complex" Includes Both Kinds

"Complex" refers to BOTH real numbers and non-real numbers — every real number is technically also a complex number.

For a polynomial with REAL coefficients, any non-real zeros always come in conjugate pairs ( $a+bi$ and $a-bi$ ).

So for such a polynomial the number of non-real zeros is always EVEN — you can never have just one. (Drop the real-coefficient condition and this fails: $x - i$ has exactly one non-real zero.)

Reading Real vs. Non-Real Zero Counts from a Graph

The graph only ever shows you the REAL zeros (where it touches or crosses the x-axis). Any zeros you can't see on the graph must be non-real — and the counts always add up to the degree.

Every quadratic and cubic here has zero counts that add up to its degree — count what you can SEE, and the rest must be non-real.

CONCEPT

How to Count

Count each visible x-intercept, using its multiplicity (a bounce still only touches once, but may represent multiplicity 2, 4, etc. — look for how flat the touch is, or use the algebra).

Subtract that total from the degree — whatever's left over must be non-real zeros (and it will always be an even number).

Step-by-Step: Finding the Number of Non-Real Zeros

Worked example

A polynomial has degree 8, with real zeros at $x=-10$ , $x=5$ (multiplicity 2), and $x=16$ . How many non-real zeros does it have?

01
Add up the multiplicities of the known real zeros ( $x=-10$ and $x=16$ are multiplicity 1 unless stated otherwise; $x=5$ is given as multiplicity 2):
$1 + 2 + 1 = 4 \text{ real zeros (with multiplicity)}$
02
Subtract from the total degree:
$8 - 4 = 4 \text{ non-real zeros}$
03
There are 4 non-real zeros (which must form 2 conjugate pairs, since non-real zeros always come in pairs).

Non-Real Zeros Come in Conjugate Pairs

KEY RULE

If $a + bi$ is a non-real zero of a polynomial with REAL coefficients, then $a - bi$ is also a zero.

The real-coefficient condition is what makes this work — without it the pairing does not hold.

Here's the good news: this is one of the easiest "tricks" in the whole unit. To find the conjugate of any complex number, you just flip the sign in front of the imaginary (i) part — the real part never changes.

Worked Example — Walking Through It Step by Step

Worked example

A polynomial (with all real coefficients) has a zero at $2 + 5i$ . What OTHER zero must it also have?

01
Identify the given zero:
$2 + 5i$
02
Split it into its real part ( $a$ ) and imaginary part ( $b$ ):
$a = 2, \quad b = 5$
03
Keep 'a' exactly the same. Flip the sign on 'b' (from +5 to −5).
04
Write the conjugate:
$2 - 5i$
05
So $2 - 5i$ must ALSO be a zero of this polynomial — guaranteed, with no further work needed.

CONCEPT

Why Does This Rule Even Exist? (The Intuition)

Think about it this way: when a polynomial has only real-number coefficients (no i's anywhere in the equation itself), all the i's that show up in the zeros have to perfectly cancel out when you multiply the factors back together.

The only way that cancellation works out perfectly is if every non-real zero is paired up with its mirror-image partner (its conjugate). If you had just ONE non-real zero all by itself with no partner, the i's wouldn't cancel, and you'd end up with an impossible polynomial that has an i floating in one of its coefficients.

So conjugate pairs aren't just a coincidence — they're mathematically required to keep the polynomial's coefficients real.

A Few More for Practice

Same trick every time: copy the real part, flip the sign on the imaginary part.

Given zero	Flip the sign on $b$ ...	Conjugate partner
$-3 + 6i$	$6i \to -6i$	$-3 - 6i$
$4 - 2i$	$-2i \to 2i$	$4 + 2i$
$-7i$ (i.e., $0 - 7i$ )	$-7i \to 7i$	$7i$ (i.e., $0 + 7i$ )

Quick check

A polynomial with real coefficients has degree $4$ and $2 + 3i$ is a zero. How many real zeros can it have?

Solving for Complex Zeros Algebraically

Using the Discriminant to Predict Zero Type

For $ax^2 + bx + c$ , the discriminant $b^2 - 4ac$ tells you the zero type before you even solve.

CONCEPT

Discriminant Rule

$b^2 - 4ac > 0$ → two distinct real zeros

$b^2 - 4ac = 0$ → exactly one real zero (repeated root)

$b^2 - 4ac < 0$ → two complex conjugate zeros (no real zeros)

Worked Example — Solving a Quadratic with Complex Zeros

Worked example

Find the zeros of $f(x) = x^2 + 4$ .

01
Set $f(x) = 0$ :
$x^2 + 4 = 0 \;\Rightarrow\; x^2 = -4$
02
Take the square root of both sides:
$x = \pm\sqrt{-4} = \pm 2i$
03
The zeros are $x = 2i$ and $x = -2i$ — a complex conjugate pair. The graph never crosses the x-axis.

Worked Example — a Cubic with One Real, Two Complex Zeros

Worked example

$f(x) = x^3 - x^2 + x - 1$ has a real zero at $x=1$ . Find the rest.

01
Divide by $(x-1)$ :
$x^3 - x^2 + x - 1 = (x - 1)(x^2 + 1)$
02
Solve the remaining factor:
$x^2 + 1 = 0 \;\Rightarrow\; x^2 = -1 \;\Rightarrow\; x = \pm i$
03
Zeros: $x=1$ (real), $x=i$ and $x=-i$ (complex conjugate pair) — matching the Fundamental Theorem's prediction of 3 total zeros.

Practice Problems

Worked example

Problem 1. Use the discriminant on $x^2 + 6x + 9 = 0$ .

01
Compute:
$b^2 - 4ac = 6^2 - 4(1)(9) = 36 - 36 = 0$
02
Discriminant = 0 → exactly one real zero (repeated root): $x = -3$ .

Worked example

Problem 2. Find the zeros of $g(x) = x^2 + 9$ .

01
Solve:
$x^2 + 9 = 0 \;\Rightarrow\; x^2 = -9 \;\Rightarrow\; x = \pm 3i$
02
Zeros: $x = 3i$ and $x = -3i$ — a complex conjugate pair.

Worked example

Problem 3. $p(x) = (x+2)^3(x-6)^2$ has degree 5. Which zero(s) make the graph bounce, and which cross?

01
$x = -2$ has multiplicity 3 (ODD) → the graph CROSSES here.
02
$x = 6$ has multiplicity 2 (EVEN) → the graph BOUNCES here.

Common slips

An even-multiplicity factor never changes sign — don't treat it like the odd ones when building your sign chart.

Don't forget isolated 'touch' points when solving ≤ 0 or ≥ 0 — a bounce point technically satisfies equality even though it doesn't create a whole interval.

Always double check whether the inequality includes equality (≤, ≥) or not (<, >) — that decides whether zeros themselves are included in the answer.

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14 cards · Polynomials, Zeros and multiplicity

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Recap card

6 lines to re-read the night before.

01
Zero, x-intercept, linear factor are three views of the same fact: $p(a)=0$ .
02
Multiplicity: odd → crosses the x-axis; even → bounces off it.
03
Sign analysis: even-multiplicity factors never flip the sign; only odd-multiplicity factors do.
04
Degree $n$ → exactly $n$ zeros total, counting multiplicity; "complex" includes both real and non-real.
05
Non-real zeros always come in conjugate pairs ( $a+bi$ , $a-bi$ ) — so their count is always even.
06
Reading a graph: count visible real zeros (with multiplicity), subtract from the degree to find the number of non-real zeros.

Practice 24 problems Next: Chapter 1.5B Even and Odd Functions

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Polynomial Functions and Complex Zeros

Zeros, x-Intercepts, and Linear Factors — Three Ways of Saying the Same Thing

Step-by-Step: Finding x-Intercepts by Factoring

Multiplicity

Why Multiplicity Controls the Sign

Sign Analysis: Solving Polynomial Inequalities

Worked Example 1 — All Odd Multiplicities

Worked Example 2 — Mixing Odd and Even Multiplicities

Complex Zeros: the Fundamental Theorem of Algebra

Reading Real vs. Non-Real Zero Counts from a Graph

Step-by-Step: Finding the Number of Non-Real Zeros

Non-Real Zeros Come in Conjugate Pairs

Worked Example — Walking Through It Step by Step

A Few More for Practice

Solving for Complex Zeros Algebraically

Using the Discriminant to Predict Zero Type

Worked Example — Solving a Quadratic with Complex Zeros

Worked Example — a Cubic with One Real, Two Complex Zeros

Practice Problems

Recap card

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