AP Precalculus 1.5 Polynomial Functions & Complex Zeros — Notes

🎯

Zeros of Polynomials & Multiplicity

When a zero repeats, the graph behaves differently

Zeros of Polynomial Functions

If p(a) = 0, then a is a zero (root) of p(x).

If a is a real number and x = a is a zero of p, then (x − a) is a linear factor of p.

The multiplicity of a zero is the degree of its factor (x − a). So if (x − a)ⁿ appears in p(x), then x = a has multiplicity n.

〰️

Odd Multiplicity

Graph passes through the x-axis at this zero. The function changes sign here. (Multiplicity 1, 3, 5, …)

↩️

Even Multiplicity

Graph is tangent to (bounces off) the x-axis at this zero. The function does NOT change sign here. (Multiplicity 2, 4, 6, …)

📌

Example: y = −.01(x+4)(x+1)³(x−3)²

x = −4: multiplicity 1 (odd) → graph passes through.
x = −1: multiplicity 3 (odd) → graph passes through (but with a flatter S-curve).
x = 3: multiplicity 2 (even) → graph bounces off the x-axis (tangent to x-axis).
Listed: x = −4, x = −1 (mult. 3), x = 3 (mult. 2)

🔢

Complex Roots & Conjugate Pairs

Imaginary roots always come in pairs — you'll never see them on the graph

✨

Complex Roots

Some polynomials have roots involving an imaginary number. Since imaginary numbers don't exist on the real number line, you will never see them on the graph. For example, f(x) = x² + 1 has zeros x = ±i (complex), and its graph never touches the x-axis.

Key Understanding: Conjugate Pairs

All imaginary roots come in conjugate pairs. If a + bi is a root of f(x), then a − bi is also a root.

The conjugate of a + bi is a − bi (flip the sign of the imaginary part only).

📌 Example 1 — Find the conjugate of each complex number.

Complex Number	Its Conjugate	Rule
4i	−4i	Flip sign of imaginary part (a=0, b=4)
−i	i	Flip sign: −1·i → +1·i
2 − 3i	2 + 3i	Keep real part (2), flip imaginary (−3→+3)
−4 + 2i	−4 − 2i	Keep real part (−4), flip imaginary (+2→−2)

Fundamental Theorem of Algebra

A polynomial of degree n has exactly n complex zeros when counting multiplicities.

📌 Example 2 — Graph of f has zeros at x = −1 (bounces — even multiplicity 2) and x = 3 (crosses). It is known that x = i√3 is also a zero. What is the least possible degree?

1

Real zeros from graph: x = −1 (mult. 2) and x = 3 (mult. 1) → contributes 3 zeros.

2

Complex zero given: x = i√3 → its conjugate x = −i√3 must also be a zero → 2 more zeros.

3

Total known zeros: x = −1 (×2), x = 3, x = i√3, x = −i√3 → at least 5 zeros.

Degree n ≥ 5 — at least 5 zeros → least possible degree is 5

📊

Polynomial Inequalities from a Graph

f(x) > 0 means ABOVE x-axis · f(x) < 0 means BELOW x-axis

Key Reminder

When we write f(x), we are referring to the y-value on the graph.

f(x) > 0 → graph is above the x-axis → write x-values where the graph is above it.

f(x) < 0 → graph is below the x-axis → write x-values where the graph is below it.

f(x) = 0 → graph is on the x-axis → the zeros of f.

📌 Reading inequalities from the graph of f(x) — zeros at x = −3, −1, 3:

a) Where is f(x) = 0?

x = −3, −1, 3

b) Where is f(x) > 0? (above x-axis)

−3 < x < −1 and x > 3

c) Where is f(x) ≤ 0? (on or below)

x ≤ −3 and −1 ≤ x ≤ 3

➕➖

Solving Polynomial Inequalities — Sign Chart

4-step method using test values in each interval

📋 4 Steps for Solving Nonlinear (Polynomial) Inequalities

1

Solve f(x) = 0 to find the zeros (the boundary points).

2

Create a sign chart — place the zeros on a number line to divide it into intervals.

3

Test a value from each interval in the original expression. Record the sign (+/−).

4

Interpret the sign chart to answer the inequality. Write in interval notation and check endpoints (include for ≤ or ≥, exclude for < or >).

📌 Example 3 — Solve (x − 3)(x + 1)(x + 4) > 0

(x − 3)(x + 1)(x + 4) > 0

1

Zeros: x = −4, x = −1, x = 3 (all multiplicity 1 — sign changes at each).

2&3

Sign chart — test a value in each region:

x<−4

−

|

x=−4

0

|

−4<x<−1

+

|

x=−1

0

|

−1<x<3

−

|

x=3

0

|

x>3

+

4

Want > 0 (positive). The positive intervals are (−4,−1) and (3,∞). Strict inequality → open endpoints.

Answer: (−4, −1) ∪ (3, ∞) → −4 < x < −1 and x > 3

📌 Example 4 — Solve (x + 2)²(x − 5) ≤ 0

(x + 2)²(x − 5) ≤ 0

1

Zeros: x = −2 (mult. 2 — even, sign does NOT change here), x = 5 (mult. 1 — sign changes).

2&3

Sign chart:

x<−2

−

|

x=−2

0

|

−2<x<5

−

|

x=5

0

|

x>5

+

4

Want ≤ 0 (negative or zero). Negative regions: (−∞,−2) and (−2,5). At x=−2: equals 0 ✓. At x=5: equals 0 ✓. Include both endpoints.

Answer: (−∞, 5] → x ≤ 5

⚠️

Even multiplicity zeros don't change sign!

In Example 4, (x+2)² is always ≥ 0. The sign of the whole expression only flips at x = 5. So the expression is negative on both sides of x = −2 — the sign stays negative. This is why the answer is the simple x ≤ 5, not a union of intervals.

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Finding Degree from a Table — Example 5

Successive differences — the number of rounds needed = the degree

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Successive Differences Method

Given a table with equal-width input intervals, subtract consecutive outputs to get 1st differences. If those aren't constant, subtract again for 2nd differences. Keep going until the differences are constant. The round number where differences become constant = the degree of the polynomial.

a) f(x) Degree 2

x	f(x)	1st Diff	2nd Diff
1	−2
3	−3	−1
5	−1	+2	+3
7	4	+5	+3
9	12	+8	+3

Degree 2 — 2nd differences are constant (+3). One more round needed than for linear.

b) g(x) Degree 3

x	g(x)	1st	2nd	3rd
0	−2
3	0	+2
6	10	+10	+8
9	27	+17	+7	−1
12	50	+23	+6	−1

Degree 3 — 3rd differences are constant (−1). Three rounds needed.

🔄

Even and Odd Functions

Symmetry about the y-axis (even) or origin (odd)

Even and Odd Functions — Symmetry

📐 Even Functions

Symmetric over the y-axis. Left and right sides mirror each other.

Test: f(−x) = f(x) for all x.

Example: f(x) = x⁴ − 8x² + 1

f(−x) = f(x) ✓

🌀 Odd Functions

Symmetric about the origin. Rotate 180° — looks the same.

Test: f(−x) = −f(x) for all x.

Example: g(x) = x³ − 9x

f(−x) = −f(x) ✓

🔑

Quick Test Method

Plug in x = −1 (or any convenient value) and x = 1. If f(−1) = f(1) → even. If f(−1) = −f(1) → odd. If neither holds → neither. (Always verify with algebraic substitution for full credit.)

📌 Example 6 — Determine if each polynomial is even, odd, or neither.

h(x) = 2x⁴ − x² + 5

h(−1) = 2(1) − 1 + 5 = 6
h(1) = 2(1) − 1 + 5 = 6
h(−1) = h(1) → f(−x) = f(x)

Even Function

k(x) = x³ + 3x − 1

k(−1) = −1 − 3 − 1 = −5
k(1) = 1 + 3 − 1 = 3
−5 ≠ 3, −5 ≠ −3 → neither

Neither

Graph c) — from graph

f(−1) = 3
f(1) = −3
f(−1) = −f(1) → f(−x) = −f(x)

Odd Function

Graph d) — from graph

f(−1) = 1
f(1) = −3
1 ≠ −3, 1 ≠ −(−3) = 3 → neither

Neither

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Quick Reference

Screenshot and save this!

🔑 Key Rules

Odd mult → passes through x-axis

Even mult → bounces (tangent to x-axis)

Imaginary roots come in conjugate pairs

Degree n → exactly n complex zeros

f(−x)=f(x) → even · f(−x)=−f(x) → odd

Successive diffs constant at round n → degree n

⚠️ Common Mistakes

❌ Forgetting to include conjugate pair

If i√3 is a zero, then −i√3 is ALSO a zero. Always add conjugate pairs when counting zeros for degree questions.

❌ Even multiplicity changes sign

Even multiplicity → graph BOUNCES. The expression does NOT change sign at that zero. Sign stays the same on both sides.

❌ Using ≤ with strict inequality answer

Check endpoints! For ≤ or ≥, include zeros (closed brackets). For < or >, exclude zeros (open brackets).

❌ Testing even/odd with just one value

Algebraically substitute f(−x) and simplify. A single test point can give misleading results — use the definition.

Polynomial Functions& Complex Zeros

Example: y = −.01(x+4)(x+1)³(x−3)²

Complex Roots

Even multiplicity zeros don't change sign!

Successive Differences Method

Quick Test Method

Polynomial Functions
& Complex Zeros