多项式函数与复零点

AP 微积分预备· AP 微积分预备 CED — 多项式与有理函数· 14 分钟阅读

1. 代数基本定理★★☆☆☆⏱ 3 min

📘 定义

代数基本定理（FTA）

Every non-constant single-variable polynomial with complex coefficients has at least one complex zero. A key AP Precalculus corollary: A degree $n \geq 1$ polynomial has exactly $n$ complex zeros when counting multiplicity.

代数基本定理证明了复数域对于求解所有多项式方程是"完备的"，而实数域无法求解所有多项式方程。AP考试经常考察总零点（计重数）、唯一零点和实零点之间的表述区分。

📐 例题

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?

1
First calculate the degree of the polynomial by adding the exponents of all factors:
2
$2x^3 \cdot x^2 \cdot 3x^2 = 6x^{3+2+2} = 6x^7$
3
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, so total zeros = 7.
4
To find unique zeros, factor $p (x)$ completely. We already have unique zeros at $x = 0$ (multiplicity 3) and $x = 1$ (multiplicity 2). Factor the remaining quadratic:
5
$3x^2 - 7x + 2 = (3x-1)(x-2)$
6
This gives two more unique zeros at $x = \frac{1}{3}$ and $x = 2$ , for a total of 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 FTA on the AP exam.

2. 复共轭根定理★★★☆☆⏱ 4 min

📘 定义

复共轭根定理

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 is the most commonly tested result for this topic on the AP exam, and it applies to all polynomials you will encounter on the test. It allows you to immediately get a second zero when given one non-real complex zero, so you can form a quadratic factor with real coefficients and avoid complex division.

📐 例题

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 $p (x)$ .

1
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.
2
Calculate the quadratic factor with these two zeros:
3
$(x - (1+2i))(x - (1-2i)) = (x-1)^2 - (2i)^2 = x^2 - 2x + 1 + 4 = x^2 - 2x + 5$
4
Divide $p (x)$ by this quadratic to get the remaining quadratic factor:
5
$\frac{x^4 - 4x^3 + 14x^2 - 28x + 45}{x^2 - 2x + 5} = x^2 - 2x + 9$
6
Solve for the roots of the remaining quadratic:
7
$x = \frac{2 \pm \sqrt{4 - 36}}{2} = 1 \pm 2\sqrt{2}i$
8
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.

3. Factoring Over Reals vs. Complex Numbers★★★☆☆⏱ 4 min

📘 定义

Linear Factorization Theorem

Any degree $n$ polynomial with leading coefficient $a_{n}$ and zeros $r_{1}, r_{2}, ..., r_{n}$ can be factored completely over the complex numbers into $n$ linear factors: $p (x) = a_{n} (x - r_{1}) (x - r_{2}) ... (x - r_{n})$

When factoring over the real numbers, non-real complex zeros produce linear factors with complex coefficients, which are not allowed. Instead, we pair conjugate non-real zeros into irreducible quadratic factors with real coefficients (irreducible means they cannot be factored further into linear terms with real coefficients). A key result: the number of non-real complex zeros (counting multiplicity) is always even, so the number of real zeros has the same parity as the degree of the polynomial. All odd-degree polynomials with real coefficients therefore have at least one real zero.

📐 例题

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.

1
By the Complex Conjugate Root Theorem, $1 - 2 i$ is also a zero, giving the quadratic factor $x^{2} - 2 x + 5$ .
2
Divide $p (x)$ by $x^{2} - 2 x + 5$ to get the remaining factor: $2 x^{2} + 2 x - 2 = 2 (x^{2} + x - 3)$ .
3
The quadratic $x^{2} + x - 3$ has discriminant $Δ = 1 + 12 = 13 > 0$ , so it has two real zeros at $x = \frac{- 1 \pm 13}{2}$ .
4
(a) Factoring over the reals, we leave the conjugate non-real zeros grouped in their irreducible quadratic:
5
$p(x) = 2(x^2 - 2x + 5)(x - \frac{-1 + \sqrt{13}}{2})(x - \frac{-1 - \sqrt{13}}{2}) = 2(x^2 - 2x + 5)(x^2 + x - 3)$
6
(b) Factoring over the complex numbers, we split all factors into linear terms:
7
$p(x) = 2(x - (1+2i))(x - (1-2i))(x - \frac{-1 + \sqrt{13}}{2})(x - \frac{-1 - \sqrt{13}}{2})$

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.

4. AP-Style Concept Check★★★★☆⏱ 3 min

✓ 快速检测

Test your understanding with these AP-style practice questions:

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
显示答案
C) 4 —
Non-real zeros come from quadratics with negative discriminant. Both $x^{2} - 5 x + 9$ ( $Δ = - 11$ ) and $x^{2} + 2 x + 5$ ( $Δ = - 16$ ) have negative discriminant, giving 2 non-real zeros each, for a total of 4.

📐 例题

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.

1
(a) By the Complex Conjugate Root Theorem, $1 - 2 i$ is also a zero. 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 gives a quotient of $x - 3$ , so the third zero is $x = 3$ . All other zeros are $1 - 2 i$ and $3$ .
2
(b) Product of linear factors:
3
$p(x) = (x - (1+2i))(x - (1-2i))(x - 3)$
4
(c) The degree of $p (x)$ is 3, which is odd. Non-real complex zeros come in pairs, so the number of non-real zeros is even. Subtracting an even number from an odd degree gives an odd number of real zeros, meaning there is at least 1 real zero (x-intercept). Thus, $p (x)$ must cross the x-axis at least once.

5. 常见陷阱

错误做法:

Claiming a 3rd degree polynomial can have 2 total complex zeros (counting multiplicity)

原因:

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.

正确做法:

First apply FTA to get total complex zeros equal to degree, then note that the number of non-real complex zeros must be even.

错误做法:

Given $2 + 3 i$ is a zero, conclude the other conjugate zero is $- 2 + 3 i$

原因:

Students confuse flipping the sign of the real part instead of the imaginary part when finding conjugates.

正确做法:

Always find the conjugate by flipping only the sign of the imaginary term: $\overline{a + bi} = a - bi$ .

错误做法:

Fails to count multiplicity when asked for the number of complex zeros

原因:

Assumes "number of zeros" always means unique zeros.

正确做法:

Always check the question prompt for wording about uniqueness; if it does not specify unique zeros, default to counting multiplicity per FTA convention.

错误做法:

Claims an odd-degree polynomial with real coefficients can have zero real zeros

原因:

Forgets non-real complex zeros come in pairs, so the number of real zeros must have the same parity as the degree.

正确做法:

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.

错误做法:

When factoring over the reals, writes the linear factors for non-real zeros, claiming they are valid over the reals

原因:

Forgets that non-real zeros have non-real coefficients in their linear factors, which are not allowed when factoring over the reals.

正确做法:

Always leave pairs of conjugate non-real zeros grouped into their irreducible quadratic with real coefficients when factoring over the reals.

6. 速查表

类别	公式 / 规则	注释
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

下一步

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. Mastery of complex zero properties is required to correctly count intercepts, simplify rational functions completely, and analyze polynomial and rational function behavior on the AP exam.