Linear-Algebra-Done-Right · Unit 10 · Video 3 · Interactive Practice

Fundamental Theorem of Algebra and Complete Factorization

IKey Formulas

Formula	Name	What it says
$p \in \mathcal{P}(\mathbf{C})$ nonconstant $\Rightarrow p(\zeta) = 0$ for some $\zeta \in \mathbf{C}$	Fundamental theorem of algebra, first version	A zero exists; the proof gives no method for finding it
$p(z) = c(z - \lambda_1)\cdots(z - \lambda_m)$	Fundamental theorem of algebra, second version	For nonconstant $p \in \mathcal{P}(\mathbf{C})$ of degree $m$ , with $c, \lambda_1, \dots, \lambda_m \in \mathbf{C}$ ; unique up to the order of the factors; $c$ is the coefficient of $z^m$ and the zeros of $p$ are exactly $\lambda_1, \dots, \lambda_m$
$\left(r^{1/k}\left(\cos\tfrac{\theta}{k} + i\sin\tfrac{\theta}{k}\right)\right)^k = r(\cos\theta + i\sin\theta)$	$k$ -th roots from De Moivre	For a positive integer $k$ , every $w = r(\cos\theta + i\sin\theta) \in \mathbf{C}$ ( $r \ge 0$ ) has a $k$ -th root: take the $k$ -th root of the length and divide the angle by $k$
$\beta^k = -\dfrac{1}{a_k}$ , $\ \|q(t\beta)\| \le 1 - t^k(1 - tB)$	The descent step	For $q(z) = 1 + a_k z^k + \cdots + a_m z^m$ with $a_k \neq 0$ and $0 < t < 1$ , where $B > 1$ is a constant with $\|a_{k+1}(t\beta)^{k+1} + \cdots + a_m(t\beta)^m\| \le B\,t^{k+1}$ ; at $t = \tfrac{1}{2B}$ the bound is $1 - \tfrac12 t^k < 1$

Key Insight: The theorem is pure existence. $x^5 - 5x^4 - 6x^3 + 17x^2 + 4x - 7$ has five zeros, approximately $-1.87$ , $-0.74$ , $0.62$ , $1.47$ , $5.51$ , yet no formula built from the coefficients will ever give them exactly.

IIEvery Complex Number Has a $k$ -th Root

If $w = r(\cos\theta + i\sin\theta)$ , the number of length $r^{1/k}$ at angle $\theta/k$ raised to the $k$ -th power is $w$ .

k 3

IIIThe Descent Step: $|q|$ Drops Below $1$

For $q(z) = 1 + z^2 + z^3$ , moving along $t\beta$ with $\beta^2 = -\tfrac{1}{a_2} = -1$ pushes $|q|$ below $1$ .

Direction t 0.25

💡 The proof's $q$ has minimum value $1$ at $0$ ; this step shows that no polynomial with $q(0) = 1$ and some $a_k \neq 0$ can, so the assumption $p(\zeta) \neq 0$ fails.

IVThe Zeros Are Exactly the $\lambda_j$

Shading darkens as $|p(z)|$ shrinks; it reaches $0$ only at the $\lambda_j$ , and each repeated $\lambda_j$ still adds one to $\deg p$ .

deg p

VQuiz Questions

Problem 1 · Factor $z^4 - 16$ Completely

Given: $p(z) = z^4 - 16 \in \mathcal{P}(\mathbf{C})$ — find its factorization $c(z - \lambda_1)\cdots(z - \lambda_4)$ from the second version of the fundamental theorem of algebra.

✅ Correct!

c = 1

and the four zeros are

2

-2

2i

-2i

: each satisfies

z^4 = 16

, and a degree-4 polynomial has four factors.

❌ True, but not complete. The product does equal

z^4 - 16

, but the second version demands degree-one factors, and over

\mathbf{C}

the factor

z^2 + 4

splits as

(z - 2i)(z + 2i)

❌ Not quite.

(z - 2)^2(z + 2)^2 = (z^2 - 4)^2 = z^4 - 8z^2 + 16

. Also

(2i)^4 = 16

, so

2i

must be a zero, and it is missing here.

❌ Check the imaginary zeros.

(z - 4i)(z + 4i) = z^2 + 16

, so this product is

(z^2 - 4)(z^2 + 16) = z^4 + 12z^2 - 64

. The zeros of

z^2 + 4

are

\pm 2i

Show solution

Difference of squares twice:

z^4 - 16 = (z^2 - 4)(z^2 + 4) = (z - 2)(z + 2)(z^2 + 4).

Over $\mathbf{C}$ , $z^2 + 4 = z^2 - (2i)^2 = (z - 2i)(z + 2i)$ , so

z^4 - 16 = (z - 2)(z + 2)(z - 2i)(z + 2i).

The coefficient of $z^4$ is $1$ , so $c = 1$ , and by uniqueness this is the only factorization apart from the order of the factors. The zeros of $p$ are exactly $2$ , $-2$ , $2i$ , $-2i$ .

Problem 2 · Counting Zeros with Repeats

Given: $p(z) = 2(z - i)(z + 1)^2(z - i)(z - 3)$ — find $\deg p$ and the number of distinct zeros of $p$ .

What is deg p?

How many distinct zeros does p have?

✅ Correct! In the form

c(z - \lambda_1)\cdots(z - \lambda_5)

the list is

i, i, -1, -1, 3

: five factors, three distinct zeros.

❌ The constant is not a factor of degree one.

c = 2

contributes nothing to the degree; only the factors

z - \lambda_j

count.

❌ That counts distinct zeros. The degree counts every factor

z - \lambda_j

, repeats included:

z - i

appears twice and

z + 1

appears twice.

❌ Count the square.

(z + 1)^2 = (z + 1)(z + 1)

is two factors, so the factors are

z - i

z + 1

z + 1

z - i

z - 3

❌ Not quite. Multiplying out, the highest power of

z

z^{1 + 2 + 1 + 1}

❌ That counts with repeats. Five is

\deg p

; the values

i

and

-1

each occur twice in the list

\lambda_1, \dots, \lambda_5

❌ The two factors $z - i$ give the same zero. Separated or not, both vanish at

z = i

; the distinct zeros are

i

-1

3

❌ A zero is missing.

p

is zero exactly when some factor is zero: at

i

, at

-1

, and at

3

❌ Not quite. The zeros of

p

are exactly the numbers

\lambda_j

in the factorization; list them without repeats.

Show solution

Collect equal factors:

p(z) = 2(z - i)^2(z + 1)^2(z - 3) = 2(z - i)(z - i)(z + 1)(z + 1)(z - 3).

This is the form $c(z - \lambda_1)\cdots(z - \lambda_m)$ with $c = 2$ and the list $\lambda_1, \dots, \lambda_5 = i, i, -1, -1, 3$ . The degree is the number of degree-one factors, $m = 5$ ; the constant $c$ adds nothing.

Since $c \neq 0$ , $p(z) = 0$ exactly when one factor is zero, so the zeros of $p$ are $i$ , $-1$ , $3$ : three distinct zeros, with $i$ and $-1$ each appearing twice in the list.

Problem 3 · A Quadratic with a Complex Coefficient

Given: $p(z) = z^2 + 2iz + 3$ — find its factorization into degree-one factors.

✅ Correct!

(z - i)(z + 3i) = z^2 + 3iz - iz - 3i^2 = z^2 + 2iz + 3

; the zeros are

i

and

-3i

❌ The signs are swapped.

(z + i)(z - 3i) = z^2 - 2iz + 3

. Its zeros

-i

3i

are the negatives of the zeros of

p

❌ Check $(2i)^2$ .

(2i)^2 = 4i^2 = -4

, not

4

. With

4

the discriminant becomes

4 - 12 = -8

, and this product expands to

z^2 + 2iz + 1

, not

p

❌ Not quite.

(z - i)(z - 3i) = z^2 - 4iz + 3i^2 = z^2 - 4iz - 3

. The two zeros must add to

-2i

and multiply to

3

Show solution

Complete the square, using $i^2 = -1$ :

z^2 + 2iz + 3 = (z + i)^2 - i^2 + 3 = (z + i)^2 + 4.

So $p(z) = 0$ exactly when $(z + i)^2 = -4 = (2i)^2$ , that is $z + i = \pm 2i$ , giving $z = i$ or $z = -3i$ . (The quadratic formula gives the same: the discriminant is $(2i)^2 - 4 \cdot 3 = -16$ , with square roots $\pm 4i$ , and $z = \frac{-2i \pm 4i}{2}$ .)

p(z) = (z - i)(z + 3i).

Check: $(z - i)(z + 3i) = z^2 + 3iz - iz - 3i^2 = z^2 + 2iz + 3$ ✓

Problem 4 · Cube Roots of $-8$ by De Moivre

Given: $p(z) = z^3 + 8$ , whose zeros are the cube roots of $-8$ — find all of them, writing $-8 = 8(\cos\theta + i\sin\theta)$ .

✅ Correct! Length

8^{1/3} = 2

and angles

\tfrac{\pi}{3}

\pi

\tfrac{5\pi}{3}

, so

z^3 + 8 = (z + 2)(z - 1 - \sqrt{3}\,i)(z - 1 + \sqrt{3}\,i)

❌ Two zeros are missing.

z^3 + 8

has degree

3

, so it has three factors

z - \lambda_j

; the other two zeros are not real.

❌ Those two are cube roots of $8$ .

-1 + \sqrt{3}\,i = 2\left(\cos\tfrac{2\pi}{3} + i\sin\tfrac{2\pi}{3}\right)

, and its cube is

8(\cos 2\pi + i\sin 2\pi) = 8

. Start from the angle

\theta = \pi

-8

❌ Not quite.

(2i)^3 = 8i^3 = -8i

, not

-8

. Divide the angle

\pi

-8

(and

\pi + 2\pi

\pi + 4\pi

) by

3

Show solution

In polar form $-8 = 8(\cos\theta + i\sin\theta)$ with $\theta = \pi$ , and equally with $\theta = 3\pi$ or $5\pi$ , since cosine and sine have period $2\pi$ . De Moivre gives a cube root of length $8^{1/3} = 2$ and angle $\theta/3$ for each choice:

\theta = \pi:\ 2\left(\cos\tfrac{\pi}{3} + i\sin\tfrac{\pi}{3}\right) = 1 + \sqrt{3}\,i,

\theta = 3\pi:\ 2(\cos\pi + i\sin\pi) = -2,

\theta = 5\pi:\ 2\left(\cos\tfrac{5\pi}{3} + i\sin\tfrac{5\pi}{3}\right) = 1 - \sqrt{3}\,i.

These are three distinct zeros of the degree-3 polynomial $z^3 + 8$ , so by the second version

z^3 + 8 = (z + 2)\big(z - (1 + \sqrt{3}\,i)\big)\big(z - (1 - \sqrt{3}\,i)\big),

and there are no other zeros. Check: $\big(z - (1 + \sqrt{3}\,i)\big)\big(z - (1 - \sqrt{3}\,i)\big) = (z - 1)^2 + 3 = z^2 - 2z + 4$ , and $(z + 2)(z^2 - 2z + 4) = z^3 + 8$ ✓

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Fundamental Theorem of Algebra and Complete Factorization

IKey Formulas

IIEvery Complex Number Has a kkk-th Root

IIIThe Descent Step: ∣q∣|q|∣q∣ Drops Below 111

IVThe Zeros Are Exactly the λj\lambda_jλj​

VQuiz Questions

IIEvery Complex Number Has a $k$ -th Root

IIIThe Descent Step: $|q|$ Drops Below $1$

IVThe Zeros Are Exactly the $\lambda_j$