Linear-Algebra-Done-Right · Unit 10 · Video 2 · Interactive Practice

Zeros, Factors and the Division Algorithm via a Basis

IKey Formulas

Formula	Name	Hypotheses
$z^k - \lambda^k = (z - \lambda)\sum_{j=1}^{k} \lambda^{j-1} z^{k-j}$	Difference of powers; the sum has degree $k - 1$ , leading term $z^{k-1}$	$k \geq 1$ , $z, \lambda \in \mathbf{F}$
$p(\lambda) = 0 \iff p(z) = (z - \lambda)\,q(z)$	Each zero gives a degree-one factor; $\deg q = m - 1$	$\deg p = m \geq 1$ , $\lambda \in \mathbf{F}$
$\deg p = m \;\Rightarrow\;$ at most $m$ zeros	Zero count; hence coefficients and degree are unique	$m \geq 1$ , induction on $m$
$p = sq + r$ , $\ \deg r < \deg s$	Division algorithm; $q$ and $r$ unique	$p, s \in \mathcal{P}(\mathbf{F})$ , $s \neq 0$

Key Insight: When $n = \deg p \geq m = \deg s$ , the list $1, z, \ldots, z^{m-1}, s, zs, \ldots, z^{n-m}s$ has exactly one polynomial of each degree $0, 1, \ldots, n$ , so it is linearly independent of length $n + 1 = \dim \mathcal{P}_n(\mathbf{F})$ , hence a basis. The coordinates of $p$ on the first $m$ entries form $r$ ; the coordinates on the $s$ -entries form $q$ .

IIEvery Difference of Powers Has the Factor $z - \lambda$

Multiplying $z - \lambda$ by $\sum_{j=1}^{k} \lambda^{j-1} z^{k-j}$ leaves only $z^k - \lambda^k$ : the middle terms cancel in pairs.

power k 3 pairs cancelled 0

IIIA Zero Splits Off a Factor

$p(z) - p(\lambda)$ always has the factor $z - \lambda$ , and it equals $p$ exactly when $p(\lambda) = 0$ .

λ 0.5

💡 Challenge: find all three zeros of $p$ . Since $\deg p = 3$ , the zero count says there is no fourth.

IVDividing by Reading Off Coordinates

In the basis $1, \ldots, z^{m-1}, s, zs, \ldots, z^{n-m}s$ , the coordinates of $p$ on the $s$ -entries form $q$ ; the rest form $r$ .

example step 0

VQuiz Questions

Problem 1 · Spot the Factor

Given: $p(z) = z^3 - 7z + 6$ — find the polynomial that is a factor of $p$ .

✅ Correct!

p(-3) = -27 + 21 + 6 = 0

, so

-3

is a zero and

z - (-3) = z + 3

is a factor:

p(z) = (z + 3)(z^2 - 3z + 2)

❌ Check the sign.

z - 3

is a factor exactly when

p(3) = 0

, but

p(3) = 27 - 21 + 6 = 12

. The zero is

-3

, and the factor it gives is

z - (-3)

❌ $-1$ is not a zero.

z + 1 = z - (-1)

is a factor exactly when

p(-1) = 0

, but

p(-1) = -1 + 7 + 6 = 12

❌ The constant term is not a zero.

z - 6

is a factor exactly when

p(6) = 0

, but

p(6) = 216 - 42 + 6 = 180

❌ Not quite. By the factor theorem,

z - \lambda

divides

p

exactly when

p(\lambda) = 0

: evaluate

p

at the

\lambda

of each option.

Show solution

Write each option as $z - \lambda$ : the values are $\lambda = 3, -1, 6, -3$ . By the factor theorem, $z - \lambda$ is a factor of $p$ if and only if $p(\lambda) = 0$ .

p(3) = 12,\quad p(-1) = 12,\quad p(6) = 180,\quad p(-3) = -27 + 21 + 6 = 0.

Only $\lambda = -3$ is a zero, so the factor is $z + 3$ , and indeed $(z + 3)(z^2 - 3z + 2) = z^3 - 7z + 6$ . (The other zeros of $p$ are $1$ and $2$ , since $z^2 - 3z + 2 = (z - 1)(z - 2)$ .)

Problem 2 · The Quotient After a Zero

Given: $p(z) = z^3 - 7z + 6$ has the zero $\lambda = 1$ — find $q$ with $p(z) = (z - 1)\,q(z)$ for all $z$ .

✅ Correct!

(z - 1)(z^2 + z - 6) = z^3 + z^2 - 6z - z^2 - z + 6 = z^3 - 7z + 6

, and

\deg q = 2 = m - 1

❌ This divides by $z + 1$ .

(z - 1)(z^2 - z - 6) = z^3 - 2z^2 - 5z + 6 \neq p(z)

. Dividing by

z - 1

uses

\lambda = 1

, not

-1

❌ The degree is wrong.

\deg q

must be

m - 1 = 2

. Skipping the coefficient

0

z^2

p

gives

z - 6

, and

(z - 1)(z - 6) = z^2 - 7z + 6

, which has degree

2

, not

3

❌ Check the constant term.

(z - 1)(z^2 + z + 6) = z^3 + 5z - 6 \neq p(z)

: the constant term of the product is

-1 \cdot 6 = -6

, but

p

has constant term

+6

❌ Not quite. Multiply

(z - 1)\,q(z)

out and compare with

z^3 + 0z^2 - 7z + 6

Show solution

Following the proof, $p(z) = p(z) - p(1) = (z^3 - 1^3) - 7(z - 1)$ . With $z^3 - 1 = (z - 1)(z^2 + z + 1)$ ,

p(z) = (z - 1)\big(z^2 + z + 1 - 7\big) = (z - 1)(z^2 + z - 6).

So $q(z) = z^2 + z - 6$ , of degree $3 - 1 = 2$ . Its leading coefficient is $a_3 = 1$ , as the proof predicts, and $q(z) = (z - 2)(z + 3)$ carries the other two zeros of $p$ .

Problem 3 · Divide in the Basis

Given: $p(z) = z^3 - 7z + 6$ and $s(z) = z^2 - 3z + 2$ , so $n = 3$ and $m = 2$ — find the basis of $\mathcal{P}_3(\mathbf{F})$ used in the proof, then $q$ and $r$ with $p = sq + r$ , $\deg r < \deg s$ .

Which list is the basis?

What are q and r?

✅ Correct!

p = 0 \cdot 1 + 0 \cdot z + 3s + 1 \cdot zs

, so

q = 3 + z

and

r = 0

s

divides

p

exactly.

❌ Two entries of degree 2.

s = z^2 - 3z + 2

lies in

\operatorname{span}(1, z, z^2)

, so this list is linearly dependent, and nothing in it has degree

3

❌ Degrees $2$ to $5$ .

z^3 s

has degree

5

, so it is not in

\mathcal{P}_3(\mathbf{F})

, and the degrees

0

and

1

are missing. The list should be

m

powers

1, \ldots, z^{m-1}

followed by

s, \ldots, z^{n-m}s

❌ Degree $1$ is missing. The degrees are

0, 2, 3, 4

;

z^2 s

has degree

4 > n

, and with

n - m = 1

the

s

-part stops at

zs

❌ Check the degrees. You need one polynomial of each degree

0, 1, 2, 3

1, \ldots, z^{m-1}

, then

s, \ldots, z^{n-m}s

❌ One step too early.

p - zs = 3z^2 - 9z + 6

has degree

2 = \deg s

, not less. It equals

3s

, so the coordinate on

s

b_0 = 3

, and the remainder is

0

❌ The coordinates are in the wrong places. The coordinates are

b_0 = 3

s

and

b_1 = 1

zs

, so

q = b_0 + b_1 z = 3 + z

. And

s(3z + 1) = 3z^3 - 8z^2 + 3z + 2 \neq p

❌ Check the sign.

s(z - 3) = z^3 - 6z^2 + 11z - 6 \neq p

; the difference

p - s(z - 3) = 6z^2 - 18z + 12

has degree

2

, not less than

2

❌ Not quite. Clear the

z^3

term with a multiple of

zs

, then the

z^2

term with a multiple of

s

; what is left is

r

Show solution

The basis. With $m = 2$ and $n - m = 1$ , the list is $1, z$ followed by $s, zs$ : degrees $0, 1, 2, 3$ , one of each, length $4 = \dim \mathcal{P}_3(\mathbf{F})$ .

The coordinates, from the top degree down. $zs = z^3 - 3z^2 + 2z$ is the only entry of degree $3$ , and $p$ has $z^3$ -coefficient $1$ , so $b_1 = 1$ :

p - zs = 3z^2 - 9z + 6 = 3(z^2 - 3z + 2) = 3s.

So $b_0 = 3$ , and nothing is left for $1$ and $z$ : $a_0 = a_1 = 0$ . Hence

p = 0 \cdot 1 + 0 \cdot z + 3s + 1 \cdot zs = s(z + 3) + 0,

giving $q = z + 3$ and $r = 0$ (degree $-\infty < 2$ ). This matches Problem 1: $s = (z - 1)(z - 2)$ , and $1$ , $2$ are both zeros of $p$ .

Problem 4 · Too Many Zeros

Given: $p \in \mathcal{P}_3(\mathbf{R})$ satisfies $p(1) = p(2) = p(-3) = p(4) = 0$ — find $p$ .

✅ Correct! A nonzero polynomial of degree

m \leq 3

has at most

m \leq 3

zeros (a nonzero constant has none), so four zeros force

p = 0

❌ That has degree 4. It expands to

z^4 - 4z^3 - 7z^2 + 34z - 24

, which is not in

\mathcal{P}_3(\mathbf{R})

❌ Check the fourth zero. This vanishes at

1

2

-3

, but

p(4) = 64 - 28 + 6 = 42 \neq 0

❌ One polynomial qualifies. The zero polynomial lies in

\mathcal{P}_3(\mathbf{R})

(its degree is

-\infty

) and vanishes everywhere.

❌ Not quite. How many zeros can a nonzero polynomial of degree at most

3

have?

Show solution

Suppose $p \neq 0$ , and let $m = \deg p \leq 3$ . If $m = 0$ , $p$ is a nonzero constant and has no zeros. If $m \geq 1$ , the zero count gives at most $m \leq 3$ zeros. Either way $p$ cannot vanish at the four distinct points $1, 2, -3, 4$ .

So $p = 0$ . The zero polynomial does lie in $\mathcal{P}_3(\mathbf{R})$ , with $\deg 0 = -\infty$ . This is the argument that makes coefficients unique: a difference of two coefficient lists that vanishes at more points than its degree must be the zero list.

Solved: 0 / 4

Zeros, Factors and the Division Algorithm via a Basis

IKey Formulas

IIEvery Difference of Powers Has the Factor z−λz - \lambdaz−λ

IIIA Zero Splits Off a Factor

IVDividing by Reading Off Coordinates

VQuiz Questions

IIEvery Difference of Powers Has the Factor $z - \lambda$