Linear-Algebra-Done-Right · Unit 10 · Video 4 · Interactive Practice

Real Polynomials: Linear and Irreducible Quadratic Factors

IKey Formulas

Formula	Name	What it says
$p(\lambda) = 0 \implies p(\bar{\lambda}) = 0$	Nonreal zeros come in pairs	For $p \in \mathcal{P}(\mathbf{C})$ with real coefficients: conjugate both sides of $a_0 + a_1\lambda + \cdots + a_m\lambda^m = 0$
$x^2 + bx + c = \left(x + \frac{b}{2}\right)^2 + \left(c - \frac{b^2}{4}\right)$	Completing the square	For $b, c \in \mathbf{R}$ : $x^2 + bx + c = (x - \lambda_1)(x - \lambda_2)$ with $\lambda_1, \lambda_2 \in \mathbf{R}$ $\iff b^2 \geq 4c$
$(x - \lambda)(x - \bar{\lambda}) = x^2 - 2(\operatorname{Re}\lambda)x + \|\lambda\|^2$	A conjugate pair gives a real quadratic	Here $b^2 - 4c = -4(\operatorname{Im}\lambda)^2 < 0$ when $\lambda \notin \mathbf{R}$ , so the quadratic has no real zeros
$p(x) = c(x - \lambda_1)\cdots(x - \lambda_m)(x^2 + b_1 x + c_1)\cdots(x^2 + b_M x + c_M)$	Factorization over $\mathbf{R}$	For nonconstant $p \in \mathcal{P}(\mathbf{R})$ : all numbers real, $b_k^2 < 4c_k$ , unique up to order; $\lambda_1, \ldots, \lambda_m$ are the real zeros and $m + 2M = \deg p$

Key Insight: Over $\mathbf{R}$ the fundamental theorem of algebra fails only by quadratics: each nonreal zero arrives with its conjugate, and the pair multiplies out to a real quadratic with $b^2 < 4c$ .

IIA Conjugate Pair Multiplies to a Real Quadratic

For nonreal $\lambda$ , the product $(x - \lambda)(x - \bar{\lambda})$ has real coefficients and stays above the $x$ -axis.

Re λ −1 Im λ 2

IIIWhen Does $x^2 + bx + c$ Factor over $\mathbf{R}$ ?

$x^2 + bx + c$ splits into real linear factors exactly when $(b, c)$ lies on or below the parabola $b^2 = 4c$ .

b −5 c 6

IVFrom the Complex Factorization to the Real One

Each real zero gives a linear factor, and each conjugate pair of nonreal zeros gives one quadratic factor.

p Step 1

💡 Since $m + 2M = \deg p$ , a real polynomial of odd degree has $m$ odd, so it always has at least one real zero.

VQuiz Questions

Problem 1 · Which Quadratic Has No Real Zeros?

Given: four real quadratics $x^2 + bx + c$ . Find the one that has no real zeros, so it cannot be written as $(x - \lambda_1)(x - \lambda_2)$ with $\lambda_1, \lambda_2 \in \mathbf{R}$ .

✅ Correct!

b^2 = 9 < 12 = 4c

, so

x^2 + 3x + 3 = \left(x + \tfrac{3}{2}\right)^2 + \tfrac{3}{4} > 0

for every real

x

❌ Not quite. A positive constant term does not prevent real zeros:

b^2 = 16 \geq 12 = 4c

, and

x^2 + 4x + 3 = (x + 1)(x + 3)

❌ Close, but the boundary counts. Here

b^2 = 36 = 4c

, so

b^2 - 4c = 0

and

x^2 - 6x + 9 = (x - 3)^2

: the repeated zero

3

is real.

❌ Not quite. With

c = -1 < 0

the condition

b^2 \geq 4c

holds automatically:

25 \geq -4

, so there are two real zeros.

❌ Not quite. Compare

b^2

with

4c

for each option.

Show solution

By the theorem on quadratics, $x^2 + bx + c$ has real linear factors exactly when $b^2 \geq 4c$ . Compute $b^2 - 4c$ for each option:

$x^2 + 4x + 3$ : $16 - 12 = 4 \geq 0$ , and $x^2 + 4x + 3 = (x + 1)(x + 3)$ .
$x^2 - 6x + 9$ : $36 - 36 = 0$ , and $x^2 - 6x + 9 = (x - 3)^2$ .
$x^2 + 3x + 3$ : $9 - 12 = -3 < 0$ .
$x^2 + 5x - 1$ : $25 + 4 = 29 > 0$ , with zeros $\frac{-5 \pm \sqrt{29}}{2}$ .

Only $x^2 + 3x + 3$ has $b^2 < 4c$ . Completing the square,

x^2 + 3x + 3 = \left(x + \frac{3}{2}\right)^2 + \left(3 - \frac{9}{4}\right) = \left(x + \frac{3}{2}\right)^2 + \frac{3}{4} > 0,

so it has no real zeros and cannot be a product of real linear factors.

Problem 2 · Factoring $x^3 - 1$ over $\mathbf{R}$

Given: the zeros of $x^3 - 1$ in $\mathbf{C}$ are $1$ and $\omega, \bar{\omega}$ with $\omega = -\tfrac{1}{2} + \tfrac{\sqrt{3}}{2}i$ . Find the factorization of $x^3 - 1$ over $\mathbf{R}$ .

✅ Correct! The real zero

1

gives

x - 1

, and the pair

\omega, \bar{\omega}

gives

x^2 - 2(\operatorname{Re}\omega)x + |\omega|^2 = x^2 + x + 1

❌ Check the sign of the $x$ -coefficient. It is

-2\operatorname{Re}\omega = -2\left(-\tfrac{1}{2}\right) = 1

, not

-1

❌ Not quite.

(x - 1)^3 = x^3 - 3x^2 + 3x - 1

; the zeros

\omega, \bar{\omega}

are nonreal, so they give a quadratic factor, not more copies of

x - 1

❌ Check the constant term.

|\omega|^2 = (\operatorname{Re}\omega)^2 + (\operatorname{Im}\omega)^2 = \tfrac{1}{4} + \tfrac{3}{4} = 1

, a sum of squares, not

\tfrac{1}{4} - \tfrac{3}{4}

❌ Not quite. Pair

\omega

with

\bar{\omega}

and use

x^2 - 2(\operatorname{Re}\omega)x + |\omega|^2

Show solution

The real zero $1$ gives the linear factor $x - 1$ . The nonreal zeros $\omega, \bar{\omega}$ form one conjugate pair, and

(x - \omega)(x - \bar{\omega}) = x^2 - 2(\operatorname{Re}\omega)x + |\omega|^2.

With $\operatorname{Re}\omega = -\tfrac{1}{2}$ and $|\omega|^2 = \tfrac{1}{4} + \tfrac{3}{4} = 1$ :

-2\operatorname{Re}\omega = 1, \qquad (x - \omega)(x - \bar{\omega}) = x^2 + x + 1.

Here $b^2 - 4c = 1 - 4 = -3 < 0$ , as it must be. So

x^3 - 1 = (x - 1)(x^2 + x + 1),

with $m = 1$ , $M = 1$ and $m + 2M = 3$ . Expanding confirms it: $x^3 + x^2 + x - x^2 - x - 1 = x^3 - 1$ .

Problem 3 · A Quartic with No Real Zeros

Given: $p(x) = x^4 + x^2 + 1$ . Substituting $y = x^2$ gives $y^2 + y + 1$ , which has no real zeros. Find the factorization of $p$ over $\mathbf{R}$ .

✅ Correct!

p(x) = (x^2 + 1)^2 - x^2

is a difference of squares, and both factors have

b^2 - 4c = -3 < 0

❌ That is the trap. No real zeros means

m = 0

, not that

p

has no factorization: the theorem gives

m + 2M = 4

, so

p

is a product of

M = 2

quadratics.

❌ Not quite.

(x^2 + 1)^2 = x^4 + 2x^2 + 1

, which has the wrong

x^2

-coefficient.

❌ Close, but expand it. This product is

(x^2 + 1)^2 - 3x^2 = x^4 - x^2 + 1

, a different polynomial.

❌ Not quite. Write

p(x)

(x^2 + 1)^2

minus a square.

Show solution

For real $x$ , $p(x) = x^4 + x^2 + 1 \geq 1$ , so $p$ has no real zeros and $m = 0$ . Since $m + 2M = \deg p = 4$ , the factorization has $M = 2$ quadratic factors. To find them, complete a square:

x^4 + x^2 + 1 = (x^4 + 2x^2 + 1) - x^2 = (x^2 + 1)^2 - x^2.

A difference of two squares factors:

x^4 + x^2 + 1 = (x^2 + 1 - x)(x^2 + 1 + x) = (x^2 - x + 1)(x^2 + x + 1).

Each factor has $b^2 - 4c = 1 - 4 = -3 < 0$ , so neither has real zeros. By uniqueness this is the factorization over $\mathbf{R}$ .

The substitution $y = x^2$ only shows that $p$ has no real zeros; a real polynomial with no linear factors still factors into quadratics whenever its degree exceeds $2$ .

Problem 4 · A Real Cubic with the Zero $2 + i$

Given: $p \in \mathcal{P}(\mathbf{R})$ has degree $3$ and leading coefficient $1$ , $p(2 + i) = 0$ , and $p(0) = 10$ . Find the real zero of $p$ .

✅ Correct!

p(x) = (x^2 - 4x + 5)(x + 2) = x^3 - 2x^2 - 3x + 10

, and indeed

p(0) = 10

❌ Check the sign. With real zero

r

p(0) = 5(0 - r) = -5r

, so

-5r = 10

❌ Not quite. The quadratic factor contributes its constant term

|2 + i|^2 = 5

p(0)

; the value

p(0)

is not just

-r

❌ Check $|2 + i|^2$ . It is

2^2 + 1^2 = 5

, a sum of squares, not

2^2 - 1^2 = 3

❌ Not quite. Use the conjugate pair

2 \pm i

to find the quadratic factor first.

Show solution

Since $p$ has real coefficients and $p(2 + i) = 0$ , also $p(2 - i) = 0$ . With $\lambda = 2 + i$ , the pair gives the quadratic factor

(x - (2 + i))(x - (2 - i)) = x^2 - 2(\operatorname{Re}\lambda)x + |\lambda|^2 = x^2 - 4x + 5,

with $b^2 - 4c = 16 - 20 = -4 < 0$ . As $m + 2M = 3$ , the factorization over $\mathbf{R}$ has $M = 1$ and $m = 1$ : one real zero $r$ , and

p(x) = (x^2 - 4x + 5)(x - r).

Then $p(0) = 5 \cdot (-r) = -5r = 10$ , so $r = -2$ and

p(x) = (x^2 - 4x + 5)(x + 2) = x^3 - 2x^2 - 3x + 10.

Check: $p(-2) = -8 - 8 + 6 + 10 = 0$ and $p(0) = 10$ .

Solved: 0 / 4

Real Polynomials: Linear and Irreducible Quadratic Factors

IKey Formulas

IIA Conjugate Pair Multiplies to a Real Quadratic

IIIWhen Does x2+bx+cx^2 + bx + cx2+bx+c Factor over R\mathbf{R}R?

IVFrom the Complex Factorization to the Real One

VQuiz Questions

IIIWhen Does $x^2 + bx + c$ Factor over $\mathbf{R}$ ?