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Polynomial Identities
in Complex Numbers

Learn to extend familiar polynomial identities into the complex number system, factor expressions like \( x^2 + 4 \), and discover how imaginary numbers unlock new factorizations.

Algebra 2

What Are Complex Numbers?

A complex number has the form \( a + bi \), where \( a \) and \( b \) are real numbers and \( i \) is the imaginary unit defined by:

\( a \)
The real part of the complex number, plotted on the horizontal axis.
\( bi \)
The imaginary part, plotted on the vertical axis.

Visualizing the Complex Plane

Every complex number \( a + bi \) maps to a point on the complex plane. The horizontal axis represents the real part, and the vertical axis represents the imaginary part.

Re Im -2 -1 1 2 3 \(i\) \(2i\) \(-i\) \(-2i\) \(3+2i\) \(-1-i\) \(2i\)

Why Extend to Complex Numbers?

Over the real numbers, the polynomial \( x^2 + 4 \) has no roots and cannot be factored. It is "irreducible." But over the complex numbers, every polynomial factors completely.

Real Numbers
\( x^2 + 4 \) is irreducible: no real roots since \( x^2 = -4 \) has no real solution.
Complex Numbers
\( x^2 + 4 = (x + 2i)(x - 2i) \): factors beautifully using \( i \).

The Difference of Squares Identity

You already know the classic real-number identity:

Now consider what happens when we write a sum of squares as a difference by using \( i^2 = -1 \):

The Sum of Squares Factorization

This is the core identity you need to master. Over the complex numbers:

Verification by FOIL:

1
First: \( a \cdot a = a^2 \)
2
Outer: \( a \cdot (-bi) = -abi \)
3
Inner: \( bi \cdot a = abi \)
4
Last: \( bi \cdot (-bi) = -b^2i^2 = b^2 \)

Difference vs. Sum of Squares

\( a^2 - b^2 \) \( a^2 + b^2 \)
Factored Form \( (a + b)(a - b) \) \( (a + bi)(a - bi) \)
Number System Real Complex
Roots of \( x^2 = c \) \( x = \pm b \) (real) \( x = \pm bi \) (imaginary)
Key Ingredient \( b^2 \) is positive \( (bi)^2 = -b^2 \)

More Extended Identities

The sum-of-squares principle extends to other polynomial patterns over the complex numbers:

Higher-Degree Sums
Expressions like \( x^4 + b^4 \) can be rewritten as \( (x^2)^2 + (b^2)^2 \) and factored using the sum-of-squares identity over \( \mathbb{C} \).
Conjugate Pairs
Complex roots of polynomials with real coefficients always come in conjugate pairs: if \( a + bi \) is a root, so is \( a - bi \).
Complete Factoring
Any polynomial with real coefficients factors into linear and irreducible quadratic factors over the reals, or fully into linear factors over \( \mathbb{C} \).

How to Factor a Sum of Squares

1
Identify the form: Confirm the expression matches \( a^2 + b^2 \) (a sum of two perfect squares).
2
Find \( a \) and \( b \): Take the square root of each term. For \( x^2 + 4 \): \( a = x \), \( b = 2 \).
3
Apply the identity: Write \( (a + bi)(a - bi) \). Here: \( (x + 2i)(x - 2i) \).
4
Verify by multiplying: Expand using FOIL to confirm the imaginary terms cancel and you get the original expression.

Example 1: Factor \( x^2 + 9 \)

1
Identify: \( x^2 + 9 = x^2 + 3^2 \). This is a sum of squares.
2
Set \( a = x \), \( b = 3 \): Apply \( a^2 + b^2 = (a + bi)(a - bi) \).
3
Factor: \( x^2 + 9 = (x + 3i)(x - 3i) \).

Example 2: Factor \( 4x^2 + 25 \)

When coefficients aren't 1, rewrite each term as a perfect square first:

1
Rewrite: \( 4x^2 + 25 = (2x)^2 + 5^2 \).
2
Set \( a = 2x \), \( b = 5 \): Apply the identity.
3
Factor: \( 4x^2 + 25 = (2x + 5i)(2x - 5i) \).

Example 3: Factor \( x^4 + 16 \)

Higher powers work the same way. Recognize the pattern:

1
Rewrite: \( x^4 + 16 = (x^2)^2 + 4^2 \). Sum of squares!
2
Set \( a = x^2 \), \( b = 4 \): Apply the identity.
3
Factor: \( x^4 + 16 = (x^2 + 4i)(x^2 - 4i) \).

Solving \( x^2 + b^2 = 0 \)

Once you can factor sums of squares, you can solve equations that have no real solutions:

1
Start: \( x^2 + 49 = 0 \)
2
Factor: \( (x + 7i)(x - 7i) = 0 \)
3
Solve: \( x + 7i = 0 \) or \( x - 7i = 0 \)
4
Solutions: \( x = -7i \) or \( x = 7i \)

Visualizing \( x^2 + 4 \)

The graph of \( y = x^2 + 4 \) never touches the \( x \)-axis: it has no real roots. The complex roots \( \pm 2i \) live on the imaginary axis.

\(x\) \(y\) -2 -1 1 2 4 8 12 vertex (0, 4) no real x-intercepts

Common Mistakes to Avoid

✕ Forgetting \( i^2 = -1 \)
Writing \( (x + 2i)(x - 2i) = x^2 - 4i^2 = x^2 - 4 \) instead of \( x^2 + 4 \). Remember: \( -4i^2 = -4(-1) = +4 \).
✕ Factoring over the reals
Claiming \( x^2 + 4 = (x + 2)(x - 2) \). That gives \( x^2 - 4 \), a difference of squares. Sums of squares need \( i \).
✕ Dropping the \( i \)
Writing \( x^2 + 9 = (x + 3)(x - 3) \). Without \( i \), you are factoring a different expression entirely.

Summary of Key Identities

Expression Factored Form Roots
\( x^2 - b^2 \) \( (x + b)(x - b) \) \( x = \pm b \)
\( x^2 + b^2 \) \( (x + bi)(x - bi) \) \( x = \pm bi \)
\( x^2 + 4 \) \( (x + 2i)(x - 2i) \) \( x = \pm 2i \)
\( 4x^2 + 25 \) \( (2x + 5i)(2x - 5i) \) \( x = \pm \frac{5}{2}i \)
\( x^4 + 16 \) \( (x^2 + 4i)(x^2 - 4i) \) 4 complex roots

Question 1 of 5

Which of the following is the correct factorization of \( x^2 + 16 \) over the complex numbers?

Question 2 of 5

What are the solutions to \( x^2 + 25 = 0 \)?

Question 3 of 5

Factor \( 9x^2 + 49 \) over the complex numbers.

Question 4 of 5

When you expand \( (x + 6i)(x - 6i) \), what do you get?

Question 5 of 5

Why can \( x^2 + b^2 \) not be factored using only real numbers?