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All courses Math III · N-CN.8 45 of 55
Extend polynomial identities to complex numbers for higher-degree polynomial work.

Factor a sum of squares over the complex numbers using conjugate imaginary factors

Problem
Factor x² + 9 completely over the complex numbers.
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Hint

Recognize x² + 9 as x² + (3)² and use the complex-number identity a² + b² = (a + bi)(a - bi).

Over the complex numbers, a sum of squares factors because i² = -1, so (a + bi)(a - bi) = a² + b².

Solution walkthrough

01

Rewrite the expression as a sum of squares

\[x^2~+~9~=~x^2~+~3^2\]

This makes the expression match the pattern a² + b².

02

Use the complex factoring pattern

\[a^2~+~b^2~=~(a~+~\text{bi})(a~-~\text{bi})\]

Over the complex numbers, this identity works because i² = -1.

03

Substitute a = x and b = 3

\[x^2~+~9~=~(x~+~3i)(x~-~3i)\]

Replacing a and b gives the factorization.

04

State the factorization

\[(x~+~3i)(x~-~3i)\]

That is the correct factorization over the complex numbers.

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Another way

  1. You can check by multiplying: (x + 3i)(x - 3i) = x² - 9i² = x² + 9.

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Common mistake

A common mistake is to write (x + 3)(x - 3), but that equals x² - 9, not x² + 9.