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All courses Algebra II · 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 \(x^{2}~+~9\) 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

Define the complex unit

\[i^2=-1\]

The imaginary unit i lets negative real numbers have square roots in the complex number system.

02

Rewrite as a difference of squares

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

Because (3i)²=9i²=-9, subtracting that square adds 9.

03

Factor the difference

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

Use u²-v²=(u+v)(u-v) with u=x and v=3i.

04

Check the product

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

The conjugate factors multiply back to the original real polynomial.

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Another way

  1. Solve x²+9=0 to get x=plus or minus 3i, then form the monic linear factors x-3i and x+3i.

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Common mistake

Do not write (x+3)(x-3), which factors x²-9. The plus 9 requires imaginary roots plus or minus 3i.

Solution walkthrough video