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All courses Algebra II · N-CN.8 6 of 55
Extend polynomial identities to complex numbers, such as factoring sums of squares over the complex numbers.

Factor a sum of squares over the complex numbers using \(a^2+b^2=(a+bi)(a-bi)\)

Problem
Factor \(x^{2}+9\) over the complex numbers by rewriting it as \(x^{2}-(3i)^{2}\) and applying the difference-of-squares pattern.
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Hint

Rewrite \(9\) as \(-(3i)^2\), since \((3i)^2=-9\).

Over the complex numbers, a sum of squares can be written as a difference of squares and then factored.

Solution walkthrough

01

Create a squared imaginary term

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

The value 3i is chosen because its square is negative 9.

02

Rewrite the sum as a difference

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

Since (3i)²=-9, subtracting it is exactly the same as adding 9.

03

Apply difference of squares

\[u^2-v^2=(u+v)(u-v)~->~(x+3i)(x-3i)\]

The identity applies with u=x and v=3i.

04

Expand to verify

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

The conjugate cross terms cancel and the product returns the original polynomial.

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Another way

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

!

Common mistake

Do not use (x+3)(x-3), which expands to x squared minus 9 rather than x squared plus 9.

Solution walkthrough video