Question

Factor 4x^2+9 over the complex numbers?

Answer 1

so, where's the complex number?

Answer 2

it just said factor it over the complex numbers and circle final solution.

Answer 3

\(\bf 4x^2+9\qquad \textit{notice that}\quad 4= 2^2\qquad 9=3^2\qquad thus\\ \quad \\
4x^2+9\implies 2^2x^2+3^2\implies (2x)^2+3^2\\ \quad \\
\textit{now keep in mind that }\qquad \color{red}{a^2-b^2 = (a-b)(a+b)}\)

Answer 4

hmmm one sec

Answer 5

okay

Answer 6

\(\bf (2x)^2\color{blue}{+3^2}\\ \quad \\
\\ \quad \\ \quad \\
\color{blue}{+3^2\implies -(-3^2)\implies -(-1\cdot 3^2)}\\ \quad \\
\textit{let us recall our good friend}\quad \sqrt{-1}\\ \quad \\
\color{blue}{i^2 = \sqrt{-1}\cdot \sqrt{-1}\implies (\sqrt{-1})^2\implies \sqrt{(-1)^2}\implies -1}\qquad thus\\ \quad \\
-(-1\cdot 3^2)\implies -(i^2\cdot 3^2)\implies -(3i)^2\qquad thus\\ \quad \\
(2x)^2+3^2\implies (2x)^2\color{blue}{-(3i)^2}\\ \quad \\
\textit{now recall }\qquad \color{red}{a^2-b^2 = (a-b)(a+b)}\)

Answer 7

so just factor it likewise

Answer 8

Here's one way to look at it


a^2-b^2 = (a-b)(a+b)


a^2-(bi)^2 = (a-bi)(a+bi) ... replace b with bi


a^2-b^2i^2 = (a-bi)(a+bi)


a^2-b^2(-1) = (a-bi)(a+bi)


a^2+b^2 = (a-bi)(a+bi)



We replaced b with bi because when you square bi, you get b^2i^2 and you take advantage of the fact that i^2 = -1 to change -b^2 to +b^2