Question

1. factor $4x^2 + 81$.

$\bigcirc$ $(4x + 9i)(4x - 9i)$
$\bigcirc$ $(2x + 9i)(2x - 9i)$
$\bigcirc$ $(2x + 18i)(2x - 18i)$
$\bigcirc$ $(x + 9i)(x - 9i)$

Explanation:

Step 1: Recall the difference of squares formula

The difference of squares formula is \(a^2 - b^2=(a + b)(a - b)\). For complex numbers, we know that \(i^2=- 1\), so we can rewrite \(4x^{2}+81\) as a difference of squares. We can rewrite \(4x^{2}+81\) as \((2x)^{2}-(- 81)\) (since \(4x^{2}=(2x)^{2}\) and \(81 = -(-81)\)). And since \(-81=-(9i)^{2}\) (because \(i^{2}=-1\), so \((9i)^{2}=9^{2}\times i^{2}=81\times(- 1)=-81\)), then \(4x^{2}+81=(2x)^{2}-(9i)^{2}\)

Step 2: Apply the difference of squares formula

Using the formula \(a^{2}-b^{2}=(a + b)(a - b)\) with \(a = 2x\) and \(b=9i\), we get \((2x)^{2}-(9i)^{2}=(2x + 9i)(2x-9i)\)

Let's check the other options:

• Option A: \((4x + 9i)(4x-9i)=(4x)^{2}-(9i)^{2}=16x^{2}+81

eq4x^{2}+81\)

• Option C: \((2x + 18i)(2x-18i)=(2x)^{2}-(18i)^{2}=4x^{2}+324

eq4x^{2}+81\)

• Option D: \((x + 9i)(x-9i)=x^{2}-(9i)^{2}=x^{2}+81

eq4x^{2}+81\)

Answer:

B. \((2x + 9i)(2x - 9i)\)