Question

Factor 4x^2+81 over the set of complex numbers

Answer 1

Answer:(2x + 9i)(2x - 9i)

Step-by-step explanation:

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

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

But -b^2 i^2 = +b^2

Answer 2

Answer:

$(4x+9i)(4x-9i)$

Step-by-step explanation:

Here, we have to apply this complex numbers property:

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

So, the given expression $4x^2+81$, can be rewrite as factor using the property:

$4x^2+81=(4x+9i)(4x-9i)$

Because, if

$a^2=x^2 \ and \ b^2=81\\\ then \ a=x \ and \ b=9$  

Therefore, the factors are $(4x+9i)(4x-9i)$