The sets of complex numbers are [tex](3x+5i) (3x-5i)[/tex]
Given equation is [tex]9x^{2} +25[/tex].
We have to find its factor over the sets of complex numbers.
Before we begin, note that 9 and 25 are both perfect squares. Adding one perfect square to another is called sum of squares or sum of two squares.
Furthermore, you cannot factor sum of squares with real numbers. However, you can solve [tex]9x^{2} +25[/tex] with a complex or imaginary number.
The sum of squares formula we will use to factor [tex]9x^{2} +25[/tex] with our imaginary number is as follows: [tex]a^{2} +b^{2} = (a+ib)\times(a-ib)[/tex].......equation 1
Now, [tex]9x^{2} +25=(3x)^{2} +5^{2}[/tex]
Let [tex]a=3x, b=5[/tex]
Putting the value of a and b in equation 1 we get,
[tex]9x^{2} +25=(3x+5i) (3x-5i)[/tex].
Hence the sets of complex numbers are [tex](3x+5i) (3x-5i)[/tex].
For more details on complex number follow the link below:
https://brainly.com/question/10251853
