If a polynomial equation p(x)=0 has 3+4i as a solution

3-i
4-3i
4+3i
3-4i

As Damon said, complex roots come in conjugate pairs.

Using properties of roots:
sum of our roots = 3+4i + 3-4i = 6
product of roots = (3+4i)(3-4i)
= 9 - 16i^2 = 25

p(x) = x^2 - 6x + 25

if 3 + 4 i is a solution

then its complex conjugate

3 - 4 i

is also a solution

because when you solve a quadratic for example
(- b +/- sqrt(b^2-4ac))/2a

if b^2-4ac is negative

then you have
-b + number*i
and
-b - same number*i
They come in complex conjugate PAIRS

Are you looking for another solution?

if so what do you think/

To find the other complex solutions to the polynomial equation p(x) = 0, given that 3 + 4i is a solution, we can use the concept of complex conjugates.

1. Start with the given solution: 3 + 4i.
2. Take the complex conjugate of the given solution by changing the sign of the imaginary part: 3 - 4i. This is one of the other complex solutions.
3. Repeat the process for the complex conjugate solution: 3 - 4i.
- Complex conjugate of 3 - 4i: 3 + 4i. This is another complex solution.
4. Finally, we have two complex solutions: 3 - 4i and 3 + 4i.

Therefore, the correct complex solutions to the polynomial equation p(x) = 0 are:
- 3 + 4i
- 3 - 4i