Math..

Complex numbers differing by just the sign of the imaginary part are said to be conjugate.

So for example 2 + 3i and 2 - 3i are conjugate, -5 - 7i and -5 + 7i are conjugate, and in general

a + bi and a - bi where a and b are real numbers are conjugate.

They have the property that their product is always a real number, so for the general case,

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

and as a particular example,

(2 + 3i)(2 - 3i) = 2^2 + 3^2 = 4 + 9 = 13.

For a polynomial equation, with real coefficients, complex roots always appear as conjugate pairs.

So, providing that the given equation does have real coefficients, it must have 3 - 4i as a second root.

The question doesn't actually state that P(x) has real coefficients, so it might have 3 - 4i as a second root, but then again it might not.

4 + 3i?

You can just multiply them together to see what results  (there should be no 'i' in the answer)

Let's look at the first and the fourth possibilities

First one

(3-i) (3+4i) =  9 +12i -3i -4i^2 =9 + 9i - 4(-1) = 9 +9i +4 = 13 + 9i

Fourth one

 (3-4i) (3+4i) = 9 +12i -12i -14i^2  = 9 - 14(-1) = 9 + 14 = 23     

So the fourth choice is good    If you try the 2nd and third ones there will be an i in the answer

If you set up the solutions and multiply them out :

Fourth one   (x-(3+4i))(x-(3-4i)) = x^2 -6x + 25    a normal polynomial expression

First one       (x-(3-i))(x-(3+4i))  =  a mess = x^2 - 6x -3ix +9i +13   Which is an imaginary polynomial

Garbled...but I hope you see it....