Assuming that the coefficients of the quartic are all real, you can say that 2 + i and 3 - 2i are also roots.

Therefore linear factors are (x - [2 - i]), (x - [2 + i]), (x - [3 + 2i]) and (x - [3 + 2i]).

So quadratic factors are (x - [2 - i])(x - [2 + i]) and (x - [3 + 2i])(x - [3 + 2i]).

Now note that

(x - [2 - i])(x - [2 + i] = ([x - 2] + i)([x - 2] - i] = [x - 2]^2 - (i)^2

using the difference of two squares formula

= (x - 2)^2 + 1 = x^2 - 4x + 5.

Do (x - [3 + 2i])(x - [3 + 2i]) the same way to get x^2 - 6x + 13 (check this).

So the quartic is Q(x) = a(x^2 - 4x + 5)(x^2 - 6x + 13).

Now sub Q(0) = 13 and solve for a:

13 = a(5)(13) => a = 1/5.