Instead of doing your homework problem for you,
I will work one exactly like yours step-by-step
with the numbers changed.  I'll do this one
instead:

Since 3+2i is a zero, so is its conjugate 3-2i

Set x = to each one:

 x = -3,    x = 1/4,    x = 3+2i,  x = 3-2i
x+3 = 0,   4x = 1  ,  x-3-2i = 0,  x-3+2i=0
           4x-1 = 0

Use the zero-factor property in reverse.
Also multiply it by a constant "a":

a(x+3)(4x-1)(x-3-2i)(x-3+2i) = 0

The polynomial function F(x) will be the left
side of this equation, since when it it is
set=0, the above is what we would get:

F(x) = a(x+3)(4x-1)(x-3-2i)(x-3+2i)
F(x) = a(x+3)(4x-1)[(x-3)-2i][(x-3)+2i]
F(x) = a(4x²+11x-3)[(x-3)²-(2i)²]
F(x) = a(4x²+11x-3)[x²-6x+9-4i²]
F(x) = a(4x²+11x-3)[x²-6x+9-4(-1)]
F(x) = a(4x²+11x-3)[x²-6x+9+4]
F(x) = a(4x²+11x-3)(x²-6x+13)
F(x) = a(4x4-24x³+52x²+11x³-66x²+143-3x²+18x-39)
F(x) = a(4x4-13x³-17x²+161x-39)
F(2) = 350 = a(4∙24-13∙2^3-17∙2^2+161∙2-39)
F(2) = 350 = a(4∙16-13∙8-17∙4+161∙2-39)
F(2) = 350 = a(175)
        = a
         2 = a

F(x) = 2(4x4-13x3-17x2+161x-39)

F(x) = 8x^4-26x^3-34x^2+322x-78

Now use this as a model to do yours by.
Yours is done exactly the same way, step by step.

Edwin