College algebra help!

You missed one of the n's:
(x-4) (x+4) (x-1/3) (2+3i) (2-3i)
For convenience multiply the fractional factor by 3 to eliminate the fraction. (This doesn't change the roots of the function):
(x-4) (x+4) (3x-1) (2+3i) (2-3i)
Multiply these factors together. It's easier to multiply the complex factors first.:
39x^3 - 13x^2 - 624x + 208
Evaluate f(1):
39(1)^3 - 13(1)^2 - 624(1) + 208 = -390
To achieve the desired value for f(1), multiply the polynomial by 100 / -390 = -10 / 39:
(-10/39) (39x^3 - 13x^2 - 624x + 208)
Maybe leave it like that, or maybe multiply through each coefficient:
-10x^3 + (10x^2)/3 + 160x - 160/3

For the fraction (1/3),
x-1/3 = 0
3x-1 = 0

f(x) = a(x+4)(3x-1)(x-2-3i)(x-?)
a is a scaling factor that will be determined at the end.
f(1) has no imaginary component, so we have to multiply on a factor with the conjugate of the complex factor to eliminate the imaginary value.
f(x) = a(x+4)(3x-1)(x-2-3i)(x-2+3i)
Let's try that.
f(1) = a(1+4)[3(1)-1](1-2-3i)(1-2+3i)
100 = a(5)(2)(-1-3i)(-1+3i)
(-1-3i)(-1+3i) = 1 - 3i + 3i - 9i^2 = 1 - (-1)(9) = 1 + 9 = 10
100 = a(10)(10)
100 = 100a
a = 1
So the complete polynomial is f(x) = (x+4)(3x-1)(x-2-3i)(x-2+3i)

you're on the right track

the last two factors are ___ [x - (2 + 3i)] and [x - (2 - 3i)]

multiply the factors together to find the polynomial ___ you may have to multiply by a constant (fudge factor) to get the correct value for f(1)