Simplify the expression. 2(x-3)+4y-2(x-y-3)+5
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If one of our zeros is 4, then the factor is x-4. If the second zero is 5i, then the conjugate root theorem says there HAS to be a root that is -5i. So our 3 factors are (x-4)(x+5i)(x-5i). We will FOIL out these factors to get the polynomial. Let's start with the ones that contain the imaginary numbers. Doing that mutliplication we get x^2-25i^2. i^2 is equal to -1, so what that expression simplifies down to is 
. Now we will multiply in that last factor of (x-4): 
. FOILing out we have 
. There you go!