Last two frustrating algebra problems

Question

Find an​ nth-degree polynomial function with real coefficients satisfying the given conditions. If you are using a graphing​ utility, use it to graph the function and verify the real zeros and the given function value.

1)

n=3

-3 and 6 + 2i are the zeros

f(2)=100

f(x)=

Type an expression using x as the variable. Simplify your​ answer.

2)

n=3

3 and 5i are the zeros

f(-1)=104

f(x)=

Type an expression using x as the variable. Simplify your​ answer.

Answer 1

#1)

then 6 - 2i is also a solution

f(x) = k(x+3) ((x-6)-2i) ((x-6)+2i)

f(2)=100

100 = k(5)(-4 -2i)(-4 + 2i)

100 = (5k)(16 - 8i + 8i+ 4)

100 = (5k)(16 + 4)

100 = 5*20*k

100 = 100k

k=1

f(x) = (x+3)((x-6)+ 2i)((x-6)-2i)

= (x+3)( (x-6)^2 + 4)

= (x+3)( x^2 - 12x+ 36 + 4)

= (x+3)(x^2 - 12x + 40)

= x^3 - 12x^2 + 40x

+3x^2 - 36x +120

= x^3 - 9x^2 + 4x + 120

==========================================