Answer:
[tex]f(x) = a(x^2+4)(x-3)^2[/tex]
Step-by-step explanation:
So I'm assuming it means that one of the zeros is at x=3 with a multiplicity of 2 and it has an imaginary solution of 4i. Anyways, imaginary solutions come in conjugate pairs meaning if you have a complex solution of [tex]a-bi[/tex] there is another complex solution which is the conjugate of that which is [tex]a+bi[/tex] but since the imaginary solution is 4i, the complex number is just [tex]0+4i[/tex] so the conjugate is [tex]0-4i[/tex] or [tex]-4i[/tex]. So since you have x=3 as a zero that can be represented as [tex](x-3)^2[/tex] since 3 would make it 0 and it has a multiplicity of 2. As for the other factors, you won't just have ([tex](x-4i)(x+4i)[/tex], you'll have a factor that is set up in a way that the solutions are: [tex]x=\pm\sqrt{-4}[/tex]. That means it'll be a quadratic. So it'll be in the form [tex]x^2+b[/tex]. Since you're moving b to the other side and it's negative. that means it has to be positive and since the value is 4, you'll have the factor [tex](x^2+4)[/tex] which when set equal to zero has the solutions sqrt(-4). So this gives you the equation
[tex]f(x) = a(x^2+4)(x-3)^2[/tex]
