Question

Need help with this question ​

Answer 1

`f(x) = a(x^2+4)(x-3)^2`

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
`a-bi` there is another complex solution which is the conjugate of that which is
`a+bi` but since the imaginary solution is 4i, the complex number is just
`0+4i` so the conjugate is
`0-4i` or
`-4i`. So since you have x=3 as a zero that can be represented as
`(x-3)^2` since 3 would make it 0 and it has a multiplicity of 2. As for the other factors, you won't just have (
`(x-4i)(x+4i)`, you'll have a factor that is set up in a way that the solutions are:
`x=\pm√(-4)`. That means it'll be a quadratic. So it'll be in the form
`x^2+b`. 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
`(x^2+4)` which when set equal to zero has the solutions sqrt(-4). So this gives you the equation

`f(x) = a(x^2+4)(x-3)^2`