Notice that the polynomial has a complex solution `x = -3-3i` , hence, the complex conjugate `x = -3+3i` is also a root of polynomial.
The problem provides the information that the solution `x= -3` has n order of multiplicity `n=2` .
You should remember that you may write the factored form of a polynomial of fourth degree such that:
`f(x) = a(x-x_1)(x-x_2)(x-x_3)(x-x_4)`
You need to substitute the given roots for `x_1,x_2,x_3,x_4` such that:
`f(x) = a(x - (-3-3i))(x - (-3+3i))(x - (-3))(x - (-3))`
`f(x) = a(x + 3 + 3i)(x + 3 - 3i)(x + 3)^2`
`f(x) = a((x + 3)^2- (3i)^2)(x + 3)^2`
Using the fact that `i^2 = -1` yields:
`f(x) = a((x + 3)^2 + 9)(x + 3)^2`
Hence, evaluating the given polynomial, under the given conditions, yields `f(x) = a((x + 3)^2 + 9)(x + 3)^2` .