Firstly, we can notice that the degree asked is 4, but we have only 2 zeros, -4-2i and 1. Since the root 1 have multiplicity 2, it counts as two zeros, but we are still missing one, because a polynomial or 4th degree has 4 comlpex zeros.
However, we can use the complex conjugate root theorem. It says that for a one variable polynomial P with real coefficients, if a + bi is a complex root, then its conjugate. a - bi, is also a root of P.
We have exactly this case, a polynomial with real coefficients, so if -4 - 2i is a zero of it, then its conjugate also is. Its conjugate is -4 + 2i.
So, we have the zeros: -4 - 2i, -4 + 2i, 1 and 1.
Given the zeros of a polynomial and its leading coefficient, a, it can be written as:
[tex]f(x)=a(x-r_1)(x-r_2)(x-r_3)\ldots[/tex]
Where the r are the zeros of the polynomial.
So, we have:
[tex]\begin{gathered} f(x)=a(x-(-4-2i))(x-(-4+2i))(x-1)(x-1) \\ f(x)=a(x+4+2i)(x+4-2i)(x-1)^2 \end{gathered}[/tex]
