If aa is a zero of the polynomial, then x−a is one of its factor. Complex conjugates come in pair so 4+4i is also a zero of f(x). Two factors of f(x) are [x−(4−4i)] and [x−(4+4i)]. Since −5 has multiplicity 2, then it corresponds to the factor

So, the polynomial is: \(f(x)=[x−(4−4i)][x−(4+4i)](x+5)^{2}\)

Expand:

\(f(x)=[x^{2}−(4+4i)x−(4−4i)x+(4−4i)(4+4i)](x^{2}+10x+25)\)

\(f(x)=[x^{2}−4x+4ix−4x+4ix+(16−16i^{2})](x^{2}+10x+25)\)

\(f(x)=[x^{2}−8x+(16+16)](x^{2}+10x+25)\)

\(f(x)=(x^{2}−8x+32)(x^{2}+10x+25)\)

\(f(x)=x^{2}(x^{2}−8x+32)+10x(x^{2}−8x+32)+25(x^{2}−8x+32)\)

\(f(x)=x^4−8x^{3}+32x^2+10x^{3}−80x^{2}+320x+25x^{2}−200x+800\)

\(f(x)=x^{4}+2x^{3}−23x^{2}+120x+800\)

So, the polynomial is: \(f(x)=[x−(4−4i)][x−(4+4i)](x+5)^{2}\)

Expand:

\(f(x)=[x^{2}−(4+4i)x−(4−4i)x+(4−4i)(4+4i)](x^{2}+10x+25)\)

\(f(x)=[x^{2}−4x+4ix−4x+4ix+(16−16i^{2})](x^{2}+10x+25)\)

\(f(x)=[x^{2}−8x+(16+16)](x^{2}+10x+25)\)

\(f(x)=(x^{2}−8x+32)(x^{2}+10x+25)\)

\(f(x)=x^{2}(x^{2}−8x+32)+10x(x^{2}−8x+32)+25(x^{2}−8x+32)\)

\(f(x)=x^4−8x^{3}+32x^2+10x^{3}−80x^{2}+320x+25x^{2}−200x+800\)

\(f(x)=x^{4}+2x^{3}−23x^{2}+120x+800\)