Since 1 and 5i are zeros, the function will have (x-1) and (x-5i) as factors. Complex zeros come in conjugate pairs so (x+5i) is also a factor. The polynomial function can be written as f(x)=a(x-1)(x-5i)(x+5i), where ain mathbb(R). Let us substitute 1 for a and multiply to get one of the functions: f(x)=(x-1)(x-5i)(x+5i) =(x-1)(x^(2)-5ix+5ix-25i^(2)) =(x-1)(x^(2)+25) Let us multiply terms in the last two pairs of parentheses: =x^(3)+25x-x^(2)-25 =x^(3)-x^(2)+25x-25 Result: x^(3)-x^(2)+25x-25