Question 1000695

{{{x[1]=5+i}}} and {{{x[2]=1+4i}}}

since given zeros are complex solutions, they always come in pairs
have {{{x[3]=5-i}}} and {{{x[4]=1-4i}}}

then, {{{f(x)=(x-x[1])(x-x[2])(x-x[3])(x-x[4])}}}

{{{f(x)=(x-(5+i))(x-(1+4i))(x-(5-i))(x-(1-4i))}}}

{{{f(x)=(x-5-i)(x-1-4i)(x-5+i)(x-1+4i)}}}

{{{f(x)=(x^2+(-6-5i)x+21i+1)(x^2+(-6+5i)x-21i+1)}}}...multiplying each term by each, and knowing that {{{i*i=i^2=-1}}}, we have

{{{f(x)=x^4-12x^3+63x^2-222x+442}}}