Question 1171293
given zeros:

{{{x[1]=-1}}} 
{{{x[2]=0}}}

 {{{x[3]=5 + i}}}=> there is also {{{x[4]=5-i}}}  (complex zeros always come in pairs)

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

{{{f(x)=(x-(-1))(x-0)(x-(5+i))(x-(5-i))}}}

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

{{{f(x)=(x^2+x)(x-5-i)(x^2-5x+cross(ix)-5x+25-cross(5i)-cross(ix) +cross(5i)-i^2)}}}

{{{f(x)=(x^2+x)(x^2 -10x + 26)}}}

{{{f(x)=x^4 - 9x^3 + 16x^2 + 26x}}}