Question 997365
a degree {{{3}}} polynomial with coefficient of {{{x^3}}} equal to {{{1}}}

{{{f(x)=x^3+bx^2+cx+d}}}

if given  zeros {{{x[1]=-1}}}, {{{x[2]=-4i}}}  and {{{x[3]=4i}}}, we will use zero product formula

{{{f(x)=(x- x[1])(x- x[2])(x- x[3])}}}....plug in given values


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

{{{f(x)=(x+1)(x+4i)(x- 4i)}}}.....recall, {{{(x+4i)(x- 4i)=x^2-(4i)^2}}}

{{{f(x)=(x+1)(x^2- (4i)^2)}}}....since {{{(4i)^2)=4^2*(i)^2=16(-1)=-16}}}, we have

{{{f(x)=(x+1)(x^2- (-16))}}}

{{{f(x)=(x+1)(x^2+16)}}}

{{{f(x)=x^3+16x+x^2+16}}}

{{{f(x)=x^3+x^2+16x+16}}}