Question 970035
Degree {{{3}}}; 
zeros: 
{{{x[1]=8}}}, 
{{{x[2]=-6-i}}}=>we also have {{{x[3]=-6+i}}} because complex zeros come always in pairs

then, using zero product rule, {{{f(x)}}} will be:

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

{{{f(x)=(x-8)(x-(-6-i))(x-(-6+i))}}}

{{{f(x)=(x-8)(x+6+i)(x+6-i)}}}

{{{f(x)=(x-8)(x+6+i)(x+6-i)}}}

{{{f(x)=(x-8)(x^2+6x-i*x+6x+36-6i+xi+6i-i^2)}}}

{{{f(x)=(x-8)(x^2+12x+36-(-1))}}}

{{{f(x)=(x-8)(x^2+12x+37)}}}

{{{f(x)=x^3+12x^2+37x-8x^2-96x-296}}}

{{{f(x)=x^3+4x^2-59x-296}}}


{{{ graph( 600, 600, -15, 15, -355, 50, x^3+4x^2-59x-296) }}}