Question 1126935

given: 

{{{x[1]=2i}}}, if that zero given,  the conjugate  {{{x[2]=-2i}}} must also be a zero
{{{x[3]=1-2i}}},  we  also have {{{x[4]=1+2i}}}

recall: complex zeros always come in pairs

{{{x[5]= -5 }}}

{{{ f(0)=200}}}


{{{f(x)=C(x-x[1])(x-x[2])(x-x[3])(x-x[4])(x-x[5])}}} where {{{C}}} is a constant, we put this for the y-intercept condition

{{{f(x)=C(x-2i)(x-(-2i))(x-(1-2i))(x-(1+2i))(x-(-5)) }}}

{{{f(x)=C(x-2i)(x+2i)(x-1+2i)(x-1-2i)(x+5)}}} 

{{{f(x)=C(x^2-4i^2)(x^2 - 2 x + 5)(x+5) }}}

{{{f(x)=C(x^2-4(-1))(x^3 + 3 x^2 - 5 x + 25) }}}

{{{f(x)=C(x^2+4)(x^3 + 3 x^2 - 5 x + 25)}}} 

{{{f(x)=C(x^5 + 3 x^4 - x^3 + 37 x^2 - 20 x + 100 )}}}


Using the fact that {{{f(0)=200}}}, we can solve for {{{C}}}.
 
{{{200 = C (0^5 + 3 *0^4 - 0^3 + 37*0^2 - 20 *0 + 100 )}}}
 
{{{200 = 100C}}}

{{{C=200/100}}}

{{{C=2}}}

so, we have

{{{f(x)=2(x^5 + 3 x^4 - x^3 + 37 x^2 - 20 x + 100 )}}}

{{{f(x)=2x^5 + 6x^4 - 2x^3 + 74x^2 - 40 x + 200 }}}