Question 956247

given:

{{{n=3}}}, {{{4}}} and {{{5i}}} are zeros;
{{{f(-1) = 260}}}


{{{f(x)=(x-3)(x-4)(x-5i)}}}....multiply all and you will get

{{{f(x)=x^3-7x^2+(-5x^2+35x-60)*i+12x }}}

to get {{{f(-1) = 260}}}, we need to find some constant {{{k}}} to multiply {{{x^3-7x^2+(-5x^2+35x-60)*i+12x }}} for {{{x=-1}}} to get {{{260}}}

{{{((-1)^3-7(-1)^2+i(-5(-1)^2+35(-1)-60)+12(-1) )k= 260}}}
{{{(-1-7+i(-5-35-60)-12 )k= 260}}}
:
:
{{{k = -1/2+(5i)/2}}}

then we have

{{{f(x)=(x^3-7x^2+i(-5x^2+35x-60)+12x)(-1/2+(5i)/2)}}}

{{{f(x)=-x^3/2+16x^2+((5x^3)/2-15x^2+(25x)/2+30)*i-(187x)/2+150}}}