Question 1177978


given:
{{{n=4}}}
{{{x[1]=2i}}}=> you also have {{{x[2]=-2i}}}
{{{x[3]=5i}}} => you also have {{{x[4]=-5i}}}


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

{{{F(x)=a(x-2i)(x+2i)(x-5i)(x+5i)}}}

{{{F(x)=a(x^4 + 29 x^2 + 100)}}}


since given {{{F(-1)=130}}}, use {{{x=-1}}} and {{{F(x)=130}}} to calculate the value of {{{a}}}

{{{130=a((-1)^4 + 29 (-1)^2 + 100)}}}

{{{130=a(1 + 29 + 100)}}}

{{{130=a(130)}}}

{{{a=1}}}

and your equation is

{{{F(x)=x^4 + 29 x^2 + 100}}}