Question 1072882
If a polynomial has real coefficients,
and a complex zero,
the conjugate complex number should also be a zero.
In other words, if {{{a+b*i}}} is a zero, so is {{{a-b*i}}} .
In this case, {{{-i}}} and {{{-3i}}} are also zeros.
With those four different zeros, a fourth degree polynomial must have the factored form
{{{f(x)=c(x-i)(x-(-i))(x-3i)(x-(-3i))}}} for some real non-zero number {{{c}}} .
Simplifying,
{{{f(x)=c(x-i)(x+i)(x-3i)(x+3i)}}}
{{{f(x)=c(x^2-i^2)(x^2-(3i)^2)}}}
{{{f(x)=c(x^2-(-1))(x^2-3^2i^2)}}}
{{{f(x)=c(x^2+1)(x^2-9(-1))}}}
{{{f(x)=c(x^2+1)(x^2+9)}}}
{{{system(f(-1)=20,f(x)=c(x^2+1)(x^2+9))}}}-->{{{system(c((-1)^2+1)((-1)^2+9)=20,f(x)=c(x^2+1)(x^2+9))}}}-->{{{system(c(1+1)(1+9)=20,f(x)=c(x^2+1)(x^2+9))}}}-->{{{system(c*2*10=20,f(x)=c(x^2+1)(x^2+9))}}}-->{{{system(c=1,f(x)=(x^2+1)(x^2+9))}}} .
So {{{highlight(f(x)=(x^2+1)(x^2+9))}}} or multiplying the factors {{{highlight(f(x)=x^4+10x^2+9)}}} .