Question 763112
Since coefficients are real this must mean that the complex roots occur in conjugate pairs. We can write:

{{{
f(x)=a(x-i)(x+i)(x-5i)(x+5i)
}}}


and given that f(-1)=52 we can determine a...


{{{
f(-1)=a(-1-i)(-1+i)(-1-5i)(1+5i)=52
}}}

{{{a((-1)^2+1^2)((-1)^2+5^2)=52}}}


so that

{{{a(2)(26)=52}}}


and a=1

this leads to

{{{
f(x)=(x^2+1)(x^2+25)=x^4+26x^2+25
}}}



{{{f(x)=x^4+26x^2+25
}}}



:)