Answer:
[tex]\boxed{\boxed{B.\ f(x)=x^4-3x^2-4}}[/tex]
Step-by-step explanation:
Complex Conjugate Root Theorem-
If [tex]a + bi[/tex] is a root of a polynomial P with a and b real numbers, then its complex conjugate [tex]a-bi[/tex] is also a root of P.
So, all roots of the polynomial function are [tex]i,-i, 2,-2[/tex]
Hence, the function will be,
[tex]f(x)=(x-i)(x+i)(x-2)(x+2)[/tex]
[tex]=[(x-i)(x+i)]\cdot[(x-2)(x+2)][/tex]
[tex]=(x^2-i^2)(x^2-2^2)[/tex]
[tex]=(x^2+1)(x^2-4)[/tex]
[tex]=x^2x^2+x^2\left(-4\right)+1\cdot \:x^2+1\cdot \left(-4\right)[/tex]
[tex]=x^2x^2-4x^2+1\cdot \:x^2-1\cdot \:4[/tex]
[tex]=x^4-3x^2-4[/tex]
The leading coefficient in this case is 1, so the function is,
[tex]f(x)=x^4-3x^2-4[/tex]
