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