Let \(n\) be a positive integer. Then \(x^2 +1\) is a factor of



  1. all \(n\);

  2. even \(n\);

  3. odd \(n\);

  4. \(n \geq 3\);

  5. no values of \(n\).

Note that the polynomial is a function of \(x^2\).

Therefore, let the polynomial \((3+x^4)^n-(x^2+3)^n(x^2-1)^n = f(x^2)\).

From the factor theorem we know that if \(x^2+1\) is a factor of \(f(x^2)\), then \(f(-1) = 0\).

This gives:

\[(3+1)^n-(-1+3)^n(-1-1)^n = 4^n - (-1)^n4^n.\]

This is zero if and only if \(n\) is even, and so the factor theorem tells us the answer is (b).

If we know about complex numbers, we can say the solution to \(x^2 +1 = 0\) is \(x = i, -i.\)

Putting either of these values into \((3+x^4)^n-(x^2+3)^n(x^2-1)^n\) for \(x\) gives us \(4^n - (-1)^n4^n\), as before.