Select all of the roots/zeros of \( f(x)=x^{4}+2 x^{3}-6 x^{2}+4 x-16 \) select 3 correct answer(s) \( \square^{-4} \) \( \square^{2} \) \( \pm i \sqrt{2} \) \( \square^{8} \) \( \square^{-2} \)

To find the roots (zeros) of the polynomial \( f(x) = x^{4} + 2x^{3} - 6x^{2} + 4x - 16 \), we can follow these steps: 1. **Identify Possible Rational Roots:** Using the Rational Root Theorem, the possible rational roots are factors of the constant term (\(-16\)) divided by factors of the leading coefficient (\(1\)). Thus, the possible rational roots are \( \pm1, \pm2, \pm4, \pm8, \pm16 \). 2. **Test the Possible Roots:** - **Testing \( x = 2 \):** \[ f(2) = 2^4 + 2(2)^3 - 6(2)^2 + 4(2) - 16 = 16 + 16 - 24 + 8 - 16 = 0 \] So, \( x = 2 \) is a root. 3. **Factor the Polynomial:** Since \( x = 2 \) is a root, we can factor \( f(x) \) as: \[ f(x) = (x - 2)(x^3 + 4x^2 + 2x + 8) \] 4. **Find Remaining Roots:** To factor the cubic polynomial \( x^3 + 4x^2 + 2x + 8 \), we test other possible roots: - **Testing \( x = -4 \):** \[ (-4)^3 + 4(-4)^2 + 2(-4) + 8 = -64 + 64 - 8 + 8 = 0 \] So, \( x = -4 \) is another root. - **Factor Further:** \[ x^3 + 4x^2 + 2x + 8 = (x + 4)(x^2 + 0x + 2) = (x + 4)(x^2 + 2) \] The quadratic equation \( x^2 + 2 = 0 \) has roots \( x = \pm i\sqrt{2} \). 5. **Summary of Roots:** The polynomial \( f(x) \) has the following roots: \[ x = 2,\quad x = -4,\quad x = i\sqrt{2},\quad x = -i\sqrt{2} \] **Correct Options:** - \( \boxed{-4} \) - \( \boxed{2} \) - \( \boxed{\pm i \sqrt{2}} \) These correspond to the first three options you provided.