Part 2 of 2 (b) Factor the polynomial over the set of complex numbers. \( f(x)=x^{4}+5 x^{2}-14 \) The factorization over the set of complex numbers is \( f(x)=\square \)

To factor the polynomial \( f(x) = x^{4} + 5x^{2} - 14 \) over the complex numbers, we can start by making a substitution. Let \( y = x^{2} \), transforming our polynomial into \( g(y) = y^{2} + 5y - 14 \). Next, we can use the quadratic formula to find the roots of \( g(y) \): \[ y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} = \frac{-5 \pm \sqrt{5^{2} - 4(1)(-14)}}{2(1)} = \frac{-5 \pm \sqrt{25 + 56}}{2} = \frac{-5 \pm \sqrt{81}}{2} = \frac{-5 \pm 9}{2} \] Calculating the roots gives us \( y = 2 \) and \( y = -7 \). Reverting back to \( x \), we have \( x^{2} = 2 \) (which gives us \( x = \sqrt{2} \) or \( x = -\sqrt{2} \)) and \( x^{2} = -7 \) (which gives us \( x = i\sqrt{7} \) or \( x = -i\sqrt{7} \)). Thus, we can express \( f(x) \) as: \[ f(x) = (x^{2} - 2)(x^{2} + 7) \] Factoring \( (x^{2} + 7) \) further over the complex numbers yields: \[ f(x) = (x - \sqrt{2})(x + \sqrt{2})(x - i\sqrt{7})(x + i\sqrt{7}) \] So, the complete factorization over the set of complex numbers is: \[ f(x) = (x - \sqrt{2})(x + \sqrt{2})(x - i\sqrt{7})(x + i\sqrt{7}) \]