Suppose that \( 5 i \) is a zero of \( f(x)=x^{3}-5 i x^{2}-25 x+125 i \) and \( g(x)=x^{3}-5 x^{2}+25 x-125 \) Part: \( 0 / 5 \) Part 1 of 5 (a) Factor \( f(x) \) into linear factors. \( f(x)= \)

To factor \( f(x) = x^3 - 5ix^2 - 25x + 125i \) given that \( 5i \) is a zero, we can use polynomial long division or synthetic division. Since \( 5i \) is a root, we can factor \( f(x) \) as follows: 1. Perform synthetic division of \( f(x) \) by \( x - 5i \). 2. After dividing, you'll find the quotient is a quadratic polynomial. 3. Use the quadratic formula or factoring to break down the quadratic part into linear factors. After completing the division, the result looks something like this: \[ f(x) = (x - 5i)(x^2 + ax + b) \] With some calculations, you'll discover that: \[ f(x) = (x - 5i)\left(x^2 + 25\right) \] Finally, you can express \( x^2 + 25 \) as: \[ x^2 + 25 = (x - 5i)(x + 5i) \] So, \( f(x) \) factors to: \[ f(x) = (x - 5i)(x - 5i)(x + 5i) = (x - 5i)^2(x + 5i) \] In conclusion: \[ f(x) = (x - 5i)^2(x + 5i) \]