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. \( f(x)= \)

We are given \[ f(x)=x^{3}-5ix^{2}-25x+125i \] and the fact that \( 5i \) is a zero of \( f(x) \); that is, \( f(5i)=0 \). **Step 1. Factor out \(\left(x-5i\right)\):** Since \( 5i \) is a zero, we have \[ f(x)=(x-5i)Q(x) \] where \( Q(x) \) is a quadratic polynomial. We can find \( Q(x) \) by polynomial division or by synthetic division. **Step 2. Polynomial division:** Divide \( f(x) \) by \( \left(x-5i\right) \). The polynomial \( f(x) \) has coefficients: \[ 1,\quad -5i,\quad -25,\quad 125i. \] Using synthetic division with \( 5i \): - Bring down the first coefficient: \(1\). - Multiply \( 1 \) by \( 5i \) to get \( 5i \); add this to the second coefficient: \[ -5i + 5i = 0. \] - Multiply the new value \( 0 \) by \( 5i \) to get \( 0 \); add to the third coefficient: \[ -25 + 0 = -25. \] - Multiply \(-25\) by \( 5i \) to get \(-125i\); add to the fourth coefficient: \[ 125i+(-125i)=0. \] Thus, the quotient is \[ Q(x)=x^2-25. \] **Step 3. Factor the quadratic:** The quadratic polynomial \( x^2-25 \) is a difference of squares: \[ x^2-25=(x-5)(x+5). \] **Step 4. Write the complete factorization:** Therefore, the factorization of \( f(x) \) into linear factors is: \[ f(x)=(x-5i)(x-5)(x+5). \]