15. Given that \( \mathrm{x}=1 \) is a zero of the polynomial: \( f(x)=x^{3}+8 x^{2}+11 x-20 \), find the other zeros.

1. Since it is given that \( x = 1 \) is a zero of the polynomial \[ f(x) = x^3 + 8x^2 + 11x - 20, \] we can factor \( f(x) \) by dividing it by \( (x - 1) \). 2. We perform synthetic division with the zero \( 1 \) using the coefficients \( 1 \), \( 8 \), \( 11 \), and \( -20 \): \[ \begin{array}{r|rrrr} 1 & 1 & 8 & 11 & -20 \\ \hline & & 1 & 9 & 20 \\ \end{array} \] - Bring down the first coefficient: \( 1 \). - Multiply \( 1 \) (the zero) by \( 1 \) to get \( 1 \); add this to \( 8 \) to get \( 9 \). - Multiply \( 1 \) by \( 9 \) to get \( 9 \); add this to \( 11 \) to get \( 20 \). - Multiply \( 1 \) by \( 20 \) to get \( 20 \); add this to \( -20 \) to obtain the remainder \( 0 \). Since the remainder is \( 0 \), the division is exact and we have: \[ f(x) = (x - 1)(x^2 + 9x + 20). \] 3. Next, we factor the quadratic \( x^2 + 9x + 20 \). We look for two numbers that add up to \( 9 \) and multiply to \( 20 \). These numbers are \( 4 \) and \( 5 \): \[ x^2 + 9x + 20 = (x + 4)(x + 5). \] 4. Therefore, the complete factorization of the polynomial is: \[ f(x) = (x - 1)(x + 4)(x + 5). \] 5. The zeros of the polynomial \( f(x) \) are the solutions to: \[ x - 1 = 0,\quad x + 4 = 0,\quad x + 5 = 0. \] Solving these equations gives: \[ x = 1,\quad x = -4,\quad x = -5. \] The other zeros of the polynomial are \( x = -4 \) and \( x = -5 \).