Answer the following questions about the equation below. \( x^{3}-17 x-4=0 \) (a) List all rational roots that are possible according to the Rational Zero Theorem. Choose the correct answer below. A. \( \pm 1 \) B. \( \pm 4 \) C. \( -1,-2,-4 \) D. \( \pm 1, \pm 2, \pm 4 \) (b) Use synthetic division to test several possible rational roots in order to identify one actual root. One rational root of the given equation is (Simplify your answer.)

To solve the equation \( x^{3}-17 x-4=0 \) using the Rational Root Theorem, we first identify the possible rational roots. ### (a) List all rational roots that are possible according to the Rational Zero Theorem. The Rational Root Theorem states that any rational solution, in the form of \( \frac{p}{q} \), must have \( p \) as a factor of the constant term (which is -4) and \( q \) as a factor of the leading coefficient (which is 1). The factors of -4 (the constant term) are: - \( \pm 1 \) - \( \pm 2 \) - \( \pm 4 \) The factors of 1 (the leading coefficient) are: - \( \pm 1 \) Thus, the possible rational roots are: \[ \pm 1, \pm 2, \pm 4 \] So the correct answer is: **D. \( \pm 1, \pm 2, \pm 4 \)** ### (b) Use synthetic division to test several possible rational roots in order to identify one actual root. Let's test the possible rational roots using synthetic division. We will start with \( x = 1 \): **Testing \( x = 1 \):** \[ \begin{array}{r|rrrr} 1 & 1 & 0 & -17 & -4 \\ & & 1 & 1 & -16 \\ \hline & 1 & 1 & -16 & -20 \\ \end{array} \] The remainder is -20, so \( x = 1 \) is not a root. **Testing \( x = -1 \):** \[ \begin{array}{r|rrrr} -1 & 1 & 0 & -17 & -4 \\ & & -1 & 1 & 16 \\ \hline & 1 & -1 & -16 & 12 \\ \end{array} \] The remainder is 12, so \( x = -1 \) is not a root. **Testing \( x = 2 \):** \[ \begin{array}{r|rrrr} 2 & 1 & 0 & -17 & -4 \\ & & 2 & 4 & -26 \\ \hline & 1 & 2 & -13 & -30 \\ \end{array} \] The remainder is -30, so \( x = 2 \) is not a root. **Testing \( x = -2 \):** \[ \begin{array}{r|rrrr} -2 & 1 & 0 & -17 & -4 \\ & & -2 & 4 & 26 \\ \hline & 1 & -2 & -13 & 22 \\ \end{array} \] The remainder is 22, so \( x = -2 \) is not a root. **Testing \( x = 4 \):** \[ \begin{array}{r|rrrr} 4 & 1 & 0 & -17 & -4 \\ & & 4 & 16 & -4 \\ \hline & 1 & 4 & -1 & -8 \\ \end{array} \] The remainder is -8, so \( x = 4 \) is not a root. **Testing \( x = -4 \):** \[ \begin{array}{r|rrrr} -4 & 1 & 0 & -17 & -4 \\ & & -4 & 16 & 4 \\ \hline & 1 & -4 & -1 & 0 \\ \end{array} \] The remainder is 0, so \( x = -4 \) is a root. Thus, one rational root of the given equation is: \[ \boxed{-4} \]