Use synthetic division to find the quotient and the remainder. \[ \left(x^{3}-3 x^{2}-9\right) \div(x+2) \]

To divide \[ x^3 - 3x^2 - 9 \] by \[ x+2, \] we first write the dividend in full polynomial form (including terms with zero coefficients): \[ x^3 - 3x^2 + 0x - 9. \] Since we are dividing by \(x+2\), we use synthetic division with \( -2 \) (the zero of \(x+2\), i.e., \(x=-2\)). **Step 1:** Write the coefficients: \[ 1 \quad -3 \quad 0 \quad -9. \] **Step 2:** Set up the synthetic division: \[ \begin{array}{cccccc} -2 & \quad | & \quad 1 & \quad -3 & \quad 0 & \quad -9 \\ & & & & & \quad \\ & & \downarrow\quad & & & \\ & & 1 & & & \\ \end{array} \] **Step 3:** Bring down the first coefficient: \[ 1. \] Multiply \(1\) by \(-2\): \[ 1 \times (-2) = -2. \] Add \(-2\) to the next coefficient \(-3\): \[ -3 + (-2) = -5. \] **Step 4:** Write \(-5\) in the row below and continue: Multiply \(-5\) by \(-2\): \[ -5 \times (-2) = 10. \] Add \(10\) to the next coefficient \(0\): \[ 0 + 10 = 10. \] **Step 5:** Write \(10\) in the row below: Multiply \(10\) by \(-2\): \[ 10 \times (-2) = -20. \] Add \(-20\) to the last coefficient \(-9\): \[ -9 + (-20) = -29. \] The synthetic division table now looks like: \[ \begin{array}{cccccc} -2 & \quad | & \quad 1 & \quad -3 & \quad 0 & \quad -9 \\ & & & -2 & 10 & -20 \\ & & \quad 1 & \quad -5 & \quad 10 & \quad -29 \\ \end{array} \] **Step 6:** Write the quotient and remainder. The quotient is given by the coefficients \(1\), \(-5\), and \(10\), which represents the polynomial: \[ x^2 - 5x + 10, \] and the remainder is \(-29\). Thus, \[ \text{Quotient} = x^2 - 5x + 10 \quad \text{and} \quad \text{Remainder} = -29. \]