Use synthetic division to divide the polynomials \[ \left(4 x^{2}+20 x+20\right) \div(x+9) \]

**Step 1.** Recognize that the divisor is in the form \( x+9 \), which can be written as \( x-(-9) \). Thus, we will use \(-9\) as our synthetic division number. **Step 2.** Write down the coefficients of the dividend polynomial \( 4x^2 + 20x + 20 \): \[ 4,\quad 20,\quad 20. \] **Step 3.** Set up the synthetic division: \[ \begin{array}{r|ccc} -9 & 4 & 20 & 20 \\ & & & \\ \hline & 4 & & \end{array} \] 1. **Bring down** the first coefficient \(4\). \[ \begin{array}{r|ccc} -9 & 4 & 20 & 20 \\ & & & \\ \hline & 4 & & \end{array} \] 2. **Multiply** \(4\) by \(-9\) to get \(-36\), and write it under the second coefficient: \[ \begin{array}{r|ccc} -9 & 4 & 20 & 20 \\ & & -36 & \\ \hline & 4 & & \end{array} \] 3. **Add** the second coefficient \(20\) and \(-36\): \[ 20 + (-36) = -16. \] Write \(-16\) below the line: \[ \begin{array}{r|ccc} -9 & 4 & 20 & 20 \\ & & -36 & \\ \hline & 4 & -16 & \end{array} \] 4. **Multiply** \(-16\) by \(-9\) to get \(144\), and write it under the third coefficient: \[ \begin{array}{r|ccc} -9 & 4 & 20 & 20 \\ & & -36 & 144 \\ \hline & 4 & -16 & \end{array} \] 5. **Add** the third coefficient \(20\) and \(144\): \[ 20 + 144 = 164. \] Write \(164\) below the line: \[ \begin{array}{r|ccc} -9 & 4 & 20 & 20 \\ & & -36 & 144 \\ \hline & 4 & -16 & 164 \end{array} \] **Step 4.** Interpret the result: - The numbers \(4\) and \(-16\) represent the coefficients of the quotient polynomial \( 4x - 16 \). - The last number, \(164\), is the remainder. Thus, the division can be expressed as: \[ 4x^2 + 20x + 20 = (x+9)(4x - 16) + 164. \]