Use synthetic division to divide the polynomials. $$ \left(3 x^{2}+7 x+9\right) \div(x+4)=\square $$

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1. Rewrite the divisor in the form $$ x - c $$. Note that
   $$
   x + 4 = x - (-4),
   $$
   so $$ c = -4 $$.

2. Write down the coefficients of the dividend $$ 3x^2 + 7x + 9 $$; they are $$ 3 $$, $$ 7 $$, and $$ 9 $$.

3. Set up the synthetic division:

$$
   \begin{array}{r|ccc}
   -4 & 3 & 7 & 9 \
   \hline
      &   &   &   \
   \end{array}
   $$

4. Bring down the leading coefficient:
   
   $$
   \begin{array}{r|ccc}
   -4 & 3 & 7 & 9 \
   \hline
      & 3 &   &   \
   \end{array}
   $$

5. Multiply $$ 3 $$ by $$ -4 $$ and write the result under the next coefficient:
   
   $$
   3 \times (-4) = -12.
   $$
   
   $$
   \begin{array}{r|ccc}
   -4 & 3 & 7   & 9 \
   \hline
      & 3 & -12 &   \
   \end{array}
   $$

6. Add the column: $$ 7 + (-12) = -5 $$. Write $$-5$$ in the bottom row:
   
   $$
   \begin{array}{r|ccc}
   -4 & 3 & 7   & 9 \
   \hline
      & 3 & -5 &   \
   \end{array}
   $$

7. Multiply $$-5$$ by $$-4$$ to get:
   
   $$
   -5 \times (-4) = 20.
   $$
   
   Write $$20$$ under the next coefficient:
   
   $$
   \begin{array}{r|ccc}
   -4 & 3 & 7   & 9  \
   \hline
      & 3 & -5  & 20 \
   \end{array}
   $$

8. Add the column: $$ 9 + 20 = 29 $$. Write $$29$$ in the bottom row:
   
   $$
   \begin{array}{r|ccc}
   -4 & 3 & 7   & 9 \
   \hline
      & 3 & -5 & 29 \
   \end{array}
   $$

9. The bottom row gives the coefficients of the quotient and the remainder. Since we started with a quadratic polynomial, the quotient will be of degree $$ 1 $$:
   
   $$
   \text{Quotient: } 3x - 5,\quad \text{Remainder: } 29.
   $$

10. Therefore, the division is expressed as:
    
    $$
    3x^2 + 7x + 9 = (x+4)(3x-5) + 29,
    $$
    
    or equivalently,
    
    $$
    \frac{3x^2+7x+9}{x+4} = 3x - 5 + \frac{29}{x+4}.
    $$

The final answer is: $$ 3x - 5 + \frac{29}{x+4} $$.
The result of dividing $$ 3x^2 + 7x + 9 $$ by $$ x + 4 $$ is $$ 3x - 5 $$ with a remainder of 29.

Use synthetic division to divide the polynomials.

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