Divide using synthetic division $$ ( 3x^{4}+x^{3}-15x^{2}-18x-16 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&3&1&-15&-18&-16\\& & -6& 10& 10& \color{black}{16} \\ \hline &\color{blue}{3}&\color{blue}{-5}&\color{blue}{-5}&\color{blue}{-8}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 3x^{4}+x^{3}-15x^{2}-18x-16 }{ x+2 } = \color{blue}{3x^{3}-5x^{2}-5x-8} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&1&-15&-18&-16\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 3 }&1&-15&-18&-16\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 3 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&1&-15&-18&-16\\& & \color{blue}{-6} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}-2&3&\color{orangered}{ 1 }&-15&-18&-16\\& & \color{orangered}{-6} & & & \\ \hline &3&\color{orangered}{-5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&1&-15&-18&-16\\& & -6& \color{blue}{10} & & \\ \hline &3&\color{blue}{-5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 10 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}-2&3&1&\color{orangered}{ -15 }&-18&-16\\& & -6& \color{orangered}{10} & & \\ \hline &3&-5&\color{orangered}{-5}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&1&-15&-18&-16\\& & -6& 10& \color{blue}{10} & \\ \hline &3&-5&\color{blue}{-5}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 10 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}-2&3&1&-15&\color{orangered}{ -18 }&-16\\& & -6& 10& \color{orangered}{10} & \\ \hline &3&-5&-5&\color{orangered}{-8}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&1&-15&-18&-16\\& & -6& 10& 10& \color{blue}{16} \\ \hline &3&-5&-5&\color{blue}{-8}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 16 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-2&3&1&-15&-18&\color{orangered}{ -16 }\\& & -6& 10& 10& \color{orangered}{16} \\ \hline &\color{blue}{3}&\color{blue}{-5}&\color{blue}{-5}&\color{blue}{-8}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}-5x^{2}-5x-8 } $ with a remainder of $ \color{red}{ 0 } $.