Divide using synthetic division $$ ( 15x^{3}+x^{2}+4 ) \div ( 3x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&15&1&0&4\\& & -30& 58& \color{black}{-116} \\ \hline &\color{blue}{15}&\color{blue}{-29}&\color{blue}{58}&\color{orangered}{-112} \end{array} $$The solution is:
$$ \frac{ 15x^{3}+x^{2}+4 }{ x+2 } = \color{blue}{15x^{2}-29x+58} \color{red}{~-~} \frac{ \color{red}{ 112 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&15&1&0&4\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 15 }&1&0&4\\& & & & \\ \hline &\color{orangered}{15}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 15 } = \color{blue}{ -30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&15&1&0&4\\& & \color{blue}{-30} & & \\ \hline &\color{blue}{15}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ -29 } $
$$ \begin{array}{c|rrrr}-2&15&\color{orangered}{ 1 }&0&4\\& & \color{orangered}{-30} & & \\ \hline &15&\color{orangered}{-29}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -29 \right) } = \color{blue}{ 58 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&15&1&0&4\\& & -30& \color{blue}{58} & \\ \hline &15&\color{blue}{-29}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 58 } = \color{orangered}{ 58 } $
$$ \begin{array}{c|rrrr}-2&15&1&\color{orangered}{ 0 }&4\\& & -30& \color{orangered}{58} & \\ \hline &15&-29&\color{orangered}{58}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 58 } = \color{blue}{ -116 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&15&1&0&4\\& & -30& 58& \color{blue}{-116} \\ \hline &15&-29&\color{blue}{58}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -116 \right) } = \color{orangered}{ -112 } $
$$ \begin{array}{c|rrrr}-2&15&1&0&\color{orangered}{ 4 }\\& & -30& 58& \color{orangered}{-116} \\ \hline &\color{blue}{15}&\color{blue}{-29}&\color{blue}{58}&\color{orangered}{-112} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 15x^{2}-29x+58 } $ with a remainder of $ \color{red}{ -112 } $.