Divide using synthetic division $$ ( 6x^{3}+5x^{2}-16x-15 ) \div ( 2x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&6&5&-16&-15\\& & -18& 39& \color{black}{-69} \\ \hline &\color{blue}{6}&\color{blue}{-13}&\color{blue}{23}&\color{orangered}{-84} \end{array} $$The solution is:
$$ \frac{ 6x^{3}+5x^{2}-16x-15 }{ x+3 } = \color{blue}{6x^{2}-13x+23} \color{red}{~-~} \frac{ \color{red}{ 84 } }{ x+3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&6&5&-16&-15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 6 }&5&-16&-15\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 6 } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&6&5&-16&-15\\& & \color{blue}{-18} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrr}-3&6&\color{orangered}{ 5 }&-16&-15\\& & \color{orangered}{-18} & & \\ \hline &6&\color{orangered}{-13}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -13 \right) } = \color{blue}{ 39 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&6&5&-16&-15\\& & -18& \color{blue}{39} & \\ \hline &6&\color{blue}{-13}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 39 } = \color{orangered}{ 23 } $
$$ \begin{array}{c|rrrr}-3&6&5&\color{orangered}{ -16 }&-15\\& & -18& \color{orangered}{39} & \\ \hline &6&-13&\color{orangered}{23}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 23 } = \color{blue}{ -69 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&6&5&-16&-15\\& & -18& 39& \color{blue}{-69} \\ \hline &6&-13&\color{blue}{23}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ \left( -69 \right) } = \color{orangered}{ -84 } $
$$ \begin{array}{c|rrrr}-3&6&5&-16&\color{orangered}{ -15 }\\& & -18& 39& \color{orangered}{-69} \\ \hline &\color{blue}{6}&\color{blue}{-13}&\color{blue}{23}&\color{orangered}{-84} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}-13x+23 } $ with a remainder of $ \color{red}{ -84 } $.