Divide using synthetic division $$ ( 6x^{3}-10x^{2}-7x-15 ) \div ( x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&6&-10&-7&-15\\& & -6& 16& \color{black}{-9} \\ \hline &\color{blue}{6}&\color{blue}{-16}&\color{blue}{9}&\color{orangered}{-24} \end{array} $$The solution is:
$$ \frac{ 6x^{3}-10x^{2}-7x-15 }{ x+1 } = \color{blue}{6x^{2}-16x+9} \color{red}{~-~} \frac{ \color{red}{ 24 } }{ x+1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 1 = 0 $ ( $ x = \color{blue}{ -1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&6&-10&-7&-15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 6 }&-10&-7&-15\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 6 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&6&-10&-7&-15\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrr}-1&6&\color{orangered}{ -10 }&-7&-15\\& & \color{orangered}{-6} & & \\ \hline &6&\color{orangered}{-16}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&6&-10&-7&-15\\& & -6& \color{blue}{16} & \\ \hline &6&\color{blue}{-16}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 16 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrr}-1&6&-10&\color{orangered}{ -7 }&-15\\& & -6& \color{orangered}{16} & \\ \hline &6&-16&\color{orangered}{9}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 9 } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&6&-10&-7&-15\\& & -6& 16& \color{blue}{-9} \\ \hline &6&-16&\color{blue}{9}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ -24 } $
$$ \begin{array}{c|rrrr}-1&6&-10&-7&\color{orangered}{ -15 }\\& & -6& 16& \color{orangered}{-9} \\ \hline &\color{blue}{6}&\color{blue}{-16}&\color{blue}{9}&\color{orangered}{-24} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}-16x+9 } $ with a remainder of $ \color{red}{ -24 } $.