Divide using synthetic division $$ ( 6x^{3}-x^{2}-x+10 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&6&-1&-1&10\\& & -18& 57& \color{black}{-168} \\ \hline &\color{blue}{6}&\color{blue}{-19}&\color{blue}{56}&\color{orangered}{-158} \end{array} $$The solution is:
$$ \frac{ 6x^{3}-x^{2}-x+10 }{ x+3 } = \color{blue}{6x^{2}-19x+56} \color{red}{~-~} \frac{ \color{red}{ 158 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&6&-1&-1&10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 6 }&-1&-1&10\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&6&-1&-1&10\\& & \color{blue}{-18} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -19 } $
$$ \begin{array}{c|rrrr}-3&6&\color{orangered}{ -1 }&-1&10\\& & \color{orangered}{-18} & & \\ \hline &6&\color{orangered}{-19}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -19 \right) } = \color{blue}{ 57 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&6&-1&-1&10\\& & -18& \color{blue}{57} & \\ \hline &6&\color{blue}{-19}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 57 } = \color{orangered}{ 56 } $
$$ \begin{array}{c|rrrr}-3&6&-1&\color{orangered}{ -1 }&10\\& & -18& \color{orangered}{57} & \\ \hline &6&-19&\color{orangered}{56}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 56 } = \color{blue}{ -168 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&6&-1&-1&10\\& & -18& 57& \color{blue}{-168} \\ \hline &6&-19&\color{blue}{56}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -168 \right) } = \color{orangered}{ -158 } $
$$ \begin{array}{c|rrrr}-3&6&-1&-1&\color{orangered}{ 10 }\\& & -18& 57& \color{orangered}{-168} \\ \hline &\color{blue}{6}&\color{blue}{-19}&\color{blue}{56}&\color{orangered}{-158} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}-19x+56 } $ with a remainder of $ \color{red}{ -158 } $.