Divide using synthetic division $$ ( 6x^{3}+7x^{2}-x+26 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&6&7&-1&26\\& & -18& 33& \color{black}{-96} \\ \hline &\color{blue}{6}&\color{blue}{-11}&\color{blue}{32}&\color{orangered}{-70} \end{array} $$The solution is:
$$ \frac{ 6x^{3}+7x^{2}-x+26 }{ x+3 } = \color{blue}{6x^{2}-11x+32} \color{red}{~-~} \frac{ \color{red}{ 70 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&6&7&-1&26\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 6 }&7&-1&26\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&6&7&-1&26\\& & \color{blue}{-18} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -11 } $
$$ \begin{array}{c|rrrr}-3&6&\color{orangered}{ 7 }&-1&26\\& & \color{orangered}{-18} & & \\ \hline &6&\color{orangered}{-11}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -11 \right) } = \color{blue}{ 33 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&6&7&-1&26\\& & -18& \color{blue}{33} & \\ \hline &6&\color{blue}{-11}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 33 } = \color{orangered}{ 32 } $
$$ \begin{array}{c|rrrr}-3&6&7&\color{orangered}{ -1 }&26\\& & -18& \color{orangered}{33} & \\ \hline &6&-11&\color{orangered}{32}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 32 } = \color{blue}{ -96 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&6&7&-1&26\\& & -18& 33& \color{blue}{-96} \\ \hline &6&-11&\color{blue}{32}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 26 } + \color{orangered}{ \left( -96 \right) } = \color{orangered}{ -70 } $
$$ \begin{array}{c|rrrr}-3&6&7&-1&\color{orangered}{ 26 }\\& & -18& 33& \color{orangered}{-96} \\ \hline &\color{blue}{6}&\color{blue}{-11}&\color{blue}{32}&\color{orangered}{-70} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}-11x+32 } $ with a remainder of $ \color{red}{ -70 } $.