Divide using synthetic division $$ ( x^{3}+17x^{2}+70x-54 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&1&17&70&-54\\& & -3& -42& \color{black}{-84} \\ \hline &\color{blue}{1}&\color{blue}{14}&\color{blue}{28}&\color{orangered}{-138} \end{array} $$The solution is:
$$ \frac{ x^{3}+17x^{2}+70x-54 }{ x+3 } = \color{blue}{x^{2}+14x+28} \color{red}{~-~} \frac{ \color{red}{ 138 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&17&70&-54\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 1 }&17&70&-54\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&17&70&-54\\& & \color{blue}{-3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 17 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrr}-3&1&\color{orangered}{ 17 }&70&-54\\& & \color{orangered}{-3} & & \\ \hline &1&\color{orangered}{14}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 14 } = \color{blue}{ -42 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&17&70&-54\\& & -3& \color{blue}{-42} & \\ \hline &1&\color{blue}{14}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 70 } + \color{orangered}{ \left( -42 \right) } = \color{orangered}{ 28 } $
$$ \begin{array}{c|rrrr}-3&1&17&\color{orangered}{ 70 }&-54\\& & -3& \color{orangered}{-42} & \\ \hline &1&14&\color{orangered}{28}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 28 } = \color{blue}{ -84 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&17&70&-54\\& & -3& -42& \color{blue}{-84} \\ \hline &1&14&\color{blue}{28}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -54 } + \color{orangered}{ \left( -84 \right) } = \color{orangered}{ -138 } $
$$ \begin{array}{c|rrrr}-3&1&17&70&\color{orangered}{ -54 }\\& & -3& -42& \color{orangered}{-84} \\ \hline &\color{blue}{1}&\color{blue}{14}&\color{blue}{28}&\color{orangered}{-138} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+14x+28 } $ with a remainder of $ \color{red}{ -138 } $.