Divide using synthetic division $$ ( x^{3}+x^{2}-18x-40 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&1&-18&-40\\& & 3& 12& \color{black}{-18} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{-6}&\color{orangered}{-58} \end{array} $$The solution is:
$$ \frac{ x^{3}+x^{2}-18x-40 }{ x-3 } = \color{blue}{x^{2}+4x-6} \color{red}{~-~} \frac{ \color{red}{ 58 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&1&-18&-40\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&1&-18&-40\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&1&-18&-40\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 3 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ 1 }&-18&-40\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&1&-18&-40\\& & 3& \color{blue}{12} & \\ \hline &1&\color{blue}{4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 12 } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}3&1&1&\color{orangered}{ -18 }&-40\\& & 3& \color{orangered}{12} & \\ \hline &1&4&\color{orangered}{-6}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&1&-18&-40\\& & 3& 12& \color{blue}{-18} \\ \hline &1&4&\color{blue}{-6}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -40 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -58 } $
$$ \begin{array}{c|rrrr}3&1&1&-18&\color{orangered}{ -40 }\\& & 3& 12& \color{orangered}{-18} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{-6}&\color{orangered}{-58} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+4x-6 } $ with a remainder of $ \color{red}{ -58 } $.