Divide using synthetic division $$ ( x^{3}-16x^{2}+83x-140 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&1&-16&83&-140\\& & -5& 105& \color{black}{-940} \\ \hline &\color{blue}{1}&\color{blue}{-21}&\color{blue}{188}&\color{orangered}{-1080} \end{array} $$The solution is:
$$ \frac{ x^{3}-16x^{2}+83x-140 }{ x+5 } = \color{blue}{x^{2}-21x+188} \color{red}{~-~} \frac{ \color{red}{ 1080 } }{ x+5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 5 = 0 $ ( $ x = \color{blue}{ -5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&-16&83&-140\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 1 }&-16&83&-140\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 1 } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&-16&83&-140\\& & \color{blue}{-5} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -21 } $
$$ \begin{array}{c|rrrr}-5&1&\color{orangered}{ -16 }&83&-140\\& & \color{orangered}{-5} & & \\ \hline &1&\color{orangered}{-21}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -21 \right) } = \color{blue}{ 105 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&-16&83&-140\\& & -5& \color{blue}{105} & \\ \hline &1&\color{blue}{-21}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 83 } + \color{orangered}{ 105 } = \color{orangered}{ 188 } $
$$ \begin{array}{c|rrrr}-5&1&-16&\color{orangered}{ 83 }&-140\\& & -5& \color{orangered}{105} & \\ \hline &1&-21&\color{orangered}{188}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 188 } = \color{blue}{ -940 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&-16&83&-140\\& & -5& 105& \color{blue}{-940} \\ \hline &1&-21&\color{blue}{188}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -140 } + \color{orangered}{ \left( -940 \right) } = \color{orangered}{ -1080 } $
$$ \begin{array}{c|rrrr}-5&1&-16&83&\color{orangered}{ -140 }\\& & -5& 105& \color{orangered}{-940} \\ \hline &\color{blue}{1}&\color{blue}{-21}&\color{blue}{188}&\color{orangered}{-1080} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-21x+188 } $ with a remainder of $ \color{red}{ -1080 } $.