Divide using synthetic division $$ ( 7x^{3}+31x^{2}-16x+4 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&7&31&-16&4\\& & -35& 20& \color{black}{-20} \\ \hline &\color{blue}{7}&\color{blue}{-4}&\color{blue}{4}&\color{orangered}{-16} \end{array} $$The solution is:
$$ \frac{ 7x^{3}+31x^{2}-16x+4 }{ x+5 } = \color{blue}{7x^{2}-4x+4} \color{red}{~-~} \frac{ \color{red}{ 16 } }{ x+5 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&7&31&-16&4\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 7 }&31&-16&4\\& & & & \\ \hline &\color{orangered}{7}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 7 } = \color{blue}{ -35 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&7&31&-16&4\\& & \color{blue}{-35} & & \\ \hline &\color{blue}{7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 31 } + \color{orangered}{ \left( -35 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}-5&7&\color{orangered}{ 31 }&-16&4\\& & \color{orangered}{-35} & & \\ \hline &7&\color{orangered}{-4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&7&31&-16&4\\& & -35& \color{blue}{20} & \\ \hline &7&\color{blue}{-4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 20 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}-5&7&31&\color{orangered}{ -16 }&4\\& & -35& \color{orangered}{20} & \\ \hline &7&-4&\color{orangered}{4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 4 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&7&31&-16&4\\& & -35& 20& \color{blue}{-20} \\ \hline &7&-4&\color{blue}{4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrr}-5&7&31&-16&\color{orangered}{ 4 }\\& & -35& 20& \color{orangered}{-20} \\ \hline &\color{blue}{7}&\color{blue}{-4}&\color{blue}{4}&\color{orangered}{-16} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{2}-4x+4 } $ with a remainder of $ \color{red}{ -16 } $.