Divide using synthetic division $$ ( 7x^{3}+x^{2}-4 ) \div ( x-5 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&7&1&0&-4\\& & 35& 180& \color{black}{900} \\ \hline &\color{blue}{7}&\color{blue}{36}&\color{blue}{180}&\color{orangered}{896} \end{array} $$The solution is:
$$ \dfrac{ 7x^{3}+x^{2}-4 }{ x-5 } = \color{blue}{7x^{2}+36x+180} ~+~ \dfrac{ \color{red}{ 896 } }{ x-5 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{5}&7&1&0&-4\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 7 }&1&0&-4\\& & & & \\ \hline &\color{orangered}{7}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&7&1&0&-4\\& & \color{blue}{35} & & \\ \hline &\color{blue}{7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 35 } = \color{orangered}{ 36 } $
$$ \begin{array}{c|rrrr}5&7&\color{orangered}{ 1 }&0&-4\\& & \color{orangered}{35} & & \\ \hline &7&\color{orangered}{36}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 36 } = \color{blue}{ 180 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&7&1&0&-4\\& & 35& \color{blue}{180} & \\ \hline &7&\color{blue}{36}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 180 } = \color{orangered}{ 180 } $
$$ \begin{array}{c|rrrr}5&7&1&\color{orangered}{ 0 }&-4\\& & 35& \color{orangered}{180} & \\ \hline &7&36&\color{orangered}{180}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 180 } = \color{blue}{ 900 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&7&1&0&-4\\& & 35& 180& \color{blue}{900} \\ \hline &7&36&\color{blue}{180}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 900 } = \color{orangered}{ 896 } $
$$ \begin{array}{c|rrrr}5&7&1&0&\color{orangered}{ -4 }\\& & 35& 180& \color{orangered}{900} \\ \hline &\color{blue}{7}&\color{blue}{36}&\color{blue}{180}&\color{orangered}{896} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{2}+36x+180 } $ with a remainder of $ \color{red}{ 896 } $.