Divide using synthetic division $$ ( 7x^{3}-5x^{2}-21x+15 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&7&-5&-21&15\\& & 35& 150& \color{black}{645} \\ \hline &\color{blue}{7}&\color{blue}{30}&\color{blue}{129}&\color{orangered}{660} \end{array} $$The solution is:
$$ \frac{ 7x^{3}-5x^{2}-21x+15 }{ x-5 } = \color{blue}{7x^{2}+30x+129} ~+~ \frac{ \color{red}{ 660 } }{ x-5 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{5}&7&-5&-21&15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 7 }&-5&-21&15\\& & & & \\ \hline &\color{orangered}{7}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&7&-5&-21&15\\& & \color{blue}{35} & & \\ \hline &\color{blue}{7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 35 } = \color{orangered}{ 30 } $
$$ \begin{array}{c|rrrr}5&7&\color{orangered}{ -5 }&-21&15\\& & \color{orangered}{35} & & \\ \hline &7&\color{orangered}{30}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 30 } = \color{blue}{ 150 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&7&-5&-21&15\\& & 35& \color{blue}{150} & \\ \hline &7&\color{blue}{30}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ 150 } = \color{orangered}{ 129 } $
$$ \begin{array}{c|rrrr}5&7&-5&\color{orangered}{ -21 }&15\\& & 35& \color{orangered}{150} & \\ \hline &7&30&\color{orangered}{129}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 129 } = \color{blue}{ 645 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&7&-5&-21&15\\& & 35& 150& \color{blue}{645} \\ \hline &7&30&\color{blue}{129}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ 645 } = \color{orangered}{ 660 } $
$$ \begin{array}{c|rrrr}5&7&-5&-21&\color{orangered}{ 15 }\\& & 35& 150& \color{orangered}{645} \\ \hline &\color{blue}{7}&\color{blue}{30}&\color{blue}{129}&\color{orangered}{660} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{2}+30x+129 } $ with a remainder of $ \color{red}{ 660 } $.