Divide using synthetic division $$ ( 7x^{3}-5x^{2}-21x+15 ) \div ( x-15 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}15&7&-5&-21&15\\& & 105& 1500& \color{black}{22185} \\ \hline &\color{blue}{7}&\color{blue}{100}&\color{blue}{1479}&\color{orangered}{22200} \end{array} $$The solution is:
$$ \frac{ 7x^{3}-5x^{2}-21x+15 }{ x-15 } = \color{blue}{7x^{2}+100x+1479} ~+~ \frac{ \color{red}{ 22200 } }{ x-15 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -15 = 0 $ ( $ x = \color{blue}{ 15 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{15}&7&-5&-21&15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}15&\color{orangered}{ 7 }&-5&-21&15\\& & & & \\ \hline &\color{orangered}{7}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 15 } \cdot \color{blue}{ 7 } = \color{blue}{ 105 } $.
$$ \begin{array}{c|rrrr}\color{blue}{15}&7&-5&-21&15\\& & \color{blue}{105} & & \\ \hline &\color{blue}{7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 105 } = \color{orangered}{ 100 } $
$$ \begin{array}{c|rrrr}15&7&\color{orangered}{ -5 }&-21&15\\& & \color{orangered}{105} & & \\ \hline &7&\color{orangered}{100}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 15 } \cdot \color{blue}{ 100 } = \color{blue}{ 1500 } $.
$$ \begin{array}{c|rrrr}\color{blue}{15}&7&-5&-21&15\\& & 105& \color{blue}{1500} & \\ \hline &7&\color{blue}{100}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ 1500 } = \color{orangered}{ 1479 } $
$$ \begin{array}{c|rrrr}15&7&-5&\color{orangered}{ -21 }&15\\& & 105& \color{orangered}{1500} & \\ \hline &7&100&\color{orangered}{1479}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 15 } \cdot \color{blue}{ 1479 } = \color{blue}{ 22185 } $.
$$ \begin{array}{c|rrrr}\color{blue}{15}&7&-5&-21&15\\& & 105& 1500& \color{blue}{22185} \\ \hline &7&100&\color{blue}{1479}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ 22185 } = \color{orangered}{ 22200 } $
$$ \begin{array}{c|rrrr}15&7&-5&-21&\color{orangered}{ 15 }\\& & 105& 1500& \color{orangered}{22185} \\ \hline &\color{blue}{7}&\color{blue}{100}&\color{blue}{1479}&\color{orangered}{22200} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{2}+100x+1479 } $ with a remainder of $ \color{red}{ 22200 } $.