Divide using synthetic division $$ ( 7x^{3}-5x^{2}-21x+15 ) \div ( x+15 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-15&7&-5&-21&15\\& & -105& 1650& \color{black}{-24435} \\ \hline &\color{blue}{7}&\color{blue}{-110}&\color{blue}{1629}&\color{orangered}{-24420} \end{array} $$The solution is:
$$ \frac{ 7x^{3}-5x^{2}-21x+15 }{ x+15 } = \color{blue}{7x^{2}-110x+1629} \color{red}{~-~} \frac{ \color{red}{ 24420 } }{ x+15 } $$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}{ \left( -105 \right) } = \color{orangered}{ -110 } $
$$ \begin{array}{c|rrrr}-15&7&\color{orangered}{ -5 }&-21&15\\& & \color{orangered}{-105} & & \\ \hline &7&\color{orangered}{-110}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -15 } \cdot \color{blue}{ \left( -110 \right) } = \color{blue}{ 1650 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-15}&7&-5&-21&15\\& & -105& \color{blue}{1650} & \\ \hline &7&\color{blue}{-110}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ 1650 } = \color{orangered}{ 1629 } $
$$ \begin{array}{c|rrrr}-15&7&-5&\color{orangered}{ -21 }&15\\& & -105& \color{orangered}{1650} & \\ \hline &7&-110&\color{orangered}{1629}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -15 } \cdot \color{blue}{ 1629 } = \color{blue}{ -24435 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-15}&7&-5&-21&15\\& & -105& 1650& \color{blue}{-24435} \\ \hline &7&-110&\color{blue}{1629}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -24435 \right) } = \color{orangered}{ -24420 } $
$$ \begin{array}{c|rrrr}-15&7&-5&-21&\color{orangered}{ 15 }\\& & -105& 1650& \color{orangered}{-24435} \\ \hline &\color{blue}{7}&\color{blue}{-110}&\color{blue}{1629}&\color{orangered}{-24420} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{2}-110x+1629 } $ with a remainder of $ \color{red}{ -24420 } $.