Divide using synthetic division $$ ( 7x^{3}-5x^{2}-21x+15 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&7&-5&-21&15\\& & -35& 200& \color{black}{-895} \\ \hline &\color{blue}{7}&\color{blue}{-40}&\color{blue}{179}&\color{orangered}{-880} \end{array} $$The solution is:
$$ \frac{ 7x^{3}-5x^{2}-21x+15 }{ x+5 } = \color{blue}{7x^{2}-40x+179} \color{red}{~-~} \frac{ \color{red}{ 880 } }{ x+5 } $$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}{ \left( -35 \right) } = \color{orangered}{ -40 } $
$$ \begin{array}{c|rrrr}-5&7&\color{orangered}{ -5 }&-21&15\\& & \color{orangered}{-35} & & \\ \hline &7&\color{orangered}{-40}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -40 \right) } = \color{blue}{ 200 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&7&-5&-21&15\\& & -35& \color{blue}{200} & \\ \hline &7&\color{blue}{-40}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ 200 } = \color{orangered}{ 179 } $
$$ \begin{array}{c|rrrr}-5&7&-5&\color{orangered}{ -21 }&15\\& & -35& \color{orangered}{200} & \\ \hline &7&-40&\color{orangered}{179}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 179 } = \color{blue}{ -895 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&7&-5&-21&15\\& & -35& 200& \color{blue}{-895} \\ \hline &7&-40&\color{blue}{179}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -895 \right) } = \color{orangered}{ -880 } $
$$ \begin{array}{c|rrrr}-5&7&-5&-21&\color{orangered}{ 15 }\\& & -35& 200& \color{orangered}{-895} \\ \hline &\color{blue}{7}&\color{blue}{-40}&\color{blue}{179}&\color{orangered}{-880} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{2}-40x+179 } $ with a remainder of $ \color{red}{ -880 } $.