Divide using synthetic division $$ ( 7x^{3}-5x^{2}-21x+15 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&7&-5&-21&15\\& & -21& 78& \color{black}{-171} \\ \hline &\color{blue}{7}&\color{blue}{-26}&\color{blue}{57}&\color{orangered}{-156} \end{array} $$The solution is:
$$ \frac{ 7x^{3}-5x^{2}-21x+15 }{ x+3 } = \color{blue}{7x^{2}-26x+57} \color{red}{~-~} \frac{ \color{red}{ 156 } }{ x+3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&7&-5&-21&15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 7 }&-5&-21&15\\& & & & \\ \hline &\color{orangered}{7}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 7 } = \color{blue}{ -21 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&7&-5&-21&15\\& & \color{blue}{-21} & & \\ \hline &\color{blue}{7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -21 \right) } = \color{orangered}{ -26 } $
$$ \begin{array}{c|rrrr}-3&7&\color{orangered}{ -5 }&-21&15\\& & \color{orangered}{-21} & & \\ \hline &7&\color{orangered}{-26}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -26 \right) } = \color{blue}{ 78 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&7&-5&-21&15\\& & -21& \color{blue}{78} & \\ \hline &7&\color{blue}{-26}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ 78 } = \color{orangered}{ 57 } $
$$ \begin{array}{c|rrrr}-3&7&-5&\color{orangered}{ -21 }&15\\& & -21& \color{orangered}{78} & \\ \hline &7&-26&\color{orangered}{57}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 57 } = \color{blue}{ -171 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&7&-5&-21&15\\& & -21& 78& \color{blue}{-171} \\ \hline &7&-26&\color{blue}{57}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -171 \right) } = \color{orangered}{ -156 } $
$$ \begin{array}{c|rrrr}-3&7&-5&-21&\color{orangered}{ 15 }\\& & -21& 78& \color{orangered}{-171} \\ \hline &\color{blue}{7}&\color{blue}{-26}&\color{blue}{57}&\color{orangered}{-156} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{2}-26x+57 } $ with a remainder of $ \color{red}{ -156 } $.