Divide using synthetic division $$ ( 4x^{3}+20x^{2}+29x+28 ) \div ( 3x+7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-7&4&20&29&28\\& & -28& 56& \color{black}{-595} \\ \hline &\color{blue}{4}&\color{blue}{-8}&\color{blue}{85}&\color{orangered}{-567} \end{array} $$The solution is:
$$ \frac{ 4x^{3}+20x^{2}+29x+28 }{ x+7 } = \color{blue}{4x^{2}-8x+85} \color{red}{~-~} \frac{ \color{red}{ 567 } }{ x+7 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 7 = 0 $ ( $ x = \color{blue}{ -7 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-7}&4&20&29&28\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-7&\color{orangered}{ 4 }&20&29&28\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ 4 } = \color{blue}{ -28 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-7}&4&20&29&28\\& & \color{blue}{-28} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -28 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}-7&4&\color{orangered}{ 20 }&29&28\\& & \color{orangered}{-28} & & \\ \hline &4&\color{orangered}{-8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 56 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-7}&4&20&29&28\\& & -28& \color{blue}{56} & \\ \hline &4&\color{blue}{-8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 29 } + \color{orangered}{ 56 } = \color{orangered}{ 85 } $
$$ \begin{array}{c|rrrr}-7&4&20&\color{orangered}{ 29 }&28\\& & -28& \color{orangered}{56} & \\ \hline &4&-8&\color{orangered}{85}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ 85 } = \color{blue}{ -595 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-7}&4&20&29&28\\& & -28& 56& \color{blue}{-595} \\ \hline &4&-8&\color{blue}{85}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 28 } + \color{orangered}{ \left( -595 \right) } = \color{orangered}{ -567 } $
$$ \begin{array}{c|rrrr}-7&4&20&29&\color{orangered}{ 28 }\\& & -28& 56& \color{orangered}{-595} \\ \hline &\color{blue}{4}&\color{blue}{-8}&\color{blue}{85}&\color{orangered}{-567} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-8x+85 } $ with a remainder of $ \color{red}{ -567 } $.