Divide using synthetic division $$ ( 4x^{3}-31x^{2}-74x+28 ) \div ( 4x+9 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-9&4&-31&-74&28\\& & -36& 603& \color{black}{-4761} \\ \hline &\color{blue}{4}&\color{blue}{-67}&\color{blue}{529}&\color{orangered}{-4733} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-31x^{2}-74x+28 }{ x+9 } = \color{blue}{4x^{2}-67x+529} \color{red}{~-~} \frac{ \color{red}{ 4733 } }{ x+9 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 9 = 0 $ ( $ x = \color{blue}{ -9 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-9}&4&-31&-74&28\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-9&\color{orangered}{ 4 }&-31&-74&28\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -9 } \cdot \color{blue}{ 4 } = \color{blue}{ -36 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-9}&4&-31&-74&28\\& & \color{blue}{-36} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -31 } + \color{orangered}{ \left( -36 \right) } = \color{orangered}{ -67 } $
$$ \begin{array}{c|rrrr}-9&4&\color{orangered}{ -31 }&-74&28\\& & \color{orangered}{-36} & & \\ \hline &4&\color{orangered}{-67}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -9 } \cdot \color{blue}{ \left( -67 \right) } = \color{blue}{ 603 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-9}&4&-31&-74&28\\& & -36& \color{blue}{603} & \\ \hline &4&\color{blue}{-67}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -74 } + \color{orangered}{ 603 } = \color{orangered}{ 529 } $
$$ \begin{array}{c|rrrr}-9&4&-31&\color{orangered}{ -74 }&28\\& & -36& \color{orangered}{603} & \\ \hline &4&-67&\color{orangered}{529}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -9 } \cdot \color{blue}{ 529 } = \color{blue}{ -4761 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-9}&4&-31&-74&28\\& & -36& 603& \color{blue}{-4761} \\ \hline &4&-67&\color{blue}{529}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 28 } + \color{orangered}{ \left( -4761 \right) } = \color{orangered}{ -4733 } $
$$ \begin{array}{c|rrrr}-9&4&-31&-74&\color{orangered}{ 28 }\\& & -36& 603& \color{orangered}{-4761} \\ \hline &\color{blue}{4}&\color{blue}{-67}&\color{blue}{529}&\color{orangered}{-4733} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-67x+529 } $ with a remainder of $ \color{red}{ -4733 } $.