Divide using synthetic division $$ ( 4x^{3}+7x^{2}-17x-7 ) \div ( 4x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&4&7&-17&-7\\& & -12& 15& \color{black}{6} \\ \hline &\color{blue}{4}&\color{blue}{-5}&\color{blue}{-2}&\color{orangered}{-1} \end{array} $$The solution is:
$$ \frac{ 4x^{3}+7x^{2}-17x-7 }{ x+3 } = \color{blue}{4x^{2}-5x-2} \color{red}{~-~} \frac{ \color{red}{ 1 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&7&-17&-7\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 4 }&7&-17&-7\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 4 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&7&-17&-7\\& & \color{blue}{-12} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}-3&4&\color{orangered}{ 7 }&-17&-7\\& & \color{orangered}{-12} & & \\ \hline &4&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&7&-17&-7\\& & -12& \color{blue}{15} & \\ \hline &4&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 15 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}-3&4&7&\color{orangered}{ -17 }&-7\\& & -12& \color{orangered}{15} & \\ \hline &4&-5&\color{orangered}{-2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&7&-17&-7\\& & -12& 15& \color{blue}{6} \\ \hline &4&-5&\color{blue}{-2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 6 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}-3&4&7&-17&\color{orangered}{ -7 }\\& & -12& 15& \color{orangered}{6} \\ \hline &\color{blue}{4}&\color{blue}{-5}&\color{blue}{-2}&\color{orangered}{-1} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-5x-2 } $ with a remainder of $ \color{red}{ -1 } $.