Divide using synthetic division $$ ( 4x^{4}+17x^{3}-4x-1 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&4&17&0&-4&-1\\& & -16& -4& 16& \color{black}{-48} \\ \hline &\color{blue}{4}&\color{blue}{1}&\color{blue}{-4}&\color{blue}{12}&\color{orangered}{-49} \end{array} $$The solution is:
$$ \frac{ 4x^{4}+17x^{3}-4x-1 }{ x+4 } = \color{blue}{4x^{3}+x^{2}-4x+12} \color{red}{~-~} \frac{ \color{red}{ 49 } }{ x+4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 4 = 0 $ ( $ x = \color{blue}{ -4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&4&17&0&-4&-1\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 4 }&17&0&-4&-1\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 4 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&4&17&0&-4&-1\\& & \color{blue}{-16} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 17 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}-4&4&\color{orangered}{ 17 }&0&-4&-1\\& & \color{orangered}{-16} & & & \\ \hline &4&\color{orangered}{1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 1 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&4&17&0&-4&-1\\& & -16& \color{blue}{-4} & & \\ \hline &4&\color{blue}{1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}-4&4&17&\color{orangered}{ 0 }&-4&-1\\& & -16& \color{orangered}{-4} & & \\ \hline &4&1&\color{orangered}{-4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&4&17&0&-4&-1\\& & -16& -4& \color{blue}{16} & \\ \hline &4&1&\color{blue}{-4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 16 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}-4&4&17&0&\color{orangered}{ -4 }&-1\\& & -16& -4& \color{orangered}{16} & \\ \hline &4&1&-4&\color{orangered}{12}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 12 } = \color{blue}{ -48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&4&17&0&-4&-1\\& & -16& -4& 16& \color{blue}{-48} \\ \hline &4&1&-4&\color{blue}{12}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -48 \right) } = \color{orangered}{ -49 } $
$$ \begin{array}{c|rrrrr}-4&4&17&0&-4&\color{orangered}{ -1 }\\& & -16& -4& 16& \color{orangered}{-48} \\ \hline &\color{blue}{4}&\color{blue}{1}&\color{blue}{-4}&\color{blue}{12}&\color{orangered}{-49} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}+x^{2}-4x+12 } $ with a remainder of $ \color{red}{ -49 } $.