Divide using synthetic division $$ ( 2x^{4}-38x^{2}-7x-4 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&2&0&-38&-7&-4\\& & 8& 32& -24& \color{black}{-124} \\ \hline &\color{blue}{2}&\color{blue}{8}&\color{blue}{-6}&\color{blue}{-31}&\color{orangered}{-128} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-38x^{2}-7x-4 }{ x-4 } = \color{blue}{2x^{3}+8x^{2}-6x-31} \color{red}{~-~} \frac{ \color{red}{ 128 } }{ x-4 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&0&-38&-7&-4\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ 2 }&0&-38&-7&-4\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 2 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&0&-38&-7&-4\\& & \color{blue}{8} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 8 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}4&2&\color{orangered}{ 0 }&-38&-7&-4\\& & \color{orangered}{8} & & & \\ \hline &2&\color{orangered}{8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 8 } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&0&-38&-7&-4\\& & 8& \color{blue}{32} & & \\ \hline &2&\color{blue}{8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -38 } + \color{orangered}{ 32 } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}4&2&0&\color{orangered}{ -38 }&-7&-4\\& & 8& \color{orangered}{32} & & \\ \hline &2&8&\color{orangered}{-6}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&0&-38&-7&-4\\& & 8& 32& \color{blue}{-24} & \\ \hline &2&8&\color{blue}{-6}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -31 } $
$$ \begin{array}{c|rrrrr}4&2&0&-38&\color{orangered}{ -7 }&-4\\& & 8& 32& \color{orangered}{-24} & \\ \hline &2&8&-6&\color{orangered}{-31}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -31 \right) } = \color{blue}{ -124 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&0&-38&-7&-4\\& & 8& 32& -24& \color{blue}{-124} \\ \hline &2&8&-6&\color{blue}{-31}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -124 \right) } = \color{orangered}{ -128 } $
$$ \begin{array}{c|rrrrr}4&2&0&-38&-7&\color{orangered}{ -4 }\\& & 8& 32& -24& \color{orangered}{-124} \\ \hline &\color{blue}{2}&\color{blue}{8}&\color{blue}{-6}&\color{blue}{-31}&\color{orangered}{-128} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+8x^{2}-6x-31 } $ with a remainder of $ \color{red}{ -128 } $.