Divide using synthetic division $$ ( 4x^{4}+17x^{3}-14x^{2}-68x-40 ) \div ( 1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}0&4&17&-14&-68&-40\\& & 0& 0& 0& \color{black}{0} \\ \hline &\color{blue}{4}&\color{blue}{17}&\color{blue}{-14}&\color{blue}{-68}&\color{orangered}{-40} \end{array} $$The solution is:
$$ \frac{ 4x^{4}+17x^{3}-14x^{2}-68x-40 }{ x } = \color{blue}{4x^{3}+17x^{2}-14x-68} \color{red}{~-~} \frac{ \color{red}{ 40 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&4&17&-14&-68&-40\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}0&\color{orangered}{ 4 }&17&-14&-68&-40\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 4 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&4&17&-14&-68&-40\\& & \color{blue}{0} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 17 } + \color{orangered}{ 0 } = \color{orangered}{ 17 } $
$$ \begin{array}{c|rrrrr}0&4&\color{orangered}{ 17 }&-14&-68&-40\\& & \color{orangered}{0} & & & \\ \hline &4&\color{orangered}{17}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 17 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&4&17&-14&-68&-40\\& & 0& \color{blue}{0} & & \\ \hline &4&\color{blue}{17}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 0 } = \color{orangered}{ -14 } $
$$ \begin{array}{c|rrrrr}0&4&17&\color{orangered}{ -14 }&-68&-40\\& & 0& \color{orangered}{0} & & \\ \hline &4&17&\color{orangered}{-14}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -14 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&4&17&-14&-68&-40\\& & 0& 0& \color{blue}{0} & \\ \hline &4&17&\color{blue}{-14}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -68 } + \color{orangered}{ 0 } = \color{orangered}{ -68 } $
$$ \begin{array}{c|rrrrr}0&4&17&-14&\color{orangered}{ -68 }&-40\\& & 0& 0& \color{orangered}{0} & \\ \hline &4&17&-14&\color{orangered}{-68}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -68 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&4&17&-14&-68&-40\\& & 0& 0& 0& \color{blue}{0} \\ \hline &4&17&-14&\color{blue}{-68}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -40 } + \color{orangered}{ 0 } = \color{orangered}{ -40 } $
$$ \begin{array}{c|rrrrr}0&4&17&-14&-68&\color{orangered}{ -40 }\\& & 0& 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{4}&\color{blue}{17}&\color{blue}{-14}&\color{blue}{-68}&\color{orangered}{-40} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}+17x^{2}-14x-68 } $ with a remainder of $ \color{red}{ -40 } $.