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