Divide using synthetic division $$ ( 2x^{4}-3x^{3}-17x^{2}+16 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&2&-3&-17&0&16\\& & 6& 9& -24& \color{black}{-72} \\ \hline &\color{blue}{2}&\color{blue}{3}&\color{blue}{-8}&\color{blue}{-24}&\color{orangered}{-56} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-3x^{3}-17x^{2}+16 }{ x-3 } = \color{blue}{2x^{3}+3x^{2}-8x-24} \color{red}{~-~} \frac{ \color{red}{ 56 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-3&-17&0&16\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 2 }&-3&-17&0&16\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-3&-17&0&16\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 6 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}3&2&\color{orangered}{ -3 }&-17&0&16\\& & \color{orangered}{6} & & & \\ \hline &2&\color{orangered}{3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-3&-17&0&16\\& & 6& \color{blue}{9} & & \\ \hline &2&\color{blue}{3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 9 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}3&2&-3&\color{orangered}{ -17 }&0&16\\& & 6& \color{orangered}{9} & & \\ \hline &2&3&\color{orangered}{-8}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-3&-17&0&16\\& & 6& 9& \color{blue}{-24} & \\ \hline &2&3&\color{blue}{-8}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -24 } $
$$ \begin{array}{c|rrrrr}3&2&-3&-17&\color{orangered}{ 0 }&16\\& & 6& 9& \color{orangered}{-24} & \\ \hline &2&3&-8&\color{orangered}{-24}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -24 \right) } = \color{blue}{ -72 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-3&-17&0&16\\& & 6& 9& -24& \color{blue}{-72} \\ \hline &2&3&-8&\color{blue}{-24}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -72 \right) } = \color{orangered}{ -56 } $
$$ \begin{array}{c|rrrrr}3&2&-3&-17&0&\color{orangered}{ 16 }\\& & 6& 9& -24& \color{orangered}{-72} \\ \hline &\color{blue}{2}&\color{blue}{3}&\color{blue}{-8}&\color{blue}{-24}&\color{orangered}{-56} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+3x^{2}-8x-24 } $ with a remainder of $ \color{red}{ -56 } $.