Divide using synthetic division $$ ( 4x^{4}-20x^{3}+23x^{2}+7x-2 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&4&-20&23&7&-2\\& & 12& -24& -3& \color{black}{12} \\ \hline &\color{blue}{4}&\color{blue}{-8}&\color{blue}{-1}&\color{blue}{4}&\color{orangered}{10} \end{array} $$The solution is:
$$ \frac{ 4x^{4}-20x^{3}+23x^{2}+7x-2 }{ x-3 } = \color{blue}{4x^{3}-8x^{2}-x+4} ~+~ \frac{ \color{red}{ 10 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&-20&23&7&-2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 4 }&-20&23&7&-2\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&-20&23&7&-2\\& & \color{blue}{12} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 12 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}3&4&\color{orangered}{ -20 }&23&7&-2\\& & \color{orangered}{12} & & & \\ \hline &4&\color{orangered}{-8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&-20&23&7&-2\\& & 12& \color{blue}{-24} & & \\ \hline &4&\color{blue}{-8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 23 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}3&4&-20&\color{orangered}{ 23 }&7&-2\\& & 12& \color{orangered}{-24} & & \\ \hline &4&-8&\color{orangered}{-1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&-20&23&7&-2\\& & 12& -24& \color{blue}{-3} & \\ \hline &4&-8&\color{blue}{-1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}3&4&-20&23&\color{orangered}{ 7 }&-2\\& & 12& -24& \color{orangered}{-3} & \\ \hline &4&-8&-1&\color{orangered}{4}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&-20&23&7&-2\\& & 12& -24& -3& \color{blue}{12} \\ \hline &4&-8&-1&\color{blue}{4}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 12 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}3&4&-20&23&7&\color{orangered}{ -2 }\\& & 12& -24& -3& \color{orangered}{12} \\ \hline &\color{blue}{4}&\color{blue}{-8}&\color{blue}{-1}&\color{blue}{4}&\color{orangered}{10} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}-8x^{2}-x+4 } $ with a remainder of $ \color{red}{ 10 } $.