Divide using synthetic division $$ ( x^{4}+2x^{3}-4x-7 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&1&2&0&-4&-7\\& & 3& 15& 45& \color{black}{123} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{15}&\color{blue}{41}&\color{orangered}{116} \end{array} $$The solution is:
$$ \frac{ x^{4}+2x^{3}-4x-7 }{ x-3 } = \color{blue}{x^{3}+5x^{2}+15x+41} ~+~ \frac{ \color{red}{ 116 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&2&0&-4&-7\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 1 }&2&0&-4&-7\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&2&0&-4&-7\\& & \color{blue}{3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 3 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}3&1&\color{orangered}{ 2 }&0&-4&-7\\& & \color{orangered}{3} & & & \\ \hline &1&\color{orangered}{5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 5 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&2&0&-4&-7\\& & 3& \color{blue}{15} & & \\ \hline &1&\color{blue}{5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 15 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrrr}3&1&2&\color{orangered}{ 0 }&-4&-7\\& & 3& \color{orangered}{15} & & \\ \hline &1&5&\color{orangered}{15}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 15 } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&2&0&-4&-7\\& & 3& 15& \color{blue}{45} & \\ \hline &1&5&\color{blue}{15}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 45 } = \color{orangered}{ 41 } $
$$ \begin{array}{c|rrrrr}3&1&2&0&\color{orangered}{ -4 }&-7\\& & 3& 15& \color{orangered}{45} & \\ \hline &1&5&15&\color{orangered}{41}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 41 } = \color{blue}{ 123 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&2&0&-4&-7\\& & 3& 15& 45& \color{blue}{123} \\ \hline &1&5&15&\color{blue}{41}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 123 } = \color{orangered}{ 116 } $
$$ \begin{array}{c|rrrrr}3&1&2&0&-4&\color{orangered}{ -7 }\\& & 3& 15& 45& \color{orangered}{123} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{15}&\color{blue}{41}&\color{orangered}{116} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+5x^{2}+15x+41 } $ with a remainder of $ \color{red}{ 116 } $.