Divide using synthetic division $$ ( -x^{4}+7x^{3}+x^{2}-2x-12 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&-1&7&1&-2&-12\\& & -3& 12& 39& \color{black}{111} \\ \hline &\color{blue}{-1}&\color{blue}{4}&\color{blue}{13}&\color{blue}{37}&\color{orangered}{99} \end{array} $$The solution is:
$$ \frac{ -x^{4}+7x^{3}+x^{2}-2x-12 }{ x-3 } = \color{blue}{-x^{3}+4x^{2}+13x+37} ~+~ \frac{ \color{red}{ 99 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&-1&7&1&-2&-12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ -1 }&7&1&-2&-12\\& & & & & \\ \hline &\color{orangered}{-1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&-1&7&1&-2&-12\\& & \color{blue}{-3} & & & \\ \hline &\color{blue}{-1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}3&-1&\color{orangered}{ 7 }&1&-2&-12\\& & \color{orangered}{-3} & & & \\ \hline &-1&\color{orangered}{4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&-1&7&1&-2&-12\\& & -3& \color{blue}{12} & & \\ \hline &-1&\color{blue}{4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 12 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrrr}3&-1&7&\color{orangered}{ 1 }&-2&-12\\& & -3& \color{orangered}{12} & & \\ \hline &-1&4&\color{orangered}{13}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 13 } = \color{blue}{ 39 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&-1&7&1&-2&-12\\& & -3& 12& \color{blue}{39} & \\ \hline &-1&4&\color{blue}{13}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 39 } = \color{orangered}{ 37 } $
$$ \begin{array}{c|rrrrr}3&-1&7&1&\color{orangered}{ -2 }&-12\\& & -3& 12& \color{orangered}{39} & \\ \hline &-1&4&13&\color{orangered}{37}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 37 } = \color{blue}{ 111 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&-1&7&1&-2&-12\\& & -3& 12& 39& \color{blue}{111} \\ \hline &-1&4&13&\color{blue}{37}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 111 } = \color{orangered}{ 99 } $
$$ \begin{array}{c|rrrrr}3&-1&7&1&-2&\color{orangered}{ -12 }\\& & -3& 12& 39& \color{orangered}{111} \\ \hline &\color{blue}{-1}&\color{blue}{4}&\color{blue}{13}&\color{blue}{37}&\color{orangered}{99} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -x^{3}+4x^{2}+13x+37 } $ with a remainder of $ \color{red}{ 99 } $.