Divide using synthetic division $$ ( 4x^{3}-x^{2}-x+3 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&4&-1&-1&3\\& & 12& 33& \color{black}{96} \\ \hline &\color{blue}{4}&\color{blue}{11}&\color{blue}{32}&\color{orangered}{99} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-x^{2}-x+3 }{ x-3 } = \color{blue}{4x^{2}+11x+32} ~+~ \frac{ \color{red}{ 99 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-1&-1&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 4 }&-1&-1&3\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-1&-1&3\\& & \color{blue}{12} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 12 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrr}3&4&\color{orangered}{ -1 }&-1&3\\& & \color{orangered}{12} & & \\ \hline &4&\color{orangered}{11}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 11 } = \color{blue}{ 33 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-1&-1&3\\& & 12& \color{blue}{33} & \\ \hline &4&\color{blue}{11}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 33 } = \color{orangered}{ 32 } $
$$ \begin{array}{c|rrrr}3&4&-1&\color{orangered}{ -1 }&3\\& & 12& \color{orangered}{33} & \\ \hline &4&11&\color{orangered}{32}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 32 } = \color{blue}{ 96 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-1&-1&3\\& & 12& 33& \color{blue}{96} \\ \hline &4&11&\color{blue}{32}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 96 } = \color{orangered}{ 99 } $
$$ \begin{array}{c|rrrr}3&4&-1&-1&\color{orangered}{ 3 }\\& & 12& 33& \color{orangered}{96} \\ \hline &\color{blue}{4}&\color{blue}{11}&\color{blue}{32}&\color{orangered}{99} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+11x+32 } $ with a remainder of $ \color{red}{ 99 } $.