Divide using synthetic division $$ ( 4x^{3}-11x^{2}-1 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&4&-11&0&-1\\& & 12& 3& \color{black}{9} \\ \hline &\color{blue}{4}&\color{blue}{1}&\color{blue}{3}&\color{orangered}{8} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-11x^{2}-1 }{ x-3 } = \color{blue}{4x^{2}+x+3} ~+~ \frac{ \color{red}{ 8 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-11&0&-1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 4 }&-11&0&-1\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-11&0&-1\\& & \color{blue}{12} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 12 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}3&4&\color{orangered}{ -11 }&0&-1\\& & \color{orangered}{12} & & \\ \hline &4&\color{orangered}{1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-11&0&-1\\& & 12& \color{blue}{3} & \\ \hline &4&\color{blue}{1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 3 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}3&4&-11&\color{orangered}{ 0 }&-1\\& & 12& \color{orangered}{3} & \\ \hline &4&1&\color{orangered}{3}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-11&0&-1\\& & 12& 3& \color{blue}{9} \\ \hline &4&1&\color{blue}{3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 9 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}3&4&-11&0&\color{orangered}{ -1 }\\& & 12& 3& \color{orangered}{9} \\ \hline &\color{blue}{4}&\color{blue}{1}&\color{blue}{3}&\color{orangered}{8} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+x+3 } $ with a remainder of $ \color{red}{ 8 } $.