Divide using synthetic division $$ ( 4x^{3}-11x+6 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&4&0&-11&6\\& & 16& 64& \color{black}{212} \\ \hline &\color{blue}{4}&\color{blue}{16}&\color{blue}{53}&\color{orangered}{218} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-11x+6 }{ x-4 } = \color{blue}{4x^{2}+16x+53} ~+~ \frac{ \color{red}{ 218 } }{ x-4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{4}&4&0&-11&6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 4 }&0&-11&6\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 4 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&4&0&-11&6\\& & \color{blue}{16} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 16 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrr}4&4&\color{orangered}{ 0 }&-11&6\\& & \color{orangered}{16} & & \\ \hline &4&\color{orangered}{16}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 16 } = \color{blue}{ 64 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&4&0&-11&6\\& & 16& \color{blue}{64} & \\ \hline &4&\color{blue}{16}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 64 } = \color{orangered}{ 53 } $
$$ \begin{array}{c|rrrr}4&4&0&\color{orangered}{ -11 }&6\\& & 16& \color{orangered}{64} & \\ \hline &4&16&\color{orangered}{53}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 53 } = \color{blue}{ 212 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&4&0&-11&6\\& & 16& 64& \color{blue}{212} \\ \hline &4&16&\color{blue}{53}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 212 } = \color{orangered}{ 218 } $
$$ \begin{array}{c|rrrr}4&4&0&-11&\color{orangered}{ 6 }\\& & 16& 64& \color{orangered}{212} \\ \hline &\color{blue}{4}&\color{blue}{16}&\color{blue}{53}&\color{orangered}{218} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+16x+53 } $ with a remainder of $ \color{red}{ 218 } $.