Divide using synthetic division $$ ( 5x^{3}-6x^{2}+8 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&5&-6&0&8\\& & 20& 56& \color{black}{224} \\ \hline &\color{blue}{5}&\color{blue}{14}&\color{blue}{56}&\color{orangered}{232} \end{array} $$The solution is:
$$ \frac{ 5x^{3}-6x^{2}+8 }{ x-4 } = \color{blue}{5x^{2}+14x+56} ~+~ \frac{ \color{red}{ 232 } }{ x-4 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{4}&5&-6&0&8\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 5 }&-6&0&8\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 5 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&5&-6&0&8\\& & \color{blue}{20} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 20 } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrr}4&5&\color{orangered}{ -6 }&0&8\\& & \color{orangered}{20} & & \\ \hline &5&\color{orangered}{14}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 14 } = \color{blue}{ 56 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&5&-6&0&8\\& & 20& \color{blue}{56} & \\ \hline &5&\color{blue}{14}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 56 } = \color{orangered}{ 56 } $
$$ \begin{array}{c|rrrr}4&5&-6&\color{orangered}{ 0 }&8\\& & 20& \color{orangered}{56} & \\ \hline &5&14&\color{orangered}{56}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 56 } = \color{blue}{ 224 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&5&-6&0&8\\& & 20& 56& \color{blue}{224} \\ \hline &5&14&\color{blue}{56}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 224 } = \color{orangered}{ 232 } $
$$ \begin{array}{c|rrrr}4&5&-6&0&\color{orangered}{ 8 }\\& & 20& 56& \color{orangered}{224} \\ \hline &\color{blue}{5}&\color{blue}{14}&\color{blue}{56}&\color{orangered}{232} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{2}+14x+56 } $ with a remainder of $ \color{red}{ 232 } $.