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