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