Divide using synthetic division $$ ( -4x^{6}+2x^{4} ) \div ( -3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrrr}0&-4&0&2&0&0&0&0\\& & 0& 0& 0& 0& 0& \color{black}{0} \\ \hline &\color{blue}{-4}&\color{blue}{0}&\color{blue}{2}&\color{blue}{0}&\color{blue}{0}&\color{blue}{0}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ -4x^{6}+2x^{4} }{ x } = \color{blue}{-4x^{5}+2x^{3}} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrrrrr}\color{blue}{0}&-4&0&2&0&0&0&0\\& & & & & & & \\ \hline &&&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrrr}0&\color{orangered}{ -4 }&0&2&0&0&0&0\\& & & & & & & \\ \hline &\color{orangered}{-4}&&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{0}&-4&0&2&0&0&0&0\\& & \color{blue}{0} & & & & & \\ \hline &\color{blue}{-4}&&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrrr}0&-4&\color{orangered}{ 0 }&2&0&0&0&0\\& & \color{orangered}{0} & & & & & \\ \hline &-4&\color{orangered}{0}&&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{0}&-4&0&2&0&0&0&0\\& & 0& \color{blue}{0} & & & & \\ \hline &-4&\color{blue}{0}&&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 0 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrrr}0&-4&0&\color{orangered}{ 2 }&0&0&0&0\\& & 0& \color{orangered}{0} & & & & \\ \hline &-4&0&\color{orangered}{2}&&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 2 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{0}&-4&0&2&0&0&0&0\\& & 0& 0& \color{blue}{0} & & & \\ \hline &-4&0&\color{blue}{2}&&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrrr}0&-4&0&2&\color{orangered}{ 0 }&0&0&0\\& & 0& 0& \color{orangered}{0} & & & \\ \hline &-4&0&2&\color{orangered}{0}&&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{0}&-4&0&2&0&0&0&0\\& & 0& 0& 0& \color{blue}{0} & & \\ \hline &-4&0&2&\color{blue}{0}&&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrrr}0&-4&0&2&0&\color{orangered}{ 0 }&0&0\\& & 0& 0& 0& \color{orangered}{0} & & \\ \hline &-4&0&2&0&\color{orangered}{0}&& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{0}&-4&0&2&0&0&0&0\\& & 0& 0& 0& 0& \color{blue}{0} & \\ \hline &-4&0&2&0&\color{blue}{0}&& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrrr}0&-4&0&2&0&0&\color{orangered}{ 0 }&0\\& & 0& 0& 0& 0& \color{orangered}{0} & \\ \hline &-4&0&2&0&0&\color{orangered}{0}& \end{array} $$Step 12 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{0}&-4&0&2&0&0&0&0\\& & 0& 0& 0& 0& 0& \color{blue}{0} \\ \hline &-4&0&2&0&0&\color{blue}{0}& \end{array} $$Step 13 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrrr}0&-4&0&2&0&0&0&\color{orangered}{ 0 }\\& & 0& 0& 0& 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{-4}&\color{blue}{0}&\color{blue}{2}&\color{blue}{0}&\color{blue}{0}&\color{blue}{0}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -4x^{5}+2x^{3} } $ with a remainder of $ \color{red}{ 0 } $.