Divide using synthetic division $$ ( 3x^{4}-10x^{3}+20x^{2}-40x+32 ) \div ( -x ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}0&3&-10&20&-40&32\\& & 0& 0& 0& \color{black}{0} \\ \hline &\color{blue}{3}&\color{blue}{-10}&\color{blue}{20}&\color{blue}{-40}&\color{orangered}{32} \end{array} $$The solution is:
$$ \frac{ 3x^{4}-10x^{3}+20x^{2}-40x+32 }{ x } = \color{blue}{3x^{3}-10x^{2}+20x-40} ~+~ \frac{ \color{red}{ 32 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&3&-10&20&-40&32\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}0&\color{orangered}{ 3 }&-10&20&-40&32\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 3 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&3&-10&20&-40&32\\& & \color{blue}{0} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 0 } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}0&3&\color{orangered}{ -10 }&20&-40&32\\& & \color{orangered}{0} & & & \\ \hline &3&\color{orangered}{-10}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&3&-10&20&-40&32\\& & 0& \color{blue}{0} & & \\ \hline &3&\color{blue}{-10}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ 0 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrrr}0&3&-10&\color{orangered}{ 20 }&-40&32\\& & 0& \color{orangered}{0} & & \\ \hline &3&-10&\color{orangered}{20}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 20 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&3&-10&20&-40&32\\& & 0& 0& \color{blue}{0} & \\ \hline &3&-10&\color{blue}{20}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -40 } + \color{orangered}{ 0 } = \color{orangered}{ -40 } $
$$ \begin{array}{c|rrrrr}0&3&-10&20&\color{orangered}{ -40 }&32\\& & 0& 0& \color{orangered}{0} & \\ \hline &3&-10&20&\color{orangered}{-40}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -40 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&3&-10&20&-40&32\\& & 0& 0& 0& \color{blue}{0} \\ \hline &3&-10&20&\color{blue}{-40}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 32 } + \color{orangered}{ 0 } = \color{orangered}{ 32 } $
$$ \begin{array}{c|rrrrr}0&3&-10&20&-40&\color{orangered}{ 32 }\\& & 0& 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{3}&\color{blue}{-10}&\color{blue}{20}&\color{blue}{-40}&\color{orangered}{32} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}-10x^{2}+20x-40 } $ with a remainder of $ \color{red}{ 32 } $.