Divide using synthetic division $$ ( 5x^{3}-10x^{2}+x-32 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&5&-10&1&-32\\& & 10& 0& \color{black}{2} \\ \hline &\color{blue}{5}&\color{blue}{0}&\color{blue}{1}&\color{orangered}{-30} \end{array} $$The solution is:
$$ \frac{ 5x^{3}-10x^{2}+x-32 }{ x-2 } = \color{blue}{5x^{2}+1} \color{red}{~-~} \frac{ \color{red}{ 30 } }{ x-2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{2}&5&-10&1&-32\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 5 }&-10&1&-32\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&5&-10&1&-32\\& & \color{blue}{10} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 10 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}2&5&\color{orangered}{ -10 }&1&-32\\& & \color{orangered}{10} & & \\ \hline &5&\color{orangered}{0}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&5&-10&1&-32\\& & 10& \color{blue}{0} & \\ \hline &5&\color{blue}{0}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 0 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}2&5&-10&\color{orangered}{ 1 }&-32\\& & 10& \color{orangered}{0} & \\ \hline &5&0&\color{orangered}{1}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&5&-10&1&-32\\& & 10& 0& \color{blue}{2} \\ \hline &5&0&\color{blue}{1}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -32 } + \color{orangered}{ 2 } = \color{orangered}{ -30 } $
$$ \begin{array}{c|rrrr}2&5&-10&1&\color{orangered}{ -32 }\\& & 10& 0& \color{orangered}{2} \\ \hline &\color{blue}{5}&\color{blue}{0}&\color{blue}{1}&\color{orangered}{-30} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{2}+1 } $ with a remainder of $ \color{red}{ -30 } $.