Divide using synthetic division $$ ( x^{4}-x^{3}+19x^{2}-25x-150 ) \div ( 6x ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}0&1&-1&19&-25&-150\\& & 0& 0& 0& \color{black}{0} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{19}&\color{blue}{-25}&\color{orangered}{-150} \end{array} $$The solution is:
$$ \frac{ x^{4}-x^{3}+19x^{2}-25x-150 }{ x } = \color{blue}{x^{3}-x^{2}+19x-25} \color{red}{~-~} \frac{ \color{red}{ 150 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&1&-1&19&-25&-150\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}0&\color{orangered}{ 1 }&-1&19&-25&-150\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 1 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&1&-1&19&-25&-150\\& & \color{blue}{0} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 0 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}0&1&\color{orangered}{ -1 }&19&-25&-150\\& & \color{orangered}{0} & & & \\ \hline &1&\color{orangered}{-1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&1&-1&19&-25&-150\\& & 0& \color{blue}{0} & & \\ \hline &1&\color{blue}{-1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ 0 } = \color{orangered}{ 19 } $
$$ \begin{array}{c|rrrrr}0&1&-1&\color{orangered}{ 19 }&-25&-150\\& & 0& \color{orangered}{0} & & \\ \hline &1&-1&\color{orangered}{19}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 19 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&1&-1&19&-25&-150\\& & 0& 0& \color{blue}{0} & \\ \hline &1&-1&\color{blue}{19}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -25 } + \color{orangered}{ 0 } = \color{orangered}{ -25 } $
$$ \begin{array}{c|rrrrr}0&1&-1&19&\color{orangered}{ -25 }&-150\\& & 0& 0& \color{orangered}{0} & \\ \hline &1&-1&19&\color{orangered}{-25}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -25 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&1&-1&19&-25&-150\\& & 0& 0& 0& \color{blue}{0} \\ \hline &1&-1&19&\color{blue}{-25}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -150 } + \color{orangered}{ 0 } = \color{orangered}{ -150 } $
$$ \begin{array}{c|rrrrr}0&1&-1&19&-25&\color{orangered}{ -150 }\\& & 0& 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{19}&\color{blue}{-25}&\color{orangered}{-150} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-x^{2}+19x-25 } $ with a remainder of $ \color{red}{ -150 } $.