Divide using synthetic division $$ ( 10x^{3}+5x^{2}+74x-40 ) \div ( 2x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&10&5&74&-40\\& & -10& 5& \color{black}{-79} \\ \hline &\color{blue}{10}&\color{blue}{-5}&\color{blue}{79}&\color{orangered}{-119} \end{array} $$The solution is:
$$ \frac{ 10x^{3}+5x^{2}+74x-40 }{ x+1 } = \color{blue}{10x^{2}-5x+79} \color{red}{~-~} \frac{ \color{red}{ 119 } }{ x+1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 1 = 0 $ ( $ x = \color{blue}{ -1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&10&5&74&-40\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 10 }&5&74&-40\\& & & & \\ \hline &\color{orangered}{10}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 10 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&10&5&74&-40\\& & \color{blue}{-10} & & \\ \hline &\color{blue}{10}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}-1&10&\color{orangered}{ 5 }&74&-40\\& & \color{orangered}{-10} & & \\ \hline &10&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&10&5&74&-40\\& & -10& \color{blue}{5} & \\ \hline &10&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 74 } + \color{orangered}{ 5 } = \color{orangered}{ 79 } $
$$ \begin{array}{c|rrrr}-1&10&5&\color{orangered}{ 74 }&-40\\& & -10& \color{orangered}{5} & \\ \hline &10&-5&\color{orangered}{79}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 79 } = \color{blue}{ -79 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&10&5&74&-40\\& & -10& 5& \color{blue}{-79} \\ \hline &10&-5&\color{blue}{79}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -40 } + \color{orangered}{ \left( -79 \right) } = \color{orangered}{ -119 } $
$$ \begin{array}{c|rrrr}-1&10&5&74&\color{orangered}{ -40 }\\& & -10& 5& \color{orangered}{-79} \\ \hline &\color{blue}{10}&\color{blue}{-5}&\color{blue}{79}&\color{orangered}{-119} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 10x^{2}-5x+79 } $ with a remainder of $ \color{red}{ -119 } $.