Divide using synthetic division $$ ( 10x^{3}+5x^{2}+74x-40 ) \div ( 2x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&10&5&74&-40\\& & 10& 15& \color{black}{89} \\ \hline &\color{blue}{10}&\color{blue}{15}&\color{blue}{89}&\color{orangered}{49} \end{array} $$The solution is:
$$ \frac{ 10x^{3}+5x^{2}+74x-40 }{ x-1 } = \color{blue}{10x^{2}+15x+89} ~+~ \frac{ \color{red}{ 49 } }{ x-1 } $$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}{ 10 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrr}1&10&\color{orangered}{ 5 }&74&-40\\& & \color{orangered}{10} & & \\ \hline &10&\color{orangered}{15}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 15 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&10&5&74&-40\\& & 10& \color{blue}{15} & \\ \hline &10&\color{blue}{15}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 74 } + \color{orangered}{ 15 } = \color{orangered}{ 89 } $
$$ \begin{array}{c|rrrr}1&10&5&\color{orangered}{ 74 }&-40\\& & 10& \color{orangered}{15} & \\ \hline &10&15&\color{orangered}{89}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 89 } = \color{blue}{ 89 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&10&5&74&-40\\& & 10& 15& \color{blue}{89} \\ \hline &10&15&\color{blue}{89}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -40 } + \color{orangered}{ 89 } = \color{orangered}{ 49 } $
$$ \begin{array}{c|rrrr}1&10&5&74&\color{orangered}{ -40 }\\& & 10& 15& \color{orangered}{89} \\ \hline &\color{blue}{10}&\color{blue}{15}&\color{blue}{89}&\color{orangered}{49} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 10x^{2}+15x+89 } $ with a remainder of $ \color{red}{ 49 } $.