Divide using synthetic division $$ ( 10x^{3}-73x^{2}+31x-6 ) \div ( 10x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&10&-73&31&-6\\& & 30& -129& \color{black}{-294} \\ \hline &\color{blue}{10}&\color{blue}{-43}&\color{blue}{-98}&\color{orangered}{-300} \end{array} $$The solution is:
$$ \frac{ 10x^{3}-73x^{2}+31x-6 }{ x-3 } = \color{blue}{10x^{2}-43x-98} \color{red}{~-~} \frac{ \color{red}{ 300 } }{ x-3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{3}&10&-73&31&-6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 10 }&-73&31&-6\\& & & & \\ \hline &\color{orangered}{10}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 10 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&10&-73&31&-6\\& & \color{blue}{30} & & \\ \hline &\color{blue}{10}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -73 } + \color{orangered}{ 30 } = \color{orangered}{ -43 } $
$$ \begin{array}{c|rrrr}3&10&\color{orangered}{ -73 }&31&-6\\& & \color{orangered}{30} & & \\ \hline &10&\color{orangered}{-43}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -43 \right) } = \color{blue}{ -129 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&10&-73&31&-6\\& & 30& \color{blue}{-129} & \\ \hline &10&\color{blue}{-43}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 31 } + \color{orangered}{ \left( -129 \right) } = \color{orangered}{ -98 } $
$$ \begin{array}{c|rrrr}3&10&-73&\color{orangered}{ 31 }&-6\\& & 30& \color{orangered}{-129} & \\ \hline &10&-43&\color{orangered}{-98}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -98 \right) } = \color{blue}{ -294 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&10&-73&31&-6\\& & 30& -129& \color{blue}{-294} \\ \hline &10&-43&\color{blue}{-98}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -294 \right) } = \color{orangered}{ -300 } $
$$ \begin{array}{c|rrrr}3&10&-73&31&\color{orangered}{ -6 }\\& & 30& -129& \color{orangered}{-294} \\ \hline &\color{blue}{10}&\color{blue}{-43}&\color{blue}{-98}&\color{orangered}{-300} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 10x^{2}-43x-98 } $ with a remainder of $ \color{red}{ -300 } $.