Divide using synthetic division $$ ( 10x^{3}-28x^{2}-29x+10 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&10&-28&-29&10\\& & 10& -18& \color{black}{-47} \\ \hline &\color{blue}{10}&\color{blue}{-18}&\color{blue}{-47}&\color{orangered}{-37} \end{array} $$The solution is:
$$ \frac{ 10x^{3}-28x^{2}-29x+10 }{ x-1 } = \color{blue}{10x^{2}-18x-47} \color{red}{~-~} \frac{ \color{red}{ 37 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&10&-28&-29&10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 10 }&-28&-29&10\\& & & & \\ \hline &\color{orangered}{10}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&10&-28&-29&10\\& & \color{blue}{10} & & \\ \hline &\color{blue}{10}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -28 } + \color{orangered}{ 10 } = \color{orangered}{ -18 } $
$$ \begin{array}{c|rrrr}1&10&\color{orangered}{ -28 }&-29&10\\& & \color{orangered}{10} & & \\ \hline &10&\color{orangered}{-18}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -18 \right) } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&10&-28&-29&10\\& & 10& \color{blue}{-18} & \\ \hline &10&\color{blue}{-18}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -29 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -47 } $
$$ \begin{array}{c|rrrr}1&10&-28&\color{orangered}{ -29 }&10\\& & 10& \color{orangered}{-18} & \\ \hline &10&-18&\color{orangered}{-47}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -47 \right) } = \color{blue}{ -47 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&10&-28&-29&10\\& & 10& -18& \color{blue}{-47} \\ \hline &10&-18&\color{blue}{-47}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -47 \right) } = \color{orangered}{ -37 } $
$$ \begin{array}{c|rrrr}1&10&-28&-29&\color{orangered}{ 10 }\\& & 10& -18& \color{orangered}{-47} \\ \hline &\color{blue}{10}&\color{blue}{-18}&\color{blue}{-47}&\color{orangered}{-37} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 10x^{2}-18x-47 } $ with a remainder of $ \color{red}{ -37 } $.