Divide using synthetic division $$ ( 15x^{3}-29x^{2}+6x+8 ) \div ( x-1 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&15&-29&6&8\\& & 15& -14& \color{black}{-8} \\ \hline &\color{blue}{15}&\color{blue}{-14}&\color{blue}{-8}&\color{orangered}{0} \end{array} $$The solution is:
$$ \dfrac{ 15x^{3}-29x^{2}+6x+8 }{ x-1 } = \color{blue}{15x^{2}-14x-8} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&15&-29&6&8\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 15 }&-29&6&8\\& & & & \\ \hline &\color{orangered}{15}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 15 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&15&-29&6&8\\& & \color{blue}{15} & & \\ \hline &\color{blue}{15}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -29 } + \color{orangered}{ 15 } = \color{orangered}{ -14 } $
$$ \begin{array}{c|rrrr}1&15&\color{orangered}{ -29 }&6&8\\& & \color{orangered}{15} & & \\ \hline &15&\color{orangered}{-14}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -14 \right) } = \color{blue}{ -14 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&15&-29&6&8\\& & 15& \color{blue}{-14} & \\ \hline &15&\color{blue}{-14}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}1&15&-29&\color{orangered}{ 6 }&8\\& & 15& \color{orangered}{-14} & \\ \hline &15&-14&\color{orangered}{-8}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&15&-29&6&8\\& & 15& -14& \color{blue}{-8} \\ \hline &15&-14&\color{blue}{-8}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}1&15&-29&6&\color{orangered}{ 8 }\\& & 15& -14& \color{orangered}{-8} \\ \hline &\color{blue}{15}&\color{blue}{-14}&\color{blue}{-8}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 15x^{2}-14x-8 } $ with a remainder of $ \color{red}{ 0 } $.