Divide using synthetic division $$ ( 8x^{3}-x^{2}-x-10 ) \div ( 3x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&8&-1&-1&-10\\& & 16& 30& \color{black}{58} \\ \hline &\color{blue}{8}&\color{blue}{15}&\color{blue}{29}&\color{orangered}{48} \end{array} $$The solution is:
$$ \frac{ 8x^{3}-x^{2}-x-10 }{ x-2 } = \color{blue}{8x^{2}+15x+29} ~+~ \frac{ \color{red}{ 48 } }{ x-2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{2}&8&-1&-1&-10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 8 }&-1&-1&-10\\& & & & \\ \hline &\color{orangered}{8}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 8 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&8&-1&-1&-10\\& & \color{blue}{16} & & \\ \hline &\color{blue}{8}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 16 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrr}2&8&\color{orangered}{ -1 }&-1&-10\\& & \color{orangered}{16} & & \\ \hline &8&\color{orangered}{15}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 15 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&8&-1&-1&-10\\& & 16& \color{blue}{30} & \\ \hline &8&\color{blue}{15}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 30 } = \color{orangered}{ 29 } $
$$ \begin{array}{c|rrrr}2&8&-1&\color{orangered}{ -1 }&-10\\& & 16& \color{orangered}{30} & \\ \hline &8&15&\color{orangered}{29}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 29 } = \color{blue}{ 58 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&8&-1&-1&-10\\& & 16& 30& \color{blue}{58} \\ \hline &8&15&\color{blue}{29}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 58 } = \color{orangered}{ 48 } $
$$ \begin{array}{c|rrrr}2&8&-1&-1&\color{orangered}{ -10 }\\& & 16& 30& \color{orangered}{58} \\ \hline &\color{blue}{8}&\color{blue}{15}&\color{blue}{29}&\color{orangered}{48} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 8x^{2}+15x+29 } $ with a remainder of $ \color{red}{ 48 } $.