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