Divide using synthetic division $$ ( 8x^{5}-5x^{2}-12 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}1&8&0&0&-5&0&-12\\& & 8& 8& 8& 3& \color{black}{3} \\ \hline &\color{blue}{8}&\color{blue}{8}&\color{blue}{8}&\color{blue}{3}&\color{blue}{3}&\color{orangered}{-9} \end{array} $$The solution is:
$$ \frac{ 8x^{5}-5x^{2}-12 }{ x-1 } = \color{blue}{8x^{4}+8x^{3}+8x^{2}+3x+3} \color{red}{~-~} \frac{ \color{red}{ 9 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&8&0&0&-5&0&-12\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}1&\color{orangered}{ 8 }&0&0&-5&0&-12\\& & & & & & \\ \hline &\color{orangered}{8}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&8&0&0&-5&0&-12\\& & \color{blue}{8} & & & & \\ \hline &\color{blue}{8}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 8 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrrr}1&8&\color{orangered}{ 0 }&0&-5&0&-12\\& & \color{orangered}{8} & & & & \\ \hline &8&\color{orangered}{8}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 8 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&8&0&0&-5&0&-12\\& & 8& \color{blue}{8} & & & \\ \hline &8&\color{blue}{8}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 8 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrrr}1&8&0&\color{orangered}{ 0 }&-5&0&-12\\& & 8& \color{orangered}{8} & & & \\ \hline &8&8&\color{orangered}{8}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&8&0&0&-5&0&-12\\& & 8& 8& \color{blue}{8} & & \\ \hline &8&8&\color{blue}{8}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 8 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrrr}1&8&0&0&\color{orangered}{ -5 }&0&-12\\& & 8& 8& \color{orangered}{8} & & \\ \hline &8&8&8&\color{orangered}{3}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 3 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&8&0&0&-5&0&-12\\& & 8& 8& 8& \color{blue}{3} & \\ \hline &8&8&8&\color{blue}{3}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 3 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrrr}1&8&0&0&-5&\color{orangered}{ 0 }&-12\\& & 8& 8& 8& \color{orangered}{3} & \\ \hline &8&8&8&3&\color{orangered}{3}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 3 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&8&0&0&-5&0&-12\\& & 8& 8& 8& 3& \color{blue}{3} \\ \hline &8&8&8&3&\color{blue}{3}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 3 } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrrrr}1&8&0&0&-5&0&\color{orangered}{ -12 }\\& & 8& 8& 8& 3& \color{orangered}{3} \\ \hline &\color{blue}{8}&\color{blue}{8}&\color{blue}{8}&\color{blue}{3}&\color{blue}{3}&\color{orangered}{-9} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 8x^{4}+8x^{3}+8x^{2}+3x+3 } $ with a remainder of $ \color{red}{ -9 } $.