Divide using synthetic division $$ ( 2x^{3}-10x^{2}-5 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&2&-10&0&-5\\& & 2& -8& \color{black}{-8} \\ \hline &\color{blue}{2}&\color{blue}{-8}&\color{blue}{-8}&\color{orangered}{-13} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-10x^{2}-5 }{ x-1 } = \color{blue}{2x^{2}-8x-8} \color{red}{~-~} \frac{ \color{red}{ 13 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&-10&0&-5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 2 }&-10&0&-5\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 2 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&-10&0&-5\\& & \color{blue}{2} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 2 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}1&2&\color{orangered}{ -10 }&0&-5\\& & \color{orangered}{2} & & \\ \hline &2&\color{orangered}{-8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&-10&0&-5\\& & 2& \color{blue}{-8} & \\ \hline &2&\color{blue}{-8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}1&2&-10&\color{orangered}{ 0 }&-5\\& & 2& \color{orangered}{-8} & \\ \hline &2&-8&\color{orangered}{-8}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&-10&0&-5\\& & 2& -8& \color{blue}{-8} \\ \hline &2&-8&\color{blue}{-8}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrr}1&2&-10&0&\color{orangered}{ -5 }\\& & 2& -8& \color{orangered}{-8} \\ \hline &\color{blue}{2}&\color{blue}{-8}&\color{blue}{-8}&\color{orangered}{-13} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-8x-8 } $ with a remainder of $ \color{red}{ -13 } $.