Divide using synthetic division $$ ( 2x^{3}-9x^{2}-6x+5 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&2&-9&-6&5\\& & 10& 5& \color{black}{-5} \\ \hline &\color{blue}{2}&\color{blue}{1}&\color{blue}{-1}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-9x^{2}-6x+5 }{ x-5 } = \color{blue}{2x^{2}+x-1} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{5}&2&-9&-6&5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 2 }&-9&-6&5\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 2 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&2&-9&-6&5\\& & \color{blue}{10} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 10 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}5&2&\color{orangered}{ -9 }&-6&5\\& & \color{orangered}{10} & & \\ \hline &2&\color{orangered}{1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&2&-9&-6&5\\& & 10& \color{blue}{5} & \\ \hline &2&\color{blue}{1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 5 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}5&2&-9&\color{orangered}{ -6 }&5\\& & 10& \color{orangered}{5} & \\ \hline &2&1&\color{orangered}{-1}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&2&-9&-6&5\\& & 10& 5& \color{blue}{-5} \\ \hline &2&1&\color{blue}{-1}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}5&2&-9&-6&\color{orangered}{ 5 }\\& & 10& 5& \color{orangered}{-5} \\ \hline &\color{blue}{2}&\color{blue}{1}&\color{blue}{-1}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}+x-1 } $ with a remainder of $ \color{red}{ 0 } $.