Divide using synthetic division $$ ( 2x^{3}-11x^{2}-63 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&2&-11&0&-63\\& & 10& -5& \color{black}{-25} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{-5}&\color{orangered}{-88} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-11x^{2}-63 }{ x-5 } = \color{blue}{2x^{2}-x-5} \color{red}{~-~} \frac{ \color{red}{ 88 } }{ x-5 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{5}&2&-11&0&-63\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 2 }&-11&0&-63\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&2&-11&0&-63\\& & \color{blue}{10} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 10 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}5&2&\color{orangered}{ -11 }&0&-63\\& & \color{orangered}{10} & & \\ \hline &2&\color{orangered}{-1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&2&-11&0&-63\\& & 10& \color{blue}{-5} & \\ \hline &2&\color{blue}{-1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}5&2&-11&\color{orangered}{ 0 }&-63\\& & 10& \color{orangered}{-5} & \\ \hline &2&-1&\color{orangered}{-5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -25 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&2&-11&0&-63\\& & 10& -5& \color{blue}{-25} \\ \hline &2&-1&\color{blue}{-5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -63 } + \color{orangered}{ \left( -25 \right) } = \color{orangered}{ -88 } $
$$ \begin{array}{c|rrrr}5&2&-11&0&\color{orangered}{ -63 }\\& & 10& -5& \color{orangered}{-25} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{-5}&\color{orangered}{-88} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-x-5 } $ with a remainder of $ \color{red}{ -88 } $.