Divide using synthetic division $$ ( 2x^{4}-12x^{3}+8x^{2}+11x+3 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&2&-12&8&11&3\\& & 10& -10& -10& \color{black}{5} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{-2}&\color{blue}{1}&\color{orangered}{8} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-12x^{3}+8x^{2}+11x+3 }{ x-5 } = \color{blue}{2x^{3}-2x^{2}-2x+1} ~+~ \frac{ \color{red}{ 8 } }{ x-5 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&2&-12&8&11&3\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 2 }&-12&8&11&3\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&2&-12&8&11&3\\& & \color{blue}{10} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 10 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}5&2&\color{orangered}{ -12 }&8&11&3\\& & \color{orangered}{10} & & & \\ \hline &2&\color{orangered}{-2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&2&-12&8&11&3\\& & 10& \color{blue}{-10} & & \\ \hline &2&\color{blue}{-2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}5&2&-12&\color{orangered}{ 8 }&11&3\\& & 10& \color{orangered}{-10} & & \\ \hline &2&-2&\color{orangered}{-2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&2&-12&8&11&3\\& & 10& -10& \color{blue}{-10} & \\ \hline &2&-2&\color{blue}{-2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}5&2&-12&8&\color{orangered}{ 11 }&3\\& & 10& -10& \color{orangered}{-10} & \\ \hline &2&-2&-2&\color{orangered}{1}& \end{array} $$Step 8 : 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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&2&-12&8&11&3\\& & 10& -10& -10& \color{blue}{5} \\ \hline &2&-2&-2&\color{blue}{1}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 5 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}5&2&-12&8&11&\color{orangered}{ 3 }\\& & 10& -10& -10& \color{orangered}{5} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{-2}&\color{blue}{1}&\color{orangered}{8} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-2x^{2}-2x+1 } $ with a remainder of $ \color{red}{ 8 } $.