Divide using synthetic division $$ ( 2x^{3}+2x^{2}-8x+3 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&2&2&-8&3\\& & 8& 40& \color{black}{128} \\ \hline &\color{blue}{2}&\color{blue}{10}&\color{blue}{32}&\color{orangered}{131} \end{array} $$The solution is:
$$ \frac{ 2x^{3}+2x^{2}-8x+3 }{ x-4 } = \color{blue}{2x^{2}+10x+32} ~+~ \frac{ \color{red}{ 131 } }{ x-4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{4}&2&2&-8&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 2 }&2&-8&3\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 2 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&2&2&-8&3\\& & \color{blue}{8} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 8 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}4&2&\color{orangered}{ 2 }&-8&3\\& & \color{orangered}{8} & & \\ \hline &2&\color{orangered}{10}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 10 } = \color{blue}{ 40 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&2&2&-8&3\\& & 8& \color{blue}{40} & \\ \hline &2&\color{blue}{10}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 40 } = \color{orangered}{ 32 } $
$$ \begin{array}{c|rrrr}4&2&2&\color{orangered}{ -8 }&3\\& & 8& \color{orangered}{40} & \\ \hline &2&10&\color{orangered}{32}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 32 } = \color{blue}{ 128 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&2&2&-8&3\\& & 8& 40& \color{blue}{128} \\ \hline &2&10&\color{blue}{32}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 128 } = \color{orangered}{ 131 } $
$$ \begin{array}{c|rrrr}4&2&2&-8&\color{orangered}{ 3 }\\& & 8& 40& \color{orangered}{128} \\ \hline &\color{blue}{2}&\color{blue}{10}&\color{blue}{32}&\color{orangered}{131} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}+10x+32 } $ with a remainder of $ \color{red}{ 131 } $.