Divide using synthetic division $$ ( 2x^{4}+5x^{3}+5x^{2}+10x+8 ) \div ( x-12 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}12&2&5&5&10&8\\& & 24& 348& 4236& \color{black}{50952} \\ \hline &\color{blue}{2}&\color{blue}{29}&\color{blue}{353}&\color{blue}{4246}&\color{orangered}{50960} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+5x^{3}+5x^{2}+10x+8 }{ x-12 } = \color{blue}{2x^{3}+29x^{2}+353x+4246} ~+~ \frac{ \color{red}{ 50960 } }{ x-12 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -12 = 0 $ ( $ x = \color{blue}{ 12 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{12}&2&5&5&10&8\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}12&\color{orangered}{ 2 }&5&5&10&8\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 12 } \cdot \color{blue}{ 2 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{12}&2&5&5&10&8\\& & \color{blue}{24} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 24 } = \color{orangered}{ 29 } $
$$ \begin{array}{c|rrrrr}12&2&\color{orangered}{ 5 }&5&10&8\\& & \color{orangered}{24} & & & \\ \hline &2&\color{orangered}{29}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 12 } \cdot \color{blue}{ 29 } = \color{blue}{ 348 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{12}&2&5&5&10&8\\& & 24& \color{blue}{348} & & \\ \hline &2&\color{blue}{29}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 348 } = \color{orangered}{ 353 } $
$$ \begin{array}{c|rrrrr}12&2&5&\color{orangered}{ 5 }&10&8\\& & 24& \color{orangered}{348} & & \\ \hline &2&29&\color{orangered}{353}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 12 } \cdot \color{blue}{ 353 } = \color{blue}{ 4236 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{12}&2&5&5&10&8\\& & 24& 348& \color{blue}{4236} & \\ \hline &2&29&\color{blue}{353}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ 4236 } = \color{orangered}{ 4246 } $
$$ \begin{array}{c|rrrrr}12&2&5&5&\color{orangered}{ 10 }&8\\& & 24& 348& \color{orangered}{4236} & \\ \hline &2&29&353&\color{orangered}{4246}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 12 } \cdot \color{blue}{ 4246 } = \color{blue}{ 50952 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{12}&2&5&5&10&8\\& & 24& 348& 4236& \color{blue}{50952} \\ \hline &2&29&353&\color{blue}{4246}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 50952 } = \color{orangered}{ 50960 } $
$$ \begin{array}{c|rrrrr}12&2&5&5&10&\color{orangered}{ 8 }\\& & 24& 348& 4236& \color{orangered}{50952} \\ \hline &\color{blue}{2}&\color{blue}{29}&\color{blue}{353}&\color{blue}{4246}&\color{orangered}{50960} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+29x^{2}+353x+4246 } $ with a remainder of $ \color{red}{ 50960 } $.