Divide using synthetic division $$ ( 2x^{4}+7x^{3}-12x^{2}+27x-17 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&2&7&-12&27&-17\\& & 6& 39& 81& \color{black}{324} \\ \hline &\color{blue}{2}&\color{blue}{13}&\color{blue}{27}&\color{blue}{108}&\color{orangered}{307} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+7x^{3}-12x^{2}+27x-17 }{ x-3 } = \color{blue}{2x^{3}+13x^{2}+27x+108} ~+~ \frac{ \color{red}{ 307 } }{ x-3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&7&-12&27&-17\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 2 }&7&-12&27&-17\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&7&-12&27&-17\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 6 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrrr}3&2&\color{orangered}{ 7 }&-12&27&-17\\& & \color{orangered}{6} & & & \\ \hline &2&\color{orangered}{13}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 13 } = \color{blue}{ 39 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&7&-12&27&-17\\& & 6& \color{blue}{39} & & \\ \hline &2&\color{blue}{13}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 39 } = \color{orangered}{ 27 } $
$$ \begin{array}{c|rrrrr}3&2&7&\color{orangered}{ -12 }&27&-17\\& & 6& \color{orangered}{39} & & \\ \hline &2&13&\color{orangered}{27}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 27 } = \color{blue}{ 81 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&7&-12&27&-17\\& & 6& 39& \color{blue}{81} & \\ \hline &2&13&\color{blue}{27}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 27 } + \color{orangered}{ 81 } = \color{orangered}{ 108 } $
$$ \begin{array}{c|rrrrr}3&2&7&-12&\color{orangered}{ 27 }&-17\\& & 6& 39& \color{orangered}{81} & \\ \hline &2&13&27&\color{orangered}{108}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 108 } = \color{blue}{ 324 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&7&-12&27&-17\\& & 6& 39& 81& \color{blue}{324} \\ \hline &2&13&27&\color{blue}{108}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 324 } = \color{orangered}{ 307 } $
$$ \begin{array}{c|rrrrr}3&2&7&-12&27&\color{orangered}{ -17 }\\& & 6& 39& 81& \color{orangered}{324} \\ \hline &\color{blue}{2}&\color{blue}{13}&\color{blue}{27}&\color{blue}{108}&\color{orangered}{307} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+13x^{2}+27x+108 } $ with a remainder of $ \color{red}{ 307 } $.