Divide using synthetic division $$ ( 2x^{4}-7x^{3}+2x-3 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&2&-7&0&2&-3\\& & -6& 39& -117& \color{black}{345} \\ \hline &\color{blue}{2}&\color{blue}{-13}&\color{blue}{39}&\color{blue}{-115}&\color{orangered}{342} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-7x^{3}+2x-3 }{ x+3 } = \color{blue}{2x^{3}-13x^{2}+39x-115} ~+~ \frac{ \color{red}{ 342 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&-7&0&2&-3\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 2 }&-7&0&2&-3\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 2 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&-7&0&2&-3\\& & \color{blue}{-6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrrr}-3&2&\color{orangered}{ -7 }&0&2&-3\\& & \color{orangered}{-6} & & & \\ \hline &2&\color{orangered}{-13}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -13 \right) } = \color{blue}{ 39 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&-7&0&2&-3\\& & -6& \color{blue}{39} & & \\ \hline &2&\color{blue}{-13}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 39 } = \color{orangered}{ 39 } $
$$ \begin{array}{c|rrrrr}-3&2&-7&\color{orangered}{ 0 }&2&-3\\& & -6& \color{orangered}{39} & & \\ \hline &2&-13&\color{orangered}{39}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 39 } = \color{blue}{ -117 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&-7&0&2&-3\\& & -6& 39& \color{blue}{-117} & \\ \hline &2&-13&\color{blue}{39}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -117 \right) } = \color{orangered}{ -115 } $
$$ \begin{array}{c|rrrrr}-3&2&-7&0&\color{orangered}{ 2 }&-3\\& & -6& 39& \color{orangered}{-117} & \\ \hline &2&-13&39&\color{orangered}{-115}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -115 \right) } = \color{blue}{ 345 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&-7&0&2&-3\\& & -6& 39& -117& \color{blue}{345} \\ \hline &2&-13&39&\color{blue}{-115}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 345 } = \color{orangered}{ 342 } $
$$ \begin{array}{c|rrrrr}-3&2&-7&0&2&\color{orangered}{ -3 }\\& & -6& 39& -117& \color{orangered}{345} \\ \hline &\color{blue}{2}&\color{blue}{-13}&\color{blue}{39}&\color{blue}{-115}&\color{orangered}{342} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-13x^{2}+39x-115 } $ with a remainder of $ \color{red}{ 342 } $.