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