Divide using synthetic division $$ ( 2x^{3}+x^{2}-8x+5 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&2&1&-8&5\\& & -6& 15& \color{black}{-21} \\ \hline &\color{blue}{2}&\color{blue}{-5}&\color{blue}{7}&\color{orangered}{-16} \end{array} $$The solution is:
$$ \frac{ 2x^{3}+x^{2}-8x+5 }{ x+3 } = \color{blue}{2x^{2}-5x+7} \color{red}{~-~} \frac{ \color{red}{ 16 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&1&-8&5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 2 }&1&-8&5\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&1&-8&5\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}-3&2&\color{orangered}{ 1 }&-8&5\\& & \color{orangered}{-6} & & \\ \hline &2&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&1&-8&5\\& & -6& \color{blue}{15} & \\ \hline &2&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 15 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}-3&2&1&\color{orangered}{ -8 }&5\\& & -6& \color{orangered}{15} & \\ \hline &2&-5&\color{orangered}{7}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 7 } = \color{blue}{ -21 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&1&-8&5\\& & -6& 15& \color{blue}{-21} \\ \hline &2&-5&\color{blue}{7}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -21 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrr}-3&2&1&-8&\color{orangered}{ 5 }\\& & -6& 15& \color{orangered}{-21} \\ \hline &\color{blue}{2}&\color{blue}{-5}&\color{blue}{7}&\color{orangered}{-16} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-5x+7 } $ with a remainder of $ \color{red}{ -16 } $.