Divide using synthetic division $$ ( x^{3}-x^{2}+2 ) \div ( 2x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&1&-1&0&2\\& & -5& 30& \color{black}{-150} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{30}&\color{orangered}{-148} \end{array} $$The solution is:
$$ \frac{ x^{3}-x^{2}+2 }{ x+5 } = \color{blue}{x^{2}-6x+30} \color{red}{~-~} \frac{ \color{red}{ 148 } }{ x+5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 5 = 0 $ ( $ x = \color{blue}{ -5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&-1&0&2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 1 }&-1&0&2\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 1 } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&-1&0&2\\& & \color{blue}{-5} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}-5&1&\color{orangered}{ -1 }&0&2\\& & \color{orangered}{-5} & & \\ \hline &1&\color{orangered}{-6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&-1&0&2\\& & -5& \color{blue}{30} & \\ \hline &1&\color{blue}{-6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 30 } = \color{orangered}{ 30 } $
$$ \begin{array}{c|rrrr}-5&1&-1&\color{orangered}{ 0 }&2\\& & -5& \color{orangered}{30} & \\ \hline &1&-6&\color{orangered}{30}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 30 } = \color{blue}{ -150 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&-1&0&2\\& & -5& 30& \color{blue}{-150} \\ \hline &1&-6&\color{blue}{30}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -150 \right) } = \color{orangered}{ -148 } $
$$ \begin{array}{c|rrrr}-5&1&-1&0&\color{orangered}{ 2 }\\& & -5& 30& \color{orangered}{-150} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{30}&\color{orangered}{-148} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-6x+30 } $ with a remainder of $ \color{red}{ -148 } $.