Divide using synthetic division $$ ( 2x^{3}+5x^{2}-28x-15 ) \div ( 2x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&2&5&-28&-15\\& & -2& -3& \color{black}{31} \\ \hline &\color{blue}{2}&\color{blue}{3}&\color{blue}{-31}&\color{orangered}{16} \end{array} $$The solution is:
$$ \frac{ 2x^{3}+5x^{2}-28x-15 }{ x+1 } = \color{blue}{2x^{2}+3x-31} ~+~ \frac{ \color{red}{ 16 } }{ x+1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 1 = 0 $ ( $ x = \color{blue}{ -1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&2&5&-28&-15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 2 }&5&-28&-15\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 2 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&2&5&-28&-15\\& & \color{blue}{-2} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}-1&2&\color{orangered}{ 5 }&-28&-15\\& & \color{orangered}{-2} & & \\ \hline &2&\color{orangered}{3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 3 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&2&5&-28&-15\\& & -2& \color{blue}{-3} & \\ \hline &2&\color{blue}{3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -28 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -31 } $
$$ \begin{array}{c|rrrr}-1&2&5&\color{orangered}{ -28 }&-15\\& & -2& \color{orangered}{-3} & \\ \hline &2&3&\color{orangered}{-31}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -31 \right) } = \color{blue}{ 31 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&2&5&-28&-15\\& & -2& -3& \color{blue}{31} \\ \hline &2&3&\color{blue}{-31}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 31 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrr}-1&2&5&-28&\color{orangered}{ -15 }\\& & -2& -3& \color{orangered}{31} \\ \hline &\color{blue}{2}&\color{blue}{3}&\color{blue}{-31}&\color{orangered}{16} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}+3x-31 } $ with a remainder of $ \color{red}{ 16 } $.