Divide using synthetic division $$ ( 2x^{4}-4x^{3}-13x^{2}+15x+28 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&2&-4&-13&15&28\\& & -4& 16& -6& \color{black}{-18} \\ \hline &\color{blue}{2}&\color{blue}{-8}&\color{blue}{3}&\color{blue}{9}&\color{orangered}{10} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-4x^{3}-13x^{2}+15x+28 }{ x+2 } = \color{blue}{2x^{3}-8x^{2}+3x+9} ~+~ \frac{ \color{red}{ 10 } }{ x+2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&-4&-13&15&28\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 2 }&-4&-13&15&28\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 2 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&-4&-13&15&28\\& & \color{blue}{-4} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}-2&2&\color{orangered}{ -4 }&-13&15&28\\& & \color{orangered}{-4} & & & \\ \hline &2&\color{orangered}{-8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&-4&-13&15&28\\& & -4& \color{blue}{16} & & \\ \hline &2&\color{blue}{-8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 16 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}-2&2&-4&\color{orangered}{ -13 }&15&28\\& & -4& \color{orangered}{16} & & \\ \hline &2&-8&\color{orangered}{3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 3 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&-4&-13&15&28\\& & -4& 16& \color{blue}{-6} & \\ \hline &2&-8&\color{blue}{3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}-2&2&-4&-13&\color{orangered}{ 15 }&28\\& & -4& 16& \color{orangered}{-6} & \\ \hline &2&-8&3&\color{orangered}{9}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 9 } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&-4&-13&15&28\\& & -4& 16& -6& \color{blue}{-18} \\ \hline &2&-8&3&\color{blue}{9}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 28 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}-2&2&-4&-13&15&\color{orangered}{ 28 }\\& & -4& 16& -6& \color{orangered}{-18} \\ \hline &\color{blue}{2}&\color{blue}{-8}&\color{blue}{3}&\color{blue}{9}&\color{orangered}{10} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-8x^{2}+3x+9 } $ with a remainder of $ \color{red}{ 10 } $.