Divide using synthetic division $$ ( 2x^{5}-10x^{3}+8x^{2}+3 ) \div ( x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-1&2&0&-10&8&0&3\\& & -2& 2& 8& -16& \color{black}{16} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{-8}&\color{blue}{16}&\color{blue}{-16}&\color{orangered}{19} \end{array} $$The solution is:
$$ \frac{ 2x^{5}-10x^{3}+8x^{2}+3 }{ x+1 } = \color{blue}{2x^{4}-2x^{3}-8x^{2}+16x-16} ~+~ \frac{ \color{red}{ 19 } }{ x+1 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&2&0&-10&8&0&3\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-1&\color{orangered}{ 2 }&0&-10&8&0&3\\& & & & & & \\ \hline &\color{orangered}{2}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&2&0&-10&8&0&3\\& & \color{blue}{-2} & & & & \\ \hline &\color{blue}{2}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}-1&2&\color{orangered}{ 0 }&-10&8&0&3\\& & \color{orangered}{-2} & & & & \\ \hline &2&\color{orangered}{-2}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&2&0&-10&8&0&3\\& & -2& \color{blue}{2} & & & \\ \hline &2&\color{blue}{-2}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 2 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrrr}-1&2&0&\color{orangered}{ -10 }&8&0&3\\& & -2& \color{orangered}{2} & & & \\ \hline &2&-2&\color{orangered}{-8}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&2&0&-10&8&0&3\\& & -2& 2& \color{blue}{8} & & \\ \hline &2&-2&\color{blue}{-8}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 8 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrrrr}-1&2&0&-10&\color{orangered}{ 8 }&0&3\\& & -2& 2& \color{orangered}{8} & & \\ \hline &2&-2&-8&\color{orangered}{16}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 16 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&2&0&-10&8&0&3\\& & -2& 2& 8& \color{blue}{-16} & \\ \hline &2&-2&-8&\color{blue}{16}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrrrr}-1&2&0&-10&8&\color{orangered}{ 0 }&3\\& & -2& 2& 8& \color{orangered}{-16} & \\ \hline &2&-2&-8&16&\color{orangered}{-16}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&2&0&-10&8&0&3\\& & -2& 2& 8& -16& \color{blue}{16} \\ \hline &2&-2&-8&16&\color{blue}{-16}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 16 } = \color{orangered}{ 19 } $
$$ \begin{array}{c|rrrrrr}-1&2&0&-10&8&0&\color{orangered}{ 3 }\\& & -2& 2& 8& -16& \color{orangered}{16} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{-8}&\color{blue}{16}&\color{blue}{-16}&\color{orangered}{19} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{4}-2x^{3}-8x^{2}+16x-16 } $ with a remainder of $ \color{red}{ 19 } $.