Divide using synthetic division $$ ( 2x^{5}+3x^{4}-27x^{3}+28x^{2}-41x+100 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-5&2&3&-27&28&-41&100\\& & -10& 35& -40& 60& \color{black}{-95} \\ \hline &\color{blue}{2}&\color{blue}{-7}&\color{blue}{8}&\color{blue}{-12}&\color{blue}{19}&\color{orangered}{5} \end{array} $$The solution is:
$$ \frac{ 2x^{5}+3x^{4}-27x^{3}+28x^{2}-41x+100 }{ x+5 } = \color{blue}{2x^{4}-7x^{3}+8x^{2}-12x+19} ~+~ \frac{ \color{red}{ 5 } }{ x+5 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&2&3&-27&28&-41&100\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-5&\color{orangered}{ 2 }&3&-27&28&-41&100\\& & & & & & \\ \hline &\color{orangered}{2}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 2 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&2&3&-27&28&-41&100\\& & \color{blue}{-10} & & & & \\ \hline &\color{blue}{2}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrrr}-5&2&\color{orangered}{ 3 }&-27&28&-41&100\\& & \color{orangered}{-10} & & & & \\ \hline &2&\color{orangered}{-7}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 35 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&2&3&-27&28&-41&100\\& & -10& \color{blue}{35} & & & \\ \hline &2&\color{blue}{-7}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -27 } + \color{orangered}{ 35 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrrr}-5&2&3&\color{orangered}{ -27 }&28&-41&100\\& & -10& \color{orangered}{35} & & & \\ \hline &2&-7&\color{orangered}{8}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 8 } = \color{blue}{ -40 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&2&3&-27&28&-41&100\\& & -10& 35& \color{blue}{-40} & & \\ \hline &2&-7&\color{blue}{8}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 28 } + \color{orangered}{ \left( -40 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrrr}-5&2&3&-27&\color{orangered}{ 28 }&-41&100\\& & -10& 35& \color{orangered}{-40} & & \\ \hline &2&-7&8&\color{orangered}{-12}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 60 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&2&3&-27&28&-41&100\\& & -10& 35& -40& \color{blue}{60} & \\ \hline &2&-7&8&\color{blue}{-12}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -41 } + \color{orangered}{ 60 } = \color{orangered}{ 19 } $
$$ \begin{array}{c|rrrrrr}-5&2&3&-27&28&\color{orangered}{ -41 }&100\\& & -10& 35& -40& \color{orangered}{60} & \\ \hline &2&-7&8&-12&\color{orangered}{19}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 19 } = \color{blue}{ -95 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&2&3&-27&28&-41&100\\& & -10& 35& -40& 60& \color{blue}{-95} \\ \hline &2&-7&8&-12&\color{blue}{19}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 100 } + \color{orangered}{ \left( -95 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrrr}-5&2&3&-27&28&-41&\color{orangered}{ 100 }\\& & -10& 35& -40& 60& \color{orangered}{-95} \\ \hline &\color{blue}{2}&\color{blue}{-7}&\color{blue}{8}&\color{blue}{-12}&\color{blue}{19}&\color{orangered}{5} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{4}-7x^{3}+8x^{2}-12x+19 } $ with a remainder of $ \color{red}{ 5 } $.