Divide using synthetic division $$ ( 2x^{5}-16x^{4}+32x^{3}-3x^{2}-35x ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}5&2&-16&32&-3&-35&0\\& & 10& -30& 10& 35& \color{black}{0} \\ \hline &\color{blue}{2}&\color{blue}{-6}&\color{blue}{2}&\color{blue}{7}&\color{blue}{0}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{5}-16x^{4}+32x^{3}-3x^{2}-35x }{ x-5 } = \color{blue}{2x^{4}-6x^{3}+2x^{2}+7x} $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{5}&2&-16&32&-3&-35&0\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}5&\color{orangered}{ 2 }&-16&32&-3&-35&0\\& & & & & & \\ \hline &\color{orangered}{2}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{5}&2&-16&32&-3&-35&0\\& & \color{blue}{10} & & & & \\ \hline &\color{blue}{2}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 10 } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrrr}5&2&\color{orangered}{ -16 }&32&-3&-35&0\\& & \color{orangered}{10} & & & & \\ \hline &2&\color{orangered}{-6}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{5}&2&-16&32&-3&-35&0\\& & 10& \color{blue}{-30} & & & \\ \hline &2&\color{blue}{-6}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 32 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrr}5&2&-16&\color{orangered}{ 32 }&-3&-35&0\\& & 10& \color{orangered}{-30} & & & \\ \hline &2&-6&\color{orangered}{2}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 2 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{5}&2&-16&32&-3&-35&0\\& & 10& -30& \color{blue}{10} & & \\ \hline &2&-6&\color{blue}{2}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 10 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrrr}5&2&-16&32&\color{orangered}{ -3 }&-35&0\\& & 10& -30& \color{orangered}{10} & & \\ \hline &2&-6&2&\color{orangered}{7}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 7 } = \color{blue}{ 35 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{5}&2&-16&32&-3&-35&0\\& & 10& -30& 10& \color{blue}{35} & \\ \hline &2&-6&2&\color{blue}{7}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -35 } + \color{orangered}{ 35 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}5&2&-16&32&-3&\color{orangered}{ -35 }&0\\& & 10& -30& 10& \color{orangered}{35} & \\ \hline &2&-6&2&7&\color{orangered}{0}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{5}&2&-16&32&-3&-35&0\\& & 10& -30& 10& 35& \color{blue}{0} \\ \hline &2&-6&2&7&\color{blue}{0}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}5&2&-16&32&-3&-35&\color{orangered}{ 0 }\\& & 10& -30& 10& 35& \color{orangered}{0} \\ \hline &\color{blue}{2}&\color{blue}{-6}&\color{blue}{2}&\color{blue}{7}&\color{blue}{0}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{4}-6x^{3}+2x^{2}+7x } $ with a remainder of $ \color{red}{ 0 } $.