Divide using synthetic division $$ ( 2x^{4}+14x^{3}-2x^{2}-14x ) \div ( x+7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-7&2&14&-2&-14&0\\& & -14& 0& 14& \color{black}{0} \\ \hline &\color{blue}{2}&\color{blue}{0}&\color{blue}{-2}&\color{blue}{0}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+14x^{3}-2x^{2}-14x }{ x+7 } = \color{blue}{2x^{3}-2x} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 7 = 0 $ ( $ x = \color{blue}{ -7 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-7}&2&14&-2&-14&0\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-7&\color{orangered}{ 2 }&14&-2&-14&0\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ 2 } = \color{blue}{ -14 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-7}&2&14&-2&-14&0\\& & \color{blue}{-14} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-7&2&\color{orangered}{ 14 }&-2&-14&0\\& & \color{orangered}{-14} & & & \\ \hline &2&\color{orangered}{0}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-7}&2&14&-2&-14&0\\& & -14& \color{blue}{0} & & \\ \hline &2&\color{blue}{0}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 0 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}-7&2&14&\color{orangered}{ -2 }&-14&0\\& & -14& \color{orangered}{0} & & \\ \hline &2&0&\color{orangered}{-2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-7}&2&14&-2&-14&0\\& & -14& 0& \color{blue}{14} & \\ \hline &2&0&\color{blue}{-2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 14 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-7&2&14&-2&\color{orangered}{ -14 }&0\\& & -14& 0& \color{orangered}{14} & \\ \hline &2&0&-2&\color{orangered}{0}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-7}&2&14&-2&-14&0\\& & -14& 0& 14& \color{blue}{0} \\ \hline &2&0&-2&\color{blue}{0}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-7&2&14&-2&-14&\color{orangered}{ 0 }\\& & -14& 0& 14& \color{orangered}{0} \\ \hline &\color{blue}{2}&\color{blue}{0}&\color{blue}{-2}&\color{blue}{0}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-2x } $ with a remainder of $ \color{red}{ 0 } $.