Divide using synthetic division $$ ( 4x^{4}+14x^{3}+13x^{2}+x-2 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&4&14&13&1&-2\\& & -8& -12& -2& \color{black}{2} \\ \hline &\color{blue}{4}&\color{blue}{6}&\color{blue}{1}&\color{blue}{-1}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 4x^{4}+14x^{3}+13x^{2}+x-2 }{ x+2 } = \color{blue}{4x^{3}+6x^{2}+x-1} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&4&14&13&1&-2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 4 }&14&13&1&-2\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 4 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&4&14&13&1&-2\\& & \color{blue}{-8} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}-2&4&\color{orangered}{ 14 }&13&1&-2\\& & \color{orangered}{-8} & & & \\ \hline &4&\color{orangered}{6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 6 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&4&14&13&1&-2\\& & -8& \color{blue}{-12} & & \\ \hline &4&\color{blue}{6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}-2&4&14&\color{orangered}{ 13 }&1&-2\\& & -8& \color{orangered}{-12} & & \\ \hline &4&6&\color{orangered}{1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 1 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&4&14&13&1&-2\\& & -8& -12& \color{blue}{-2} & \\ \hline &4&6&\color{blue}{1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}-2&4&14&13&\color{orangered}{ 1 }&-2\\& & -8& -12& \color{orangered}{-2} & \\ \hline &4&6&1&\color{orangered}{-1}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&4&14&13&1&-2\\& & -8& -12& -2& \color{blue}{2} \\ \hline &4&6&1&\color{blue}{-1}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 2 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-2&4&14&13&1&\color{orangered}{ -2 }\\& & -8& -12& -2& \color{orangered}{2} \\ \hline &\color{blue}{4}&\color{blue}{6}&\color{blue}{1}&\color{blue}{-1}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}+6x^{2}+x-1 } $ with a remainder of $ \color{red}{ 0 } $.