Divide using synthetic division $$ ( 5x^{4}-6x+1 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&5&0&0&-6&1\\& & -10& 20& -40& \color{black}{92} \\ \hline &\color{blue}{5}&\color{blue}{-10}&\color{blue}{20}&\color{blue}{-46}&\color{orangered}{93} \end{array} $$The solution is:
$$ \frac{ 5x^{4}-6x+1 }{ x+2 } = \color{blue}{5x^{3}-10x^{2}+20x-46} ~+~ \frac{ \color{red}{ 93 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&5&0&0&-6&1\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 5 }&0&0&-6&1\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 5 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&5&0&0&-6&1\\& & \color{blue}{-10} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}-2&5&\color{orangered}{ 0 }&0&-6&1\\& & \color{orangered}{-10} & & & \\ \hline &5&\color{orangered}{-10}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&5&0&0&-6&1\\& & -10& \color{blue}{20} & & \\ \hline &5&\color{blue}{-10}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 20 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrrr}-2&5&0&\color{orangered}{ 0 }&-6&1\\& & -10& \color{orangered}{20} & & \\ \hline &5&-10&\color{orangered}{20}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 20 } = \color{blue}{ -40 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&5&0&0&-6&1\\& & -10& 20& \color{blue}{-40} & \\ \hline &5&-10&\color{blue}{20}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -40 \right) } = \color{orangered}{ -46 } $
$$ \begin{array}{c|rrrrr}-2&5&0&0&\color{orangered}{ -6 }&1\\& & -10& 20& \color{orangered}{-40} & \\ \hline &5&-10&20&\color{orangered}{-46}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -46 \right) } = \color{blue}{ 92 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&5&0&0&-6&1\\& & -10& 20& -40& \color{blue}{92} \\ \hline &5&-10&20&\color{blue}{-46}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 92 } = \color{orangered}{ 93 } $
$$ \begin{array}{c|rrrrr}-2&5&0&0&-6&\color{orangered}{ 1 }\\& & -10& 20& -40& \color{orangered}{92} \\ \hline &\color{blue}{5}&\color{blue}{-10}&\color{blue}{20}&\color{blue}{-46}&\color{orangered}{93} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{3}-10x^{2}+20x-46 } $ with a remainder of $ \color{red}{ 93 } $.