Divide using synthetic division $$ ( 5x^{4}+2x^{2}-15x+10 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&5&0&2&-15&10\\& & -10& 20& -44& \color{black}{118} \\ \hline &\color{blue}{5}&\color{blue}{-10}&\color{blue}{22}&\color{blue}{-59}&\color{orangered}{128} \end{array} $$The solution is:
$$ \frac{ 5x^{4}+2x^{2}-15x+10 }{ x+2 } = \color{blue}{5x^{3}-10x^{2}+22x-59} ~+~ \frac{ \color{red}{ 128 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&5&0&2&-15&10\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 5 }&0&2&-15&10\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&5&0&2&-15&10\\& & \color{blue}{-10} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}-2&5&\color{orangered}{ 0 }&2&-15&10\\& & \color{orangered}{-10} & & & \\ \hline &5&\color{orangered}{-10}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&5&0&2&-15&10\\& & -10& \color{blue}{20} & & \\ \hline &5&\color{blue}{-10}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 20 } = \color{orangered}{ 22 } $
$$ \begin{array}{c|rrrrr}-2&5&0&\color{orangered}{ 2 }&-15&10\\& & -10& \color{orangered}{20} & & \\ \hline &5&-10&\color{orangered}{22}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 22 } = \color{blue}{ -44 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&5&0&2&-15&10\\& & -10& 20& \color{blue}{-44} & \\ \hline &5&-10&\color{blue}{22}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ \left( -44 \right) } = \color{orangered}{ -59 } $
$$ \begin{array}{c|rrrrr}-2&5&0&2&\color{orangered}{ -15 }&10\\& & -10& 20& \color{orangered}{-44} & \\ \hline &5&-10&22&\color{orangered}{-59}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -59 \right) } = \color{blue}{ 118 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&5&0&2&-15&10\\& & -10& 20& -44& \color{blue}{118} \\ \hline &5&-10&22&\color{blue}{-59}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ 118 } = \color{orangered}{ 128 } $
$$ \begin{array}{c|rrrrr}-2&5&0&2&-15&\color{orangered}{ 10 }\\& & -10& 20& -44& \color{orangered}{118} \\ \hline &\color{blue}{5}&\color{blue}{-10}&\color{blue}{22}&\color{blue}{-59}&\color{orangered}{128} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{3}-10x^{2}+22x-59 } $ with a remainder of $ \color{red}{ 128 } $.