Divide using synthetic division $$ ( 15x^{4}+22x^{3}-17x^{2}+4x+12 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&15&22&-17&4&12\\& & -30& 16& 2& \color{black}{-12} \\ \hline &\color{blue}{15}&\color{blue}{-8}&\color{blue}{-1}&\color{blue}{6}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 15x^{4}+22x^{3}-17x^{2}+4x+12 }{ x+2 } = \color{blue}{15x^{3}-8x^{2}-x+6} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&15&22&-17&4&12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 15 }&22&-17&4&12\\& & & & & \\ \hline &\color{orangered}{15}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 15 } = \color{blue}{ -30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&15&22&-17&4&12\\& & \color{blue}{-30} & & & \\ \hline &\color{blue}{15}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 22 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}-2&15&\color{orangered}{ 22 }&-17&4&12\\& & \color{orangered}{-30} & & & \\ \hline &15&\color{orangered}{-8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&15&22&-17&4&12\\& & -30& \color{blue}{16} & & \\ \hline &15&\color{blue}{-8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 16 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}-2&15&22&\color{orangered}{ -17 }&4&12\\& & -30& \color{orangered}{16} & & \\ \hline &15&-8&\color{orangered}{-1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&15&22&-17&4&12\\& & -30& 16& \color{blue}{2} & \\ \hline &15&-8&\color{blue}{-1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 2 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}-2&15&22&-17&\color{orangered}{ 4 }&12\\& & -30& 16& \color{orangered}{2} & \\ \hline &15&-8&-1&\color{orangered}{6}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 6 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&15&22&-17&4&12\\& & -30& 16& 2& \color{blue}{-12} \\ \hline &15&-8&-1&\color{blue}{6}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-2&15&22&-17&4&\color{orangered}{ 12 }\\& & -30& 16& 2& \color{orangered}{-12} \\ \hline &\color{blue}{15}&\color{blue}{-8}&\color{blue}{-1}&\color{blue}{6}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 15x^{3}-8x^{2}-x+6 } $ with a remainder of $ \color{red}{ 0 } $.