Divide using synthetic division $$ ( 15x^{4}+23x^{3}-11x^{2}-12 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&15&23&-11&0&-12\\& & -30& 14& -6& \color{black}{12} \\ \hline &\color{blue}{15}&\color{blue}{-7}&\color{blue}{3}&\color{blue}{-6}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 15x^{4}+23x^{3}-11x^{2}-12 }{ x+2 } = \color{blue}{15x^{3}-7x^{2}+3x-6} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&15&23&-11&0&-12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 15 }&23&-11&0&-12\\& & & & & \\ \hline &\color{orangered}{15}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&15&23&-11&0&-12\\& & \color{blue}{-30} & & & \\ \hline &\color{blue}{15}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 23 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}-2&15&\color{orangered}{ 23 }&-11&0&-12\\& & \color{orangered}{-30} & & & \\ \hline &15&\color{orangered}{-7}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&15&23&-11&0&-12\\& & -30& \color{blue}{14} & & \\ \hline &15&\color{blue}{-7}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 14 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}-2&15&23&\color{orangered}{ -11 }&0&-12\\& & -30& \color{orangered}{14} & & \\ \hline &15&-7&\color{orangered}{3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 3 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&15&23&-11&0&-12\\& & -30& 14& \color{blue}{-6} & \\ \hline &15&-7&\color{blue}{3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}-2&15&23&-11&\color{orangered}{ 0 }&-12\\& & -30& 14& \color{orangered}{-6} & \\ \hline &15&-7&3&\color{orangered}{-6}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&15&23&-11&0&-12\\& & -30& 14& -6& \color{blue}{12} \\ \hline &15&-7&3&\color{blue}{-6}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 12 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-2&15&23&-11&0&\color{orangered}{ -12 }\\& & -30& 14& -6& \color{orangered}{12} \\ \hline &\color{blue}{15}&\color{blue}{-7}&\color{blue}{3}&\color{blue}{-6}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 15x^{3}-7x^{2}+3x-6 } $ with a remainder of $ \color{red}{ 0 } $.