Divide using synthetic division $$ ( 2x^{4}+15x^{3}+18x^{2}-64x-96 ) \div ( x+4 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&2&15&18&-64&-96\\& & -8& -28& 40& \color{black}{96} \\ \hline &\color{blue}{2}&\color{blue}{7}&\color{blue}{-10}&\color{blue}{-24}&\color{orangered}{0} \end{array} $$The solution is:
$$ \dfrac{ 2x^{4}+15x^{3}+18x^{2}-64x-96 }{ x+4 } = \color{blue}{2x^{3}+7x^{2}-10x-24} $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 4 = 0 $ ( $ x = \color{blue}{ -4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&15&18&-64&-96\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 2 }&15&18&-64&-96\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 2 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&15&18&-64&-96\\& & \color{blue}{-8} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrr}-4&2&\color{orangered}{ 15 }&18&-64&-96\\& & \color{orangered}{-8} & & & \\ \hline &2&\color{orangered}{7}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 7 } = \color{blue}{ -28 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&15&18&-64&-96\\& & -8& \color{blue}{-28} & & \\ \hline &2&\color{blue}{7}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ \left( -28 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}-4&2&15&\color{orangered}{ 18 }&-64&-96\\& & -8& \color{orangered}{-28} & & \\ \hline &2&7&\color{orangered}{-10}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 40 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&15&18&-64&-96\\& & -8& -28& \color{blue}{40} & \\ \hline &2&7&\color{blue}{-10}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -64 } + \color{orangered}{ 40 } = \color{orangered}{ -24 } $
$$ \begin{array}{c|rrrrr}-4&2&15&18&\color{orangered}{ -64 }&-96\\& & -8& -28& \color{orangered}{40} & \\ \hline &2&7&-10&\color{orangered}{-24}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -24 \right) } = \color{blue}{ 96 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&15&18&-64&-96\\& & -8& -28& 40& \color{blue}{96} \\ \hline &2&7&-10&\color{blue}{-24}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -96 } + \color{orangered}{ 96 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-4&2&15&18&-64&\color{orangered}{ -96 }\\& & -8& -28& 40& \color{orangered}{96} \\ \hline &\color{blue}{2}&\color{blue}{7}&\color{blue}{-10}&\color{blue}{-24}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+7x^{2}-10x-24 } $ with a remainder of $ \color{red}{ 0 } $.