Divide using synthetic division $$ ( x^{4}+7x^{3}-20x^{2}+96 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&1&7&-20&0&96\\& & -3& -12& 96& \color{black}{-288} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{-32}&\color{blue}{96}&\color{orangered}{-192} \end{array} $$The solution is:
$$ \frac{ x^{4}+7x^{3}-20x^{2}+96 }{ x+3 } = \color{blue}{x^{3}+4x^{2}-32x+96} \color{red}{~-~} \frac{ \color{red}{ 192 } }{ x+3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&7&-20&0&96\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 1 }&7&-20&0&96\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&7&-20&0&96\\& & \color{blue}{-3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}-3&1&\color{orangered}{ 7 }&-20&0&96\\& & \color{orangered}{-3} & & & \\ \hline &1&\color{orangered}{4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 4 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&7&-20&0&96\\& & -3& \color{blue}{-12} & & \\ \hline &1&\color{blue}{4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -32 } $
$$ \begin{array}{c|rrrrr}-3&1&7&\color{orangered}{ -20 }&0&96\\& & -3& \color{orangered}{-12} & & \\ \hline &1&4&\color{orangered}{-32}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -32 \right) } = \color{blue}{ 96 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&7&-20&0&96\\& & -3& -12& \color{blue}{96} & \\ \hline &1&4&\color{blue}{-32}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 96 } = \color{orangered}{ 96 } $
$$ \begin{array}{c|rrrrr}-3&1&7&-20&\color{orangered}{ 0 }&96\\& & -3& -12& \color{orangered}{96} & \\ \hline &1&4&-32&\color{orangered}{96}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 96 } = \color{blue}{ -288 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&7&-20&0&96\\& & -3& -12& 96& \color{blue}{-288} \\ \hline &1&4&-32&\color{blue}{96}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 96 } + \color{orangered}{ \left( -288 \right) } = \color{orangered}{ -192 } $
$$ \begin{array}{c|rrrrr}-3&1&7&-20&0&\color{orangered}{ 96 }\\& & -3& -12& 96& \color{orangered}{-288} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{-32}&\color{blue}{96}&\color{orangered}{-192} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+4x^{2}-32x+96 } $ with a remainder of $ \color{red}{ -192 } $.