Divide using synthetic division $$ ( x^{3}+x^{2}+3 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-4&1&1&0&3\\& & -4& 12& \color{black}{-48} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{12}&\color{orangered}{-45} \end{array} $$The solution is:
$$ \frac{ x^{3}+x^{2}+3 }{ x+4 } = \color{blue}{x^{2}-3x+12} \color{red}{~-~} \frac{ \color{red}{ 45 } }{ x+4 } $$Explanation
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.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&1&0&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-4&\color{orangered}{ 1 }&1&0&3\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 1 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&1&0&3\\& & \color{blue}{-4} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}-4&1&\color{orangered}{ 1 }&0&3\\& & \color{orangered}{-4} & & \\ \hline &1&\color{orangered}{-3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&1&0&3\\& & -4& \color{blue}{12} & \\ \hline &1&\color{blue}{-3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrr}-4&1&1&\color{orangered}{ 0 }&3\\& & -4& \color{orangered}{12} & \\ \hline &1&-3&\color{orangered}{12}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 12 } = \color{blue}{ -48 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&1&0&3\\& & -4& 12& \color{blue}{-48} \\ \hline &1&-3&\color{blue}{12}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -48 \right) } = \color{orangered}{ -45 } $
$$ \begin{array}{c|rrrr}-4&1&1&0&\color{orangered}{ 3 }\\& & -4& 12& \color{orangered}{-48} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{12}&\color{orangered}{-45} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-3x+12 } $ with a remainder of $ \color{red}{ -45 } $.