Divide using synthetic division $$ ( 4x^{4}+x^{3}+x^{2}-4 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&4&1&1&0&-4\\& & -12& 33& -102& \color{black}{306} \\ \hline &\color{blue}{4}&\color{blue}{-11}&\color{blue}{34}&\color{blue}{-102}&\color{orangered}{302} \end{array} $$The solution is:
$$ \frac{ 4x^{4}+x^{3}+x^{2}-4 }{ x+3 } = \color{blue}{4x^{3}-11x^{2}+34x-102} ~+~ \frac{ \color{red}{ 302 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&1&1&0&-4\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 4 }&1&1&0&-4\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 4 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&1&1&0&-4\\& & \color{blue}{-12} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -11 } $
$$ \begin{array}{c|rrrrr}-3&4&\color{orangered}{ 1 }&1&0&-4\\& & \color{orangered}{-12} & & & \\ \hline &4&\color{orangered}{-11}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -11 \right) } = \color{blue}{ 33 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&1&1&0&-4\\& & -12& \color{blue}{33} & & \\ \hline &4&\color{blue}{-11}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 33 } = \color{orangered}{ 34 } $
$$ \begin{array}{c|rrrrr}-3&4&1&\color{orangered}{ 1 }&0&-4\\& & -12& \color{orangered}{33} & & \\ \hline &4&-11&\color{orangered}{34}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 34 } = \color{blue}{ -102 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&1&1&0&-4\\& & -12& 33& \color{blue}{-102} & \\ \hline &4&-11&\color{blue}{34}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -102 \right) } = \color{orangered}{ -102 } $
$$ \begin{array}{c|rrrrr}-3&4&1&1&\color{orangered}{ 0 }&-4\\& & -12& 33& \color{orangered}{-102} & \\ \hline &4&-11&34&\color{orangered}{-102}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -102 \right) } = \color{blue}{ 306 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&1&1&0&-4\\& & -12& 33& -102& \color{blue}{306} \\ \hline &4&-11&34&\color{blue}{-102}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 306 } = \color{orangered}{ 302 } $
$$ \begin{array}{c|rrrrr}-3&4&1&1&0&\color{orangered}{ -4 }\\& & -12& 33& -102& \color{orangered}{306} \\ \hline &\color{blue}{4}&\color{blue}{-11}&\color{blue}{34}&\color{blue}{-102}&\color{orangered}{302} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}-11x^{2}+34x-102 } $ with a remainder of $ \color{red}{ 302 } $.