Divide using synthetic division $$ ( 9x^{4}+x^{3}-12x+21 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&9&1&0&-12&21\\& & -36& 140& -560& \color{black}{2288} \\ \hline &\color{blue}{9}&\color{blue}{-35}&\color{blue}{140}&\color{blue}{-572}&\color{orangered}{2309} \end{array} $$The solution is:
$$ \frac{ 9x^{4}+x^{3}-12x+21 }{ x+4 } = \color{blue}{9x^{3}-35x^{2}+140x-572} ~+~ \frac{ \color{red}{ 2309 } }{ x+4 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&9&1&0&-12&21\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 9 }&1&0&-12&21\\& & & & & \\ \hline &\color{orangered}{9}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 9 } = \color{blue}{ -36 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&9&1&0&-12&21\\& & \color{blue}{-36} & & & \\ \hline &\color{blue}{9}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -36 \right) } = \color{orangered}{ -35 } $
$$ \begin{array}{c|rrrrr}-4&9&\color{orangered}{ 1 }&0&-12&21\\& & \color{orangered}{-36} & & & \\ \hline &9&\color{orangered}{-35}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -35 \right) } = \color{blue}{ 140 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&9&1&0&-12&21\\& & -36& \color{blue}{140} & & \\ \hline &9&\color{blue}{-35}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 140 } = \color{orangered}{ 140 } $
$$ \begin{array}{c|rrrrr}-4&9&1&\color{orangered}{ 0 }&-12&21\\& & -36& \color{orangered}{140} & & \\ \hline &9&-35&\color{orangered}{140}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 140 } = \color{blue}{ -560 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&9&1&0&-12&21\\& & -36& 140& \color{blue}{-560} & \\ \hline &9&-35&\color{blue}{140}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ \left( -560 \right) } = \color{orangered}{ -572 } $
$$ \begin{array}{c|rrrrr}-4&9&1&0&\color{orangered}{ -12 }&21\\& & -36& 140& \color{orangered}{-560} & \\ \hline &9&-35&140&\color{orangered}{-572}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -572 \right) } = \color{blue}{ 2288 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&9&1&0&-12&21\\& & -36& 140& -560& \color{blue}{2288} \\ \hline &9&-35&140&\color{blue}{-572}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ 2288 } = \color{orangered}{ 2309 } $
$$ \begin{array}{c|rrrrr}-4&9&1&0&-12&\color{orangered}{ 21 }\\& & -36& 140& -560& \color{orangered}{2288} \\ \hline &\color{blue}{9}&\color{blue}{-35}&\color{blue}{140}&\color{blue}{-572}&\color{orangered}{2309} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 9x^{3}-35x^{2}+140x-572 } $ with a remainder of $ \color{red}{ 2309 } $.