Divide using synthetic division $$ ( 4x^{3}-23x-21 ) \div ( -x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&4&0&-23&-21\\& & -12& 36& \color{black}{-39} \\ \hline &\color{blue}{4}&\color{blue}{-12}&\color{blue}{13}&\color{orangered}{-60} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-23x-21 }{ x+3 } = \color{blue}{4x^{2}-12x+13} \color{red}{~-~} \frac{ \color{red}{ 60 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&0&-23&-21\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 4 }&0&-23&-21\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&0&-23&-21\\& & \color{blue}{-12} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrr}-3&4&\color{orangered}{ 0 }&-23&-21\\& & \color{orangered}{-12} & & \\ \hline &4&\color{orangered}{-12}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&0&-23&-21\\& & -12& \color{blue}{36} & \\ \hline &4&\color{blue}{-12}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -23 } + \color{orangered}{ 36 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrr}-3&4&0&\color{orangered}{ -23 }&-21\\& & -12& \color{orangered}{36} & \\ \hline &4&-12&\color{orangered}{13}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 13 } = \color{blue}{ -39 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&0&-23&-21\\& & -12& 36& \color{blue}{-39} \\ \hline &4&-12&\color{blue}{13}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ \left( -39 \right) } = \color{orangered}{ -60 } $
$$ \begin{array}{c|rrrr}-3&4&0&-23&\color{orangered}{ -21 }\\& & -12& 36& \color{orangered}{-39} \\ \hline &\color{blue}{4}&\color{blue}{-12}&\color{blue}{13}&\color{orangered}{-60} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-12x+13 } $ with a remainder of $ \color{red}{ -60 } $.