Divide using synthetic division $$ ( 4x^{4}-8x^{3}-21x^{2}+10x+13 ) \div ( 2x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&4&-8&-21&10&13\\& & -12& 60& -117& \color{black}{321} \\ \hline &\color{blue}{4}&\color{blue}{-20}&\color{blue}{39}&\color{blue}{-107}&\color{orangered}{334} \end{array} $$The solution is:
$$ \frac{ 4x^{4}-8x^{3}-21x^{2}+10x+13 }{ x+3 } = \color{blue}{4x^{3}-20x^{2}+39x-107} ~+~ \frac{ \color{red}{ 334 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&-8&-21&10&13\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 4 }&-8&-21&10&13\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&-8&-21&10&13\\& & \color{blue}{-12} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrrrr}-3&4&\color{orangered}{ -8 }&-21&10&13\\& & \color{orangered}{-12} & & & \\ \hline &4&\color{orangered}{-20}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -20 \right) } = \color{blue}{ 60 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&-8&-21&10&13\\& & -12& \color{blue}{60} & & \\ \hline &4&\color{blue}{-20}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ 60 } = \color{orangered}{ 39 } $
$$ \begin{array}{c|rrrrr}-3&4&-8&\color{orangered}{ -21 }&10&13\\& & -12& \color{orangered}{60} & & \\ \hline &4&-20&\color{orangered}{39}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 39 } = \color{blue}{ -117 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&-8&-21&10&13\\& & -12& 60& \color{blue}{-117} & \\ \hline &4&-20&\color{blue}{39}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -117 \right) } = \color{orangered}{ -107 } $
$$ \begin{array}{c|rrrrr}-3&4&-8&-21&\color{orangered}{ 10 }&13\\& & -12& 60& \color{orangered}{-117} & \\ \hline &4&-20&39&\color{orangered}{-107}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -107 \right) } = \color{blue}{ 321 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&-8&-21&10&13\\& & -12& 60& -117& \color{blue}{321} \\ \hline &4&-20&39&\color{blue}{-107}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ 321 } = \color{orangered}{ 334 } $
$$ \begin{array}{c|rrrrr}-3&4&-8&-21&10&\color{orangered}{ 13 }\\& & -12& 60& -117& \color{orangered}{321} \\ \hline &\color{blue}{4}&\color{blue}{-20}&\color{blue}{39}&\color{blue}{-107}&\color{orangered}{334} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}-20x^{2}+39x-107 } $ with a remainder of $ \color{red}{ 334 } $.