Divide using synthetic division $$ ( 3x^{4}+4x^{3}+15x+10 ) \div ( 3x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&3&4&0&15&10\\& & -12& 32& -128& \color{black}{452} \\ \hline &\color{blue}{3}&\color{blue}{-8}&\color{blue}{32}&\color{blue}{-113}&\color{orangered}{462} \end{array} $$The solution is:
$$ \frac{ 3x^{4}+4x^{3}+15x+10 }{ x+4 } = \color{blue}{3x^{3}-8x^{2}+32x-113} ~+~ \frac{ \color{red}{ 462 } }{ x+4 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&3&4&0&15&10\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 3 }&4&0&15&10\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 3 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&3&4&0&15&10\\& & \color{blue}{-12} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}-4&3&\color{orangered}{ 4 }&0&15&10\\& & \color{orangered}{-12} & & & \\ \hline &3&\color{orangered}{-8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&3&4&0&15&10\\& & -12& \color{blue}{32} & & \\ \hline &3&\color{blue}{-8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 32 } = \color{orangered}{ 32 } $
$$ \begin{array}{c|rrrrr}-4&3&4&\color{orangered}{ 0 }&15&10\\& & -12& \color{orangered}{32} & & \\ \hline &3&-8&\color{orangered}{32}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 32 } = \color{blue}{ -128 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&3&4&0&15&10\\& & -12& 32& \color{blue}{-128} & \\ \hline &3&-8&\color{blue}{32}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -128 \right) } = \color{orangered}{ -113 } $
$$ \begin{array}{c|rrrrr}-4&3&4&0&\color{orangered}{ 15 }&10\\& & -12& 32& \color{orangered}{-128} & \\ \hline &3&-8&32&\color{orangered}{-113}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -113 \right) } = \color{blue}{ 452 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&3&4&0&15&10\\& & -12& 32& -128& \color{blue}{452} \\ \hline &3&-8&32&\color{blue}{-113}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ 452 } = \color{orangered}{ 462 } $
$$ \begin{array}{c|rrrrr}-4&3&4&0&15&\color{orangered}{ 10 }\\& & -12& 32& -128& \color{orangered}{452} \\ \hline &\color{blue}{3}&\color{blue}{-8}&\color{blue}{32}&\color{blue}{-113}&\color{orangered}{462} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}-8x^{2}+32x-113 } $ with a remainder of $ \color{red}{ 462 } $.