Divide using synthetic division $$ ( 4x^{4}+8x^{3}-15x^{2}-19x-30 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&4&8&-15&-19&-30\\& & -12& 12& 9& \color{black}{30} \\ \hline &\color{blue}{4}&\color{blue}{-4}&\color{blue}{-3}&\color{blue}{-10}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 4x^{4}+8x^{3}-15x^{2}-19x-30 }{ x+3 } = \color{blue}{4x^{3}-4x^{2}-3x-10} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&8&-15&-19&-30\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 4 }&8&-15&-19&-30\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&8&-15&-19&-30\\& & \color{blue}{-12} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}-3&4&\color{orangered}{ 8 }&-15&-19&-30\\& & \color{orangered}{-12} & & & \\ \hline &4&\color{orangered}{-4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&8&-15&-19&-30\\& & -12& \color{blue}{12} & & \\ \hline &4&\color{blue}{-4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 12 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-3&4&8&\color{orangered}{ -15 }&-19&-30\\& & -12& \color{orangered}{12} & & \\ \hline &4&-4&\color{orangered}{-3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&8&-15&-19&-30\\& & -12& 12& \color{blue}{9} & \\ \hline &4&-4&\color{blue}{-3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ 9 } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}-3&4&8&-15&\color{orangered}{ -19 }&-30\\& & -12& 12& \color{orangered}{9} & \\ \hline &4&-4&-3&\color{orangered}{-10}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&8&-15&-19&-30\\& & -12& 12& 9& \color{blue}{30} \\ \hline &4&-4&-3&\color{blue}{-10}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -30 } + \color{orangered}{ 30 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-3&4&8&-15&-19&\color{orangered}{ -30 }\\& & -12& 12& 9& \color{orangered}{30} \\ \hline &\color{blue}{4}&\color{blue}{-4}&\color{blue}{-3}&\color{blue}{-10}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}-4x^{2}-3x-10 } $ with a remainder of $ \color{red}{ 0 } $.