Divide using synthetic division $$ ( 4x^{3}+8x^{2}-x-15 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&4&8&-1&-15\\& & -20& 60& \color{black}{-295} \\ \hline &\color{blue}{4}&\color{blue}{-12}&\color{blue}{59}&\color{orangered}{-310} \end{array} $$The solution is:
$$ \frac{ 4x^{3}+8x^{2}-x-15 }{ x+5 } = \color{blue}{4x^{2}-12x+59} \color{red}{~-~} \frac{ \color{red}{ 310 } }{ x+5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 5 = 0 $ ( $ x = \color{blue}{ -5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&4&8&-1&-15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 4 }&8&-1&-15\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 4 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&4&8&-1&-15\\& & \color{blue}{-20} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrr}-5&4&\color{orangered}{ 8 }&-1&-15\\& & \color{orangered}{-20} & & \\ \hline &4&\color{orangered}{-12}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 60 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&4&8&-1&-15\\& & -20& \color{blue}{60} & \\ \hline &4&\color{blue}{-12}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 60 } = \color{orangered}{ 59 } $
$$ \begin{array}{c|rrrr}-5&4&8&\color{orangered}{ -1 }&-15\\& & -20& \color{orangered}{60} & \\ \hline &4&-12&\color{orangered}{59}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 59 } = \color{blue}{ -295 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&4&8&-1&-15\\& & -20& 60& \color{blue}{-295} \\ \hline &4&-12&\color{blue}{59}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ \left( -295 \right) } = \color{orangered}{ -310 } $
$$ \begin{array}{c|rrrr}-5&4&8&-1&\color{orangered}{ -15 }\\& & -20& 60& \color{orangered}{-295} \\ \hline &\color{blue}{4}&\color{blue}{-12}&\color{blue}{59}&\color{orangered}{-310} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-12x+59 } $ with a remainder of $ \color{red}{ -310 } $.