Divide using synthetic division $$ ( 4x^{3}+10x^{2}-25x-4 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-4&4&10&-25&-4\\& & -16& 24& \color{black}{4} \\ \hline &\color{blue}{4}&\color{blue}{-6}&\color{blue}{-1}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 4x^{3}+10x^{2}-25x-4 }{ x+4 } = \color{blue}{4x^{2}-6x-1} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&4&10&-25&-4\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-4&\color{orangered}{ 4 }&10&-25&-4\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 4 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&4&10&-25&-4\\& & \color{blue}{-16} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}-4&4&\color{orangered}{ 10 }&-25&-4\\& & \color{orangered}{-16} & & \\ \hline &4&\color{orangered}{-6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&4&10&-25&-4\\& & -16& \color{blue}{24} & \\ \hline &4&\color{blue}{-6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -25 } + \color{orangered}{ 24 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}-4&4&10&\color{orangered}{ -25 }&-4\\& & -16& \color{orangered}{24} & \\ \hline &4&-6&\color{orangered}{-1}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&4&10&-25&-4\\& & -16& 24& \color{blue}{4} \\ \hline &4&-6&\color{blue}{-1}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 4 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-4&4&10&-25&\color{orangered}{ -4 }\\& & -16& 24& \color{orangered}{4} \\ \hline &\color{blue}{4}&\color{blue}{-6}&\color{blue}{-1}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-6x-1 } $ with a remainder of $ \color{red}{ 0 } $.