Divide using synthetic division $$ ( 4x^{3}+25x^{2}-58x-16 ) \div ( x+8 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-8&4&25&-58&-16\\& & -32& 56& \color{black}{16} \\ \hline &\color{blue}{4}&\color{blue}{-7}&\color{blue}{-2}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 4x^{3}+25x^{2}-58x-16 }{ x+8 } = \color{blue}{4x^{2}-7x-2} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 8 = 0 $ ( $ x = \color{blue}{ -8 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-8}&4&25&-58&-16\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-8&\color{orangered}{ 4 }&25&-58&-16\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ 4 } = \color{blue}{ -32 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-8}&4&25&-58&-16\\& & \color{blue}{-32} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 25 } + \color{orangered}{ \left( -32 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}-8&4&\color{orangered}{ 25 }&-58&-16\\& & \color{orangered}{-32} & & \\ \hline &4&\color{orangered}{-7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 56 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-8}&4&25&-58&-16\\& & -32& \color{blue}{56} & \\ \hline &4&\color{blue}{-7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -58 } + \color{orangered}{ 56 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}-8&4&25&\color{orangered}{ -58 }&-16\\& & -32& \color{orangered}{56} & \\ \hline &4&-7&\color{orangered}{-2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-8}&4&25&-58&-16\\& & -32& 56& \color{blue}{16} \\ \hline &4&-7&\color{blue}{-2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 16 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-8&4&25&-58&\color{orangered}{ -16 }\\& & -32& 56& \color{orangered}{16} \\ \hline &\color{blue}{4}&\color{blue}{-7}&\color{blue}{-2}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-7x-2 } $ with a remainder of $ \color{red}{ 0 } $.