Divide using synthetic division $$ ( 4x^{3}+25x^{2}-58x-16 ) \div ( x-8 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}8&4&25&-58&-16\\& & 32& 456& \color{black}{3184} \\ \hline &\color{blue}{4}&\color{blue}{57}&\color{blue}{398}&\color{orangered}{3168} \end{array} $$The solution is:
$$ \frac{ 4x^{3}+25x^{2}-58x-16 }{ x-8 } = \color{blue}{4x^{2}+57x+398} ~+~ \frac{ \color{red}{ 3168 } }{ x-8 } $$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}{ 32 } = \color{orangered}{ 57 } $
$$ \begin{array}{c|rrrr}8&4&\color{orangered}{ 25 }&-58&-16\\& & \color{orangered}{32} & & \\ \hline &4&\color{orangered}{57}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 57 } = \color{blue}{ 456 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&4&25&-58&-16\\& & 32& \color{blue}{456} & \\ \hline &4&\color{blue}{57}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -58 } + \color{orangered}{ 456 } = \color{orangered}{ 398 } $
$$ \begin{array}{c|rrrr}8&4&25&\color{orangered}{ -58 }&-16\\& & 32& \color{orangered}{456} & \\ \hline &4&57&\color{orangered}{398}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 398 } = \color{blue}{ 3184 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&4&25&-58&-16\\& & 32& 456& \color{blue}{3184} \\ \hline &4&57&\color{blue}{398}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 3184 } = \color{orangered}{ 3168 } $
$$ \begin{array}{c|rrrr}8&4&25&-58&\color{orangered}{ -16 }\\& & 32& 456& \color{orangered}{3184} \\ \hline &\color{blue}{4}&\color{blue}{57}&\color{blue}{398}&\color{orangered}{3168} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+57x+398 } $ with a remainder of $ \color{red}{ 3168 } $.