Divide using synthetic division $$ ( 4x^{3}-16x^{2}+25x-100 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&4&-16&25&-100\\& & 16& 0& \color{black}{100} \\ \hline &\color{blue}{4}&\color{blue}{0}&\color{blue}{25}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-16x^{2}+25x-100 }{ x-4 } = \color{blue}{4x^{2}+25} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{4}&4&-16&25&-100\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 4 }&-16&25&-100\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&4&-16&25&-100\\& & \color{blue}{16} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 16 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}4&4&\color{orangered}{ -16 }&25&-100\\& & \color{orangered}{16} & & \\ \hline &4&\color{orangered}{0}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&4&-16&25&-100\\& & 16& \color{blue}{0} & \\ \hline &4&\color{blue}{0}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 25 } + \color{orangered}{ 0 } = \color{orangered}{ 25 } $
$$ \begin{array}{c|rrrr}4&4&-16&\color{orangered}{ 25 }&-100\\& & 16& \color{orangered}{0} & \\ \hline &4&0&\color{orangered}{25}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 25 } = \color{blue}{ 100 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&4&-16&25&-100\\& & 16& 0& \color{blue}{100} \\ \hline &4&0&\color{blue}{25}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -100 } + \color{orangered}{ 100 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}4&4&-16&25&\color{orangered}{ -100 }\\& & 16& 0& \color{orangered}{100} \\ \hline &\color{blue}{4}&\color{blue}{0}&\color{blue}{25}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+25 } $ with a remainder of $ \color{red}{ 0 } $.