Divide using synthetic division $$ ( 4x^{3}-25x^{2}-154x+40 ) \div ( x-10 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}10&4&-25&-154&40\\& & 40& 150& \color{black}{-40} \\ \hline &\color{blue}{4}&\color{blue}{15}&\color{blue}{-4}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-25x^{2}-154x+40 }{ x-10 } = \color{blue}{4x^{2}+15x-4} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -10 = 0 $ ( $ x = \color{blue}{ 10 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{10}&4&-25&-154&40\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}10&\color{orangered}{ 4 }&-25&-154&40\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 10 } \cdot \color{blue}{ 4 } = \color{blue}{ 40 } $.
$$ \begin{array}{c|rrrr}\color{blue}{10}&4&-25&-154&40\\& & \color{blue}{40} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -25 } + \color{orangered}{ 40 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrr}10&4&\color{orangered}{ -25 }&-154&40\\& & \color{orangered}{40} & & \\ \hline &4&\color{orangered}{15}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 10 } \cdot \color{blue}{ 15 } = \color{blue}{ 150 } $.
$$ \begin{array}{c|rrrr}\color{blue}{10}&4&-25&-154&40\\& & 40& \color{blue}{150} & \\ \hline &4&\color{blue}{15}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -154 } + \color{orangered}{ 150 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}10&4&-25&\color{orangered}{ -154 }&40\\& & 40& \color{orangered}{150} & \\ \hline &4&15&\color{orangered}{-4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 10 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -40 } $.
$$ \begin{array}{c|rrrr}\color{blue}{10}&4&-25&-154&40\\& & 40& 150& \color{blue}{-40} \\ \hline &4&15&\color{blue}{-4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 40 } + \color{orangered}{ \left( -40 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}10&4&-25&-154&\color{orangered}{ 40 }\\& & 40& 150& \color{orangered}{-40} \\ \hline &\color{blue}{4}&\color{blue}{15}&\color{blue}{-4}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+15x-4 } $ with a remainder of $ \color{red}{ 0 } $.