Divide using synthetic division $$ ( 50x^{3}+10x^{2}-35x-7 ) \div ( 5x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&50&10&-35&-7\\& & 200& 840& \color{black}{3220} \\ \hline &\color{blue}{50}&\color{blue}{210}&\color{blue}{805}&\color{orangered}{3213} \end{array} $$The solution is:
$$ \frac{ 50x^{3}+10x^{2}-35x-7 }{ x-4 } = \color{blue}{50x^{2}+210x+805} ~+~ \frac{ \color{red}{ 3213 } }{ x-4 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{4}&50&10&-35&-7\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 50 }&10&-35&-7\\& & & & \\ \hline &\color{orangered}{50}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 50 } = \color{blue}{ 200 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&50&10&-35&-7\\& & \color{blue}{200} & & \\ \hline &\color{blue}{50}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ 200 } = \color{orangered}{ 210 } $
$$ \begin{array}{c|rrrr}4&50&\color{orangered}{ 10 }&-35&-7\\& & \color{orangered}{200} & & \\ \hline &50&\color{orangered}{210}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 210 } = \color{blue}{ 840 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&50&10&-35&-7\\& & 200& \color{blue}{840} & \\ \hline &50&\color{blue}{210}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -35 } + \color{orangered}{ 840 } = \color{orangered}{ 805 } $
$$ \begin{array}{c|rrrr}4&50&10&\color{orangered}{ -35 }&-7\\& & 200& \color{orangered}{840} & \\ \hline &50&210&\color{orangered}{805}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 805 } = \color{blue}{ 3220 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&50&10&-35&-7\\& & 200& 840& \color{blue}{3220} \\ \hline &50&210&\color{blue}{805}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 3220 } = \color{orangered}{ 3213 } $
$$ \begin{array}{c|rrrr}4&50&10&-35&\color{orangered}{ -7 }\\& & 200& 840& \color{orangered}{3220} \\ \hline &\color{blue}{50}&\color{blue}{210}&\color{blue}{805}&\color{orangered}{3213} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 50x^{2}+210x+805 } $ with a remainder of $ \color{red}{ 3213 } $.