Divide using synthetic division $$ ( 10x^{3}+60x^{2}+52x+19 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&10&60&52&19\\& & -50& -50& \color{black}{-10} \\ \hline &\color{blue}{10}&\color{blue}{10}&\color{blue}{2}&\color{orangered}{9} \end{array} $$The solution is:
$$ \frac{ 10x^{3}+60x^{2}+52x+19 }{ x+5 } = \color{blue}{10x^{2}+10x+2} ~+~ \frac{ \color{red}{ 9 } }{ x+5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 5 = 0 $ ( $ x = \color{blue}{ -5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&10&60&52&19\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 10 }&60&52&19\\& & & & \\ \hline &\color{orangered}{10}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 10 } = \color{blue}{ -50 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&10&60&52&19\\& & \color{blue}{-50} & & \\ \hline &\color{blue}{10}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 60 } + \color{orangered}{ \left( -50 \right) } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}-5&10&\color{orangered}{ 60 }&52&19\\& & \color{orangered}{-50} & & \\ \hline &10&\color{orangered}{10}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 10 } = \color{blue}{ -50 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&10&60&52&19\\& & -50& \color{blue}{-50} & \\ \hline &10&\color{blue}{10}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 52 } + \color{orangered}{ \left( -50 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}-5&10&60&\color{orangered}{ 52 }&19\\& & -50& \color{orangered}{-50} & \\ \hline &10&10&\color{orangered}{2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 2 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&10&60&52&19\\& & -50& -50& \color{blue}{-10} \\ \hline &10&10&\color{blue}{2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrr}-5&10&60&52&\color{orangered}{ 19 }\\& & -50& -50& \color{orangered}{-10} \\ \hline &\color{blue}{10}&\color{blue}{10}&\color{blue}{2}&\color{orangered}{9} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 10x^{2}+10x+2 } $ with a remainder of $ \color{red}{ 9 } $.