Divide using synthetic division $$ ( 5x^{3}-53x^{2}+24x+57 ) \div ( x-10 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}10&5&-53&24&57\\& & 50& -30& \color{black}{-60} \\ \hline &\color{blue}{5}&\color{blue}{-3}&\color{blue}{-6}&\color{orangered}{-3} \end{array} $$The solution is:
$$ \frac{ 5x^{3}-53x^{2}+24x+57 }{ x-10 } = \color{blue}{5x^{2}-3x-6} \color{red}{~-~} \frac{ \color{red}{ 3 } }{ x-10 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{10}&5&-53&24&57\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}10&\color{orangered}{ 5 }&-53&24&57\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 10 } \cdot \color{blue}{ 5 } = \color{blue}{ 50 } $.
$$ \begin{array}{c|rrrr}\color{blue}{10}&5&-53&24&57\\& & \color{blue}{50} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -53 } + \color{orangered}{ 50 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}10&5&\color{orangered}{ -53 }&24&57\\& & \color{orangered}{50} & & \\ \hline &5&\color{orangered}{-3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 10 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{10}&5&-53&24&57\\& & 50& \color{blue}{-30} & \\ \hline &5&\color{blue}{-3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 24 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}10&5&-53&\color{orangered}{ 24 }&57\\& & 50& \color{orangered}{-30} & \\ \hline &5&-3&\color{orangered}{-6}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 10 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -60 } $.
$$ \begin{array}{c|rrrr}\color{blue}{10}&5&-53&24&57\\& & 50& -30& \color{blue}{-60} \\ \hline &5&-3&\color{blue}{-6}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 57 } + \color{orangered}{ \left( -60 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}10&5&-53&24&\color{orangered}{ 57 }\\& & 50& -30& \color{orangered}{-60} \\ \hline &\color{blue}{5}&\color{blue}{-3}&\color{blue}{-6}&\color{orangered}{-3} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{2}-3x-6 } $ with a remainder of $ \color{red}{ -3 } $.