Divide using synthetic division $$ ( 4x^{3}-13x^{2}-37x+10 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&4&-13&-37&10\\& & 20& 35& \color{black}{-10} \\ \hline &\color{blue}{4}&\color{blue}{7}&\color{blue}{-2}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-13x^{2}-37x+10 }{ x-5 } = \color{blue}{4x^{2}+7x-2} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{5}&4&-13&-37&10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 4 }&-13&-37&10\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 4 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&4&-13&-37&10\\& & \color{blue}{20} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 20 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}5&4&\color{orangered}{ -13 }&-37&10\\& & \color{orangered}{20} & & \\ \hline &4&\color{orangered}{7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 7 } = \color{blue}{ 35 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&4&-13&-37&10\\& & 20& \color{blue}{35} & \\ \hline &4&\color{blue}{7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -37 } + \color{orangered}{ 35 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}5&4&-13&\color{orangered}{ -37 }&10\\& & 20& \color{orangered}{35} & \\ \hline &4&7&\color{orangered}{-2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&4&-13&-37&10\\& & 20& 35& \color{blue}{-10} \\ \hline &4&7&\color{blue}{-2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}5&4&-13&-37&\color{orangered}{ 10 }\\& & 20& 35& \color{orangered}{-10} \\ \hline &\color{blue}{4}&\color{blue}{7}&\color{blue}{-2}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+7x-2 } $ with a remainder of $ \color{red}{ 0 } $.