Divide using synthetic division $$ ( 10x^{3}-37x^{2}+15x+18 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&10&-37&15&18\\& & 30& -21& \color{black}{-18} \\ \hline &\color{blue}{10}&\color{blue}{-7}&\color{blue}{-6}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 10x^{3}-37x^{2}+15x+18 }{ x-3 } = \color{blue}{10x^{2}-7x-6} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{3}&10&-37&15&18\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 10 }&-37&15&18\\& & & & \\ \hline &\color{orangered}{10}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 10 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&10&-37&15&18\\& & \color{blue}{30} & & \\ \hline &\color{blue}{10}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -37 } + \color{orangered}{ 30 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}3&10&\color{orangered}{ -37 }&15&18\\& & \color{orangered}{30} & & \\ \hline &10&\color{orangered}{-7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -21 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&10&-37&15&18\\& & 30& \color{blue}{-21} & \\ \hline &10&\color{blue}{-7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -21 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}3&10&-37&\color{orangered}{ 15 }&18\\& & 30& \color{orangered}{-21} & \\ \hline &10&-7&\color{orangered}{-6}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&10&-37&15&18\\& & 30& -21& \color{blue}{-18} \\ \hline &10&-7&\color{blue}{-6}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}3&10&-37&15&\color{orangered}{ 18 }\\& & 30& -21& \color{orangered}{-18} \\ \hline &\color{blue}{10}&\color{blue}{-7}&\color{blue}{-6}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 10x^{2}-7x-6 } $ with a remainder of $ \color{red}{ 0 } $.