Divide using synthetic division $$ ( 2x^{3}+23x^{2}+64x+18 ) \div ( x+7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-7&2&23&64&18\\& & -14& -63& \color{black}{-7} \\ \hline &\color{blue}{2}&\color{blue}{9}&\color{blue}{1}&\color{orangered}{11} \end{array} $$The solution is:
$$ \frac{ 2x^{3}+23x^{2}+64x+18 }{ x+7 } = \color{blue}{2x^{2}+9x+1} ~+~ \frac{ \color{red}{ 11 } }{ x+7 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 7 = 0 $ ( $ x = \color{blue}{ -7 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-7}&2&23&64&18\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-7&\color{orangered}{ 2 }&23&64&18\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ 2 } = \color{blue}{ -14 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-7}&2&23&64&18\\& & \color{blue}{-14} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 23 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrr}-7&2&\color{orangered}{ 23 }&64&18\\& & \color{orangered}{-14} & & \\ \hline &2&\color{orangered}{9}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ 9 } = \color{blue}{ -63 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-7}&2&23&64&18\\& & -14& \color{blue}{-63} & \\ \hline &2&\color{blue}{9}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 64 } + \color{orangered}{ \left( -63 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}-7&2&23&\color{orangered}{ 64 }&18\\& & -14& \color{orangered}{-63} & \\ \hline &2&9&\color{orangered}{1}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ 1 } = \color{blue}{ -7 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-7}&2&23&64&18\\& & -14& -63& \color{blue}{-7} \\ \hline &2&9&\color{blue}{1}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ \left( -7 \right) } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrr}-7&2&23&64&\color{orangered}{ 18 }\\& & -14& -63& \color{orangered}{-7} \\ \hline &\color{blue}{2}&\color{blue}{9}&\color{blue}{1}&\color{orangered}{11} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}+9x+1 } $ with a remainder of $ \color{red}{ 11 } $.