Find remainder when $ x^{3}+15x^{2}+68x+84 $ is divided by $ x+7 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-7&1&15&68&84\\& & -7& -56& \color{black}{-84} \\ \hline &\color{blue}{1}&\color{blue}{8}&\color{blue}{12}&\color{orangered}{0} \end{array} $$The remainder when $ x^{3}+15x^{2}+68x+84 $ is divided by $ x+7 $ is $ \, \color{red}{ 0 } $.
Explanation
We can find remainder using synthetic division method.
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.
$$ \begin{array}{c|rrrr}\color{blue}{-7}&1&15&68&84\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-7&\color{orangered}{ 1 }&15&68&84\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ 1 } = \color{blue}{ -7 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-7}&1&15&68&84\\& & \color{blue}{-7} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -7 \right) } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}-7&1&\color{orangered}{ 15 }&68&84\\& & \color{orangered}{-7} & & \\ \hline &1&\color{orangered}{8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ 8 } = \color{blue}{ -56 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-7}&1&15&68&84\\& & -7& \color{blue}{-56} & \\ \hline &1&\color{blue}{8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 68 } + \color{orangered}{ \left( -56 \right) } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrr}-7&1&15&\color{orangered}{ 68 }&84\\& & -7& \color{orangered}{-56} & \\ \hline &1&8&\color{orangered}{12}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ 12 } = \color{blue}{ -84 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-7}&1&15&68&84\\& & -7& -56& \color{blue}{-84} \\ \hline &1&8&\color{blue}{12}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 84 } + \color{orangered}{ \left( -84 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-7&1&15&68&\color{orangered}{ 84 }\\& & -7& -56& \color{orangered}{-84} \\ \hline &\color{blue}{1}&\color{blue}{8}&\color{blue}{12}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right) $.