Find remainder when $ 7x^{3}-12x^{2}+3x+5 $ is divided by $ 7x+2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&7&-12&3&5\\& & -14& 52& \color{black}{-110} \\ \hline &\color{blue}{7}&\color{blue}{-26}&\color{blue}{55}&\color{orangered}{-105} \end{array} $$The remainder when $ 7x^{3}-12x^{2}+3x+5 $ is divided by $ x+2 $ is $ \, \color{red}{ -105 } $.
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 + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&7&-12&3&5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 7 }&-12&3&5\\& & & & \\ \hline &\color{orangered}{7}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 7 } = \color{blue}{ -14 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&7&-12&3&5\\& & \color{blue}{-14} & & \\ \hline &\color{blue}{7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ -26 } $
$$ \begin{array}{c|rrrr}-2&7&\color{orangered}{ -12 }&3&5\\& & \color{orangered}{-14} & & \\ \hline &7&\color{orangered}{-26}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -26 \right) } = \color{blue}{ 52 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&7&-12&3&5\\& & -14& \color{blue}{52} & \\ \hline &7&\color{blue}{-26}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 52 } = \color{orangered}{ 55 } $
$$ \begin{array}{c|rrrr}-2&7&-12&\color{orangered}{ 3 }&5\\& & -14& \color{orangered}{52} & \\ \hline &7&-26&\color{orangered}{55}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 55 } = \color{blue}{ -110 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&7&-12&3&5\\& & -14& 52& \color{blue}{-110} \\ \hline &7&-26&\color{blue}{55}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -110 \right) } = \color{orangered}{ -105 } $
$$ \begin{array}{c|rrrr}-2&7&-12&3&\color{orangered}{ 5 }\\& & -14& 52& \color{orangered}{-110} \\ \hline &\color{blue}{7}&\color{blue}{-26}&\color{blue}{55}&\color{orangered}{-105} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -105 }\right) $.