Find remainder when $ 6x^{2}-10x+7 $ is divided by $ 3x+1 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrr}-1&6&-10&7\\& & -6& \color{black}{16} \\ \hline &\color{blue}{6}&\color{blue}{-16}&\color{orangered}{23} \end{array} $$The remainder when $ 6x^{2}-10x+7 $ is divided by $ x+1 $ is $ \, \color{red}{ 23 } $.
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 + 1 = 0 $ ( $ x = \color{blue}{ -1 } $ ) at the left.
$$ \begin{array}{c|rrr}\color{blue}{-1}&6&-10&7\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}-1&\color{orangered}{ 6 }&-10&7\\& & & \\ \hline &\color{orangered}{6}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 6 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrr}\color{blue}{-1}&6&-10&7\\& & \color{blue}{-6} & \\ \hline &\color{blue}{6}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrr}-1&6&\color{orangered}{ -10 }&7\\& & \color{orangered}{-6} & \\ \hline &6&\color{orangered}{-16}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrr}\color{blue}{-1}&6&-10&7\\& & -6& \color{blue}{16} \\ \hline &6&\color{blue}{-16}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 16 } = \color{orangered}{ 23 } $
$$ \begin{array}{c|rrr}-1&6&-10&\color{orangered}{ 7 }\\& & -6& \color{orangered}{16} \\ \hline &\color{blue}{6}&\color{blue}{-16}&\color{orangered}{23} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 23 }\right) $.