Find remainder when $ 6x^{3}-x^{2}+4x+3 $ is divided by $ x+3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&6&-1&4&3\\& & -18& 57& \color{black}{-183} \\ \hline &\color{blue}{6}&\color{blue}{-19}&\color{blue}{61}&\color{orangered}{-180} \end{array} $$The remainder when $ 6x^{3}-x^{2}+4x+3 $ is divided by $ x+3 $ is $ \, \color{red}{ -180 } $.
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 + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&6&-1&4&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 6 }&-1&4&3\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 6 } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&6&-1&4&3\\& & \color{blue}{-18} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -19 } $
$$ \begin{array}{c|rrrr}-3&6&\color{orangered}{ -1 }&4&3\\& & \color{orangered}{-18} & & \\ \hline &6&\color{orangered}{-19}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -19 \right) } = \color{blue}{ 57 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&6&-1&4&3\\& & -18& \color{blue}{57} & \\ \hline &6&\color{blue}{-19}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 57 } = \color{orangered}{ 61 } $
$$ \begin{array}{c|rrrr}-3&6&-1&\color{orangered}{ 4 }&3\\& & -18& \color{orangered}{57} & \\ \hline &6&-19&\color{orangered}{61}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 61 } = \color{blue}{ -183 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&6&-1&4&3\\& & -18& 57& \color{blue}{-183} \\ \hline &6&-19&\color{blue}{61}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -183 \right) } = \color{orangered}{ -180 } $
$$ \begin{array}{c|rrrr}-3&6&-1&4&\color{orangered}{ 3 }\\& & -18& 57& \color{orangered}{-183} \\ \hline &\color{blue}{6}&\color{blue}{-19}&\color{blue}{61}&\color{orangered}{-180} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -180 }\right) $.