Find remainder when $ 4x^{3}-3x^{2}+5x-6 $ is divided by $ x+3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&4&-3&5&-6\\& & -12& 45& \color{black}{-150} \\ \hline &\color{blue}{4}&\color{blue}{-15}&\color{blue}{50}&\color{orangered}{-156} \end{array} $$The remainder when $ 4x^{3}-3x^{2}+5x-6 $ is divided by $ x+3 $ is $ \, \color{red}{ -156 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&-3&5&-6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 4 }&-3&5&-6\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 4 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&-3&5&-6\\& & \color{blue}{-12} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -15 } $
$$ \begin{array}{c|rrrr}-3&4&\color{orangered}{ -3 }&5&-6\\& & \color{orangered}{-12} & & \\ \hline &4&\color{orangered}{-15}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -15 \right) } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&-3&5&-6\\& & -12& \color{blue}{45} & \\ \hline &4&\color{blue}{-15}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 45 } = \color{orangered}{ 50 } $
$$ \begin{array}{c|rrrr}-3&4&-3&\color{orangered}{ 5 }&-6\\& & -12& \color{orangered}{45} & \\ \hline &4&-15&\color{orangered}{50}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 50 } = \color{blue}{ -150 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&-3&5&-6\\& & -12& 45& \color{blue}{-150} \\ \hline &4&-15&\color{blue}{50}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -150 \right) } = \color{orangered}{ -156 } $
$$ \begin{array}{c|rrrr}-3&4&-3&5&\color{orangered}{ -6 }\\& & -12& 45& \color{orangered}{-150} \\ \hline &\color{blue}{4}&\color{blue}{-15}&\color{blue}{50}&\color{orangered}{-156} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -156 }\right) $.