Find remainder when $ 4x^{4}+2x^{3}-24x^{2}+17x-3 $ is divided by $ x+3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&4&2&-24&17&-3\\& & -12& 30& -18& \color{black}{3} \\ \hline &\color{blue}{4}&\color{blue}{-10}&\color{blue}{6}&\color{blue}{-1}&\color{orangered}{0} \end{array} $$The remainder when $ 4x^{4}+2x^{3}-24x^{2}+17x-3 $ is divided by $ x+3 $ 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 + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&2&-24&17&-3\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 4 }&2&-24&17&-3\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&2&-24&17&-3\\& & \color{blue}{-12} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}-3&4&\color{orangered}{ 2 }&-24&17&-3\\& & \color{orangered}{-12} & & & \\ \hline &4&\color{orangered}{-10}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&2&-24&17&-3\\& & -12& \color{blue}{30} & & \\ \hline &4&\color{blue}{-10}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ 30 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}-3&4&2&\color{orangered}{ -24 }&17&-3\\& & -12& \color{orangered}{30} & & \\ \hline &4&-10&\color{orangered}{6}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 6 } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&2&-24&17&-3\\& & -12& 30& \color{blue}{-18} & \\ \hline &4&-10&\color{blue}{6}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 17 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}-3&4&2&-24&\color{orangered}{ 17 }&-3\\& & -12& 30& \color{orangered}{-18} & \\ \hline &4&-10&6&\color{orangered}{-1}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&2&-24&17&-3\\& & -12& 30& -18& \color{blue}{3} \\ \hline &4&-10&6&\color{blue}{-1}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 3 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-3&4&2&-24&17&\color{orangered}{ -3 }\\& & -12& 30& -18& \color{orangered}{3} \\ \hline &\color{blue}{4}&\color{blue}{-10}&\color{blue}{6}&\color{blue}{-1}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right) $.