Find remainder when $ 4x^{4}+10x^{3}-14x^{2}-22x+6 $ is divided by $ x+3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&4&10&-14&-22&6\\& & -12& 6& 24& \color{black}{-6} \\ \hline &\color{blue}{4}&\color{blue}{-2}&\color{blue}{-8}&\color{blue}{2}&\color{orangered}{0} \end{array} $$The remainder when $ 4x^{4}+10x^{3}-14x^{2}-22x+6 $ 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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&10&-14&-22&6\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 4 }&10&-14&-22&6\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&10&-14&-22&6\\& & \color{blue}{-12} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}-3&4&\color{orangered}{ 10 }&-14&-22&6\\& & \color{orangered}{-12} & & & \\ \hline &4&\color{orangered}{-2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&10&-14&-22&6\\& & -12& \color{blue}{6} & & \\ \hline &4&\color{blue}{-2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 6 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}-3&4&10&\color{orangered}{ -14 }&-22&6\\& & -12& \color{orangered}{6} & & \\ \hline &4&-2&\color{orangered}{-8}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&10&-14&-22&6\\& & -12& 6& \color{blue}{24} & \\ \hline &4&-2&\color{blue}{-8}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -22 } + \color{orangered}{ 24 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-3&4&10&-14&\color{orangered}{ -22 }&6\\& & -12& 6& \color{orangered}{24} & \\ \hline &4&-2&-8&\color{orangered}{2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 2 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&10&-14&-22&6\\& & -12& 6& 24& \color{blue}{-6} \\ \hline &4&-2&-8&\color{blue}{2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-3&4&10&-14&-22&\color{orangered}{ 6 }\\& & -12& 6& 24& \color{orangered}{-6} \\ \hline &\color{blue}{4}&\color{blue}{-2}&\color{blue}{-8}&\color{blue}{2}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right) $.