Find remainder when $ 4x^{4}+4x^{3}-14x^{2}+25x-15 $ is divided by $ x+3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&4&4&-14&25&-15\\& & -12& 24& -30& \color{black}{15} \\ \hline &\color{blue}{4}&\color{blue}{-8}&\color{blue}{10}&\color{blue}{-5}&\color{orangered}{0} \end{array} $$The remainder when $ 4x^{4}+4x^{3}-14x^{2}+25x-15 $ 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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&4&-14&25&-15\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 4 }&4&-14&25&-15\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&4&-14&25&-15\\& & \color{blue}{-12} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}-3&4&\color{orangered}{ 4 }&-14&25&-15\\& & \color{orangered}{-12} & & & \\ \hline &4&\color{orangered}{-8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&4&-14&25&-15\\& & -12& \color{blue}{24} & & \\ \hline &4&\color{blue}{-8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 24 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}-3&4&4&\color{orangered}{ -14 }&25&-15\\& & -12& \color{orangered}{24} & & \\ \hline &4&-8&\color{orangered}{10}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 10 } = \color{blue}{ -30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&4&-14&25&-15\\& & -12& 24& \color{blue}{-30} & \\ \hline &4&-8&\color{blue}{10}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 25 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}-3&4&4&-14&\color{orangered}{ 25 }&-15\\& & -12& 24& \color{orangered}{-30} & \\ \hline &4&-8&10&\color{orangered}{-5}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&4&-14&25&-15\\& & -12& 24& -30& \color{blue}{15} \\ \hline &4&-8&10&\color{blue}{-5}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 15 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-3&4&4&-14&25&\color{orangered}{ -15 }\\& & -12& 24& -30& \color{orangered}{15} \\ \hline &\color{blue}{4}&\color{blue}{-8}&\color{blue}{10}&\color{blue}{-5}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right) $.