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