Find remainder when $ x^{3}+x^{2}-10x+13 $ is divided by $ x+4 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-4&1&1&-10&13\\& & -4& 12& \color{black}{-8} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{2}&\color{orangered}{5} \end{array} $$The remainder when $ x^{3}+x^{2}-10x+13 $ is divided by $ x+4 $ is $ \, \color{red}{ 5 } $.
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 + 4 = 0 $ ( $ x = \color{blue}{ -4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&1&-10&13\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-4&\color{orangered}{ 1 }&1&-10&13\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 1 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&1&-10&13\\& & \color{blue}{-4} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}-4&1&\color{orangered}{ 1 }&-10&13\\& & \color{orangered}{-4} & & \\ \hline &1&\color{orangered}{-3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&1&-10&13\\& & -4& \color{blue}{12} & \\ \hline &1&\color{blue}{-3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 12 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}-4&1&1&\color{orangered}{ -10 }&13\\& & -4& \color{orangered}{12} & \\ \hline &1&-3&\color{orangered}{2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 2 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&1&-10&13\\& & -4& 12& \color{blue}{-8} \\ \hline &1&-3&\color{blue}{2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}-4&1&1&-10&\color{orangered}{ 13 }\\& & -4& 12& \color{orangered}{-8} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{2}&\color{orangered}{5} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 5 }\right) $.