Find remainder when $ 7x^{3}+32x^{2}-11x+10 $ is divided by $ x+5 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&7&32&-11&10\\& & -35& 15& \color{black}{-20} \\ \hline &\color{blue}{7}&\color{blue}{-3}&\color{blue}{4}&\color{orangered}{-10} \end{array} $$The remainder when $ 7x^{3}+32x^{2}-11x+10 $ is divided by $ x+5 $ is $ \, \color{red}{ -10 } $.
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 + 5 = 0 $ ( $ x = \color{blue}{ -5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&7&32&-11&10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 7 }&32&-11&10\\& & & & \\ \hline &\color{orangered}{7}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 7 } = \color{blue}{ -35 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&7&32&-11&10\\& & \color{blue}{-35} & & \\ \hline &\color{blue}{7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 32 } + \color{orangered}{ \left( -35 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}-5&7&\color{orangered}{ 32 }&-11&10\\& & \color{orangered}{-35} & & \\ \hline &7&\color{orangered}{-3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&7&32&-11&10\\& & -35& \color{blue}{15} & \\ \hline &7&\color{blue}{-3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 15 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}-5&7&32&\color{orangered}{ -11 }&10\\& & -35& \color{orangered}{15} & \\ \hline &7&-3&\color{orangered}{4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 4 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&7&32&-11&10\\& & -35& 15& \color{blue}{-20} \\ \hline &7&-3&\color{blue}{4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrr}-5&7&32&-11&\color{orangered}{ 10 }\\& & -35& 15& \color{orangered}{-20} \\ \hline &\color{blue}{7}&\color{blue}{-3}&\color{blue}{4}&\color{orangered}{-10} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -10 }\right) $.