Find remainder when $ 4x^{3}-17x^{2}-14x+27 $ is divided by $ x-5 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&4&-17&-14&27\\& & 20& 15& \color{black}{5} \\ \hline &\color{blue}{4}&\color{blue}{3}&\color{blue}{1}&\color{orangered}{32} \end{array} $$The remainder when $ 4x^{3}-17x^{2}-14x+27 $ is divided by $ x-5 $ is $ \, \color{red}{ 32 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{5}&4&-17&-14&27\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 4 }&-17&-14&27\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 4 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&4&-17&-14&27\\& & \color{blue}{20} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 20 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}5&4&\color{orangered}{ -17 }&-14&27\\& & \color{orangered}{20} & & \\ \hline &4&\color{orangered}{3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 3 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&4&-17&-14&27\\& & 20& \color{blue}{15} & \\ \hline &4&\color{blue}{3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 15 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}5&4&-17&\color{orangered}{ -14 }&27\\& & 20& \color{orangered}{15} & \\ \hline &4&3&\color{orangered}{1}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&4&-17&-14&27\\& & 20& 15& \color{blue}{5} \\ \hline &4&3&\color{blue}{1}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 27 } + \color{orangered}{ 5 } = \color{orangered}{ 32 } $
$$ \begin{array}{c|rrrr}5&4&-17&-14&\color{orangered}{ 27 }\\& & 20& 15& \color{orangered}{5} \\ \hline &\color{blue}{4}&\color{blue}{3}&\color{blue}{1}&\color{orangered}{32} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 32 }\right) $.