Divide using synthetic division $$ ( 4x^{5}-x^{3}+2x+3 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}2&4&0&-1&0&2&3\\& & 8& 16& 30& 60& \color{black}{124} \\ \hline &\color{blue}{4}&\color{blue}{8}&\color{blue}{15}&\color{blue}{30}&\color{blue}{62}&\color{orangered}{127} \end{array} $$The solution is:
$$ \frac{ 4x^{5}-x^{3}+2x+3 }{ x-2 } = \color{blue}{4x^{4}+8x^{3}+15x^{2}+30x+62} ~+~ \frac{ \color{red}{ 127 } }{ x-2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&4&0&-1&0&2&3\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}2&\color{orangered}{ 4 }&0&-1&0&2&3\\& & & & & & \\ \hline &\color{orangered}{4}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 4 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&4&0&-1&0&2&3\\& & \color{blue}{8} & & & & \\ \hline &\color{blue}{4}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 8 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrrr}2&4&\color{orangered}{ 0 }&-1&0&2&3\\& & \color{orangered}{8} & & & & \\ \hline &4&\color{orangered}{8}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 8 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&4&0&-1&0&2&3\\& & 8& \color{blue}{16} & & & \\ \hline &4&\color{blue}{8}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 16 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrrrr}2&4&0&\color{orangered}{ -1 }&0&2&3\\& & 8& \color{orangered}{16} & & & \\ \hline &4&8&\color{orangered}{15}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 15 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&4&0&-1&0&2&3\\& & 8& 16& \color{blue}{30} & & \\ \hline &4&8&\color{blue}{15}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 30 } = \color{orangered}{ 30 } $
$$ \begin{array}{c|rrrrrr}2&4&0&-1&\color{orangered}{ 0 }&2&3\\& & 8& 16& \color{orangered}{30} & & \\ \hline &4&8&15&\color{orangered}{30}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 30 } = \color{blue}{ 60 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&4&0&-1&0&2&3\\& & 8& 16& 30& \color{blue}{60} & \\ \hline &4&8&15&\color{blue}{30}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 60 } = \color{orangered}{ 62 } $
$$ \begin{array}{c|rrrrrr}2&4&0&-1&0&\color{orangered}{ 2 }&3\\& & 8& 16& 30& \color{orangered}{60} & \\ \hline &4&8&15&30&\color{orangered}{62}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 62 } = \color{blue}{ 124 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&4&0&-1&0&2&3\\& & 8& 16& 30& 60& \color{blue}{124} \\ \hline &4&8&15&30&\color{blue}{62}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 124 } = \color{orangered}{ 127 } $
$$ \begin{array}{c|rrrrrr}2&4&0&-1&0&2&\color{orangered}{ 3 }\\& & 8& 16& 30& 60& \color{orangered}{124} \\ \hline &\color{blue}{4}&\color{blue}{8}&\color{blue}{15}&\color{blue}{30}&\color{blue}{62}&\color{orangered}{127} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{4}+8x^{3}+15x^{2}+30x+62 } $ with a remainder of $ \color{red}{ 127 } $.