Divide using synthetic division $$ ( 4x^{3}-10x^{2}+x+8 ) \div ( 2x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&4&-10&1&8\\& & 12& 6& \color{black}{21} \\ \hline &\color{blue}{4}&\color{blue}{2}&\color{blue}{7}&\color{orangered}{29} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-10x^{2}+x+8 }{ x-3 } = \color{blue}{4x^{2}+2x+7} ~+~ \frac{ \color{red}{ 29 } }{ x-3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-10&1&8\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 4 }&-10&1&8\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-10&1&8\\& & \color{blue}{12} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 12 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}3&4&\color{orangered}{ -10 }&1&8\\& & \color{orangered}{12} & & \\ \hline &4&\color{orangered}{2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-10&1&8\\& & 12& \color{blue}{6} & \\ \hline &4&\color{blue}{2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 6 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}3&4&-10&\color{orangered}{ 1 }&8\\& & 12& \color{orangered}{6} & \\ \hline &4&2&\color{orangered}{7}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 7 } = \color{blue}{ 21 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-10&1&8\\& & 12& 6& \color{blue}{21} \\ \hline &4&2&\color{blue}{7}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 21 } = \color{orangered}{ 29 } $
$$ \begin{array}{c|rrrr}3&4&-10&1&\color{orangered}{ 8 }\\& & 12& 6& \color{orangered}{21} \\ \hline &\color{blue}{4}&\color{blue}{2}&\color{blue}{7}&\color{orangered}{29} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+2x+7 } $ with a remainder of $ \color{red}{ 29 } $.