Divide using synthetic division $$ ( 8x^{3}-14x^{2}+x ) \div ( 2x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&8&-14&1&0\\& & 24& 30& \color{black}{93} \\ \hline &\color{blue}{8}&\color{blue}{10}&\color{blue}{31}&\color{orangered}{93} \end{array} $$The solution is:
$$ \frac{ 8x^{3}-14x^{2}+x }{ x-3 } = \color{blue}{8x^{2}+10x+31} ~+~ \frac{ \color{red}{ 93 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&8&-14&1&0\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 8 }&-14&1&0\\& & & & \\ \hline &\color{orangered}{8}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 8 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&8&-14&1&0\\& & \color{blue}{24} & & \\ \hline &\color{blue}{8}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 24 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}3&8&\color{orangered}{ -14 }&1&0\\& & \color{orangered}{24} & & \\ \hline &8&\color{orangered}{10}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 10 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&8&-14&1&0\\& & 24& \color{blue}{30} & \\ \hline &8&\color{blue}{10}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 30 } = \color{orangered}{ 31 } $
$$ \begin{array}{c|rrrr}3&8&-14&\color{orangered}{ 1 }&0\\& & 24& \color{orangered}{30} & \\ \hline &8&10&\color{orangered}{31}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 31 } = \color{blue}{ 93 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&8&-14&1&0\\& & 24& 30& \color{blue}{93} \\ \hline &8&10&\color{blue}{31}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 93 } = \color{orangered}{ 93 } $
$$ \begin{array}{c|rrrr}3&8&-14&1&\color{orangered}{ 0 }\\& & 24& 30& \color{orangered}{93} \\ \hline &\color{blue}{8}&\color{blue}{10}&\color{blue}{31}&\color{orangered}{93} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 8x^{2}+10x+31 } $ with a remainder of $ \color{red}{ 93 } $.