Divide using synthetic division $$ ( 46x^{3}+11x^{2}-30x+15 ) \div ( 2x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&46&11&-30&15\\& & 138& 447& \color{black}{1251} \\ \hline &\color{blue}{46}&\color{blue}{149}&\color{blue}{417}&\color{orangered}{1266} \end{array} $$The solution is:
$$ \frac{ 46x^{3}+11x^{2}-30x+15 }{ x-3 } = \color{blue}{46x^{2}+149x+417} ~+~ \frac{ \color{red}{ 1266 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&46&11&-30&15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 46 }&11&-30&15\\& & & & \\ \hline &\color{orangered}{46}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 46 } = \color{blue}{ 138 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&46&11&-30&15\\& & \color{blue}{138} & & \\ \hline &\color{blue}{46}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ 138 } = \color{orangered}{ 149 } $
$$ \begin{array}{c|rrrr}3&46&\color{orangered}{ 11 }&-30&15\\& & \color{orangered}{138} & & \\ \hline &46&\color{orangered}{149}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 149 } = \color{blue}{ 447 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&46&11&-30&15\\& & 138& \color{blue}{447} & \\ \hline &46&\color{blue}{149}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -30 } + \color{orangered}{ 447 } = \color{orangered}{ 417 } $
$$ \begin{array}{c|rrrr}3&46&11&\color{orangered}{ -30 }&15\\& & 138& \color{orangered}{447} & \\ \hline &46&149&\color{orangered}{417}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 417 } = \color{blue}{ 1251 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&46&11&-30&15\\& & 138& 447& \color{blue}{1251} \\ \hline &46&149&\color{blue}{417}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ 1251 } = \color{orangered}{ 1266 } $
$$ \begin{array}{c|rrrr}3&46&11&-30&\color{orangered}{ 15 }\\& & 138& 447& \color{orangered}{1251} \\ \hline &\color{blue}{46}&\color{blue}{149}&\color{blue}{417}&\color{orangered}{1266} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 46x^{2}+149x+417 } $ with a remainder of $ \color{red}{ 1266 } $.