Divide using synthetic division $$ ( 3x^{4}+10x^{3}-7x^{2}+4x ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&3&10&-7&4&0\\& & 9& 57& 150& \color{black}{462} \\ \hline &\color{blue}{3}&\color{blue}{19}&\color{blue}{50}&\color{blue}{154}&\color{orangered}{462} \end{array} $$The solution is:
$$ \frac{ 3x^{4}+10x^{3}-7x^{2}+4x }{ x-3 } = \color{blue}{3x^{3}+19x^{2}+50x+154} ~+~ \frac{ \color{red}{ 462 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&3&10&-7&4&0\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 3 }&10&-7&4&0\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&3&10&-7&4&0\\& & \color{blue}{9} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ 9 } = \color{orangered}{ 19 } $
$$ \begin{array}{c|rrrrr}3&3&\color{orangered}{ 10 }&-7&4&0\\& & \color{orangered}{9} & & & \\ \hline &3&\color{orangered}{19}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 19 } = \color{blue}{ 57 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&3&10&-7&4&0\\& & 9& \color{blue}{57} & & \\ \hline &3&\color{blue}{19}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 57 } = \color{orangered}{ 50 } $
$$ \begin{array}{c|rrrrr}3&3&10&\color{orangered}{ -7 }&4&0\\& & 9& \color{orangered}{57} & & \\ \hline &3&19&\color{orangered}{50}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 50 } = \color{blue}{ 150 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&3&10&-7&4&0\\& & 9& 57& \color{blue}{150} & \\ \hline &3&19&\color{blue}{50}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 150 } = \color{orangered}{ 154 } $
$$ \begin{array}{c|rrrrr}3&3&10&-7&\color{orangered}{ 4 }&0\\& & 9& 57& \color{orangered}{150} & \\ \hline &3&19&50&\color{orangered}{154}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 154 } = \color{blue}{ 462 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&3&10&-7&4&0\\& & 9& 57& 150& \color{blue}{462} \\ \hline &3&19&50&\color{blue}{154}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 462 } = \color{orangered}{ 462 } $
$$ \begin{array}{c|rrrrr}3&3&10&-7&4&\color{orangered}{ 0 }\\& & 9& 57& 150& \color{orangered}{462} \\ \hline &\color{blue}{3}&\color{blue}{19}&\color{blue}{50}&\color{blue}{154}&\color{orangered}{462} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}+19x^{2}+50x+154 } $ with a remainder of $ \color{red}{ 462 } $.