Divide using synthetic division $$ ( 4x^{3}+23x^{2}+34x-10 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&4&23&34&-10\\& & 12& 105& \color{black}{417} \\ \hline &\color{blue}{4}&\color{blue}{35}&\color{blue}{139}&\color{orangered}{407} \end{array} $$The solution is:
$$ \frac{ 4x^{3}+23x^{2}+34x-10 }{ x-3 } = \color{blue}{4x^{2}+35x+139} ~+~ \frac{ \color{red}{ 407 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&23&34&-10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 4 }&23&34&-10\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&23&34&-10\\& & \color{blue}{12} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 23 } + \color{orangered}{ 12 } = \color{orangered}{ 35 } $
$$ \begin{array}{c|rrrr}3&4&\color{orangered}{ 23 }&34&-10\\& & \color{orangered}{12} & & \\ \hline &4&\color{orangered}{35}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 35 } = \color{blue}{ 105 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&23&34&-10\\& & 12& \color{blue}{105} & \\ \hline &4&\color{blue}{35}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 34 } + \color{orangered}{ 105 } = \color{orangered}{ 139 } $
$$ \begin{array}{c|rrrr}3&4&23&\color{orangered}{ 34 }&-10\\& & 12& \color{orangered}{105} & \\ \hline &4&35&\color{orangered}{139}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 139 } = \color{blue}{ 417 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&23&34&-10\\& & 12& 105& \color{blue}{417} \\ \hline &4&35&\color{blue}{139}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 417 } = \color{orangered}{ 407 } $
$$ \begin{array}{c|rrrr}3&4&23&34&\color{orangered}{ -10 }\\& & 12& 105& \color{orangered}{417} \\ \hline &\color{blue}{4}&\color{blue}{35}&\color{blue}{139}&\color{orangered}{407} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+35x+139 } $ with a remainder of $ \color{red}{ 407 } $.