Divide using synthetic division $$ ( 12x^{4}-22x^{3}+23x^{2}-21x+8 ) \div ( 3x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&12&-22&23&-21&8\\& & 48& 104& 508& \color{black}{1948} \\ \hline &\color{blue}{12}&\color{blue}{26}&\color{blue}{127}&\color{blue}{487}&\color{orangered}{1956} \end{array} $$The solution is:
$$ \frac{ 12x^{4}-22x^{3}+23x^{2}-21x+8 }{ x-4 } = \color{blue}{12x^{3}+26x^{2}+127x+487} ~+~ \frac{ \color{red}{ 1956 } }{ x-4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&12&-22&23&-21&8\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ 12 }&-22&23&-21&8\\& & & & & \\ \hline &\color{orangered}{12}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 12 } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&12&-22&23&-21&8\\& & \color{blue}{48} & & & \\ \hline &\color{blue}{12}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -22 } + \color{orangered}{ 48 } = \color{orangered}{ 26 } $
$$ \begin{array}{c|rrrrr}4&12&\color{orangered}{ -22 }&23&-21&8\\& & \color{orangered}{48} & & & \\ \hline &12&\color{orangered}{26}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 26 } = \color{blue}{ 104 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&12&-22&23&-21&8\\& & 48& \color{blue}{104} & & \\ \hline &12&\color{blue}{26}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 23 } + \color{orangered}{ 104 } = \color{orangered}{ 127 } $
$$ \begin{array}{c|rrrrr}4&12&-22&\color{orangered}{ 23 }&-21&8\\& & 48& \color{orangered}{104} & & \\ \hline &12&26&\color{orangered}{127}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 127 } = \color{blue}{ 508 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&12&-22&23&-21&8\\& & 48& 104& \color{blue}{508} & \\ \hline &12&26&\color{blue}{127}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ 508 } = \color{orangered}{ 487 } $
$$ \begin{array}{c|rrrrr}4&12&-22&23&\color{orangered}{ -21 }&8\\& & 48& 104& \color{orangered}{508} & \\ \hline &12&26&127&\color{orangered}{487}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 487 } = \color{blue}{ 1948 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&12&-22&23&-21&8\\& & 48& 104& 508& \color{blue}{1948} \\ \hline &12&26&127&\color{blue}{487}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 1948 } = \color{orangered}{ 1956 } $
$$ \begin{array}{c|rrrrr}4&12&-22&23&-21&\color{orangered}{ 8 }\\& & 48& 104& 508& \color{orangered}{1948} \\ \hline &\color{blue}{12}&\color{blue}{26}&\color{blue}{127}&\color{blue}{487}&\color{orangered}{1956} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 12x^{3}+26x^{2}+127x+487 } $ with a remainder of $ \color{red}{ 1956 } $.