Divide using synthetic division $$ ( 6x^{3}+x^{2}-12x+5 ) \div ( 3x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&6&1&-12&5\\& & 24& 100& \color{black}{352} \\ \hline &\color{blue}{6}&\color{blue}{25}&\color{blue}{88}&\color{orangered}{357} \end{array} $$The solution is:
$$ \frac{ 6x^{3}+x^{2}-12x+5 }{ x-4 } = \color{blue}{6x^{2}+25x+88} ~+~ \frac{ \color{red}{ 357 } }{ x-4 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{4}&6&1&-12&5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 6 }&1&-12&5\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 6 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&6&1&-12&5\\& & \color{blue}{24} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 24 } = \color{orangered}{ 25 } $
$$ \begin{array}{c|rrrr}4&6&\color{orangered}{ 1 }&-12&5\\& & \color{orangered}{24} & & \\ \hline &6&\color{orangered}{25}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 25 } = \color{blue}{ 100 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&6&1&-12&5\\& & 24& \color{blue}{100} & \\ \hline &6&\color{blue}{25}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 100 } = \color{orangered}{ 88 } $
$$ \begin{array}{c|rrrr}4&6&1&\color{orangered}{ -12 }&5\\& & 24& \color{orangered}{100} & \\ \hline &6&25&\color{orangered}{88}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 88 } = \color{blue}{ 352 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&6&1&-12&5\\& & 24& 100& \color{blue}{352} \\ \hline &6&25&\color{blue}{88}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 352 } = \color{orangered}{ 357 } $
$$ \begin{array}{c|rrrr}4&6&1&-12&\color{orangered}{ 5 }\\& & 24& 100& \color{orangered}{352} \\ \hline &\color{blue}{6}&\color{blue}{25}&\color{blue}{88}&\color{orangered}{357} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}+25x+88 } $ with a remainder of $ \color{red}{ 357 } $.