Divide using synthetic division $$ ( 12x^{3}+38x^{2}-16x-20 ) \div ( 6x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&12&38&-16&-20\\& & 60& 490& \color{black}{2370} \\ \hline &\color{blue}{12}&\color{blue}{98}&\color{blue}{474}&\color{orangered}{2350} \end{array} $$The solution is:
$$ \frac{ 12x^{3}+38x^{2}-16x-20 }{ x-5 } = \color{blue}{12x^{2}+98x+474} ~+~ \frac{ \color{red}{ 2350 } }{ x-5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{5}&12&38&-16&-20\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 12 }&38&-16&-20\\& & & & \\ \hline &\color{orangered}{12}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 12 } = \color{blue}{ 60 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&12&38&-16&-20\\& & \color{blue}{60} & & \\ \hline &\color{blue}{12}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 38 } + \color{orangered}{ 60 } = \color{orangered}{ 98 } $
$$ \begin{array}{c|rrrr}5&12&\color{orangered}{ 38 }&-16&-20\\& & \color{orangered}{60} & & \\ \hline &12&\color{orangered}{98}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 98 } = \color{blue}{ 490 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&12&38&-16&-20\\& & 60& \color{blue}{490} & \\ \hline &12&\color{blue}{98}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 490 } = \color{orangered}{ 474 } $
$$ \begin{array}{c|rrrr}5&12&38&\color{orangered}{ -16 }&-20\\& & 60& \color{orangered}{490} & \\ \hline &12&98&\color{orangered}{474}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 474 } = \color{blue}{ 2370 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&12&38&-16&-20\\& & 60& 490& \color{blue}{2370} \\ \hline &12&98&\color{blue}{474}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 2370 } = \color{orangered}{ 2350 } $
$$ \begin{array}{c|rrrr}5&12&38&-16&\color{orangered}{ -20 }\\& & 60& 490& \color{orangered}{2370} \\ \hline &\color{blue}{12}&\color{blue}{98}&\color{blue}{474}&\color{orangered}{2350} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 12x^{2}+98x+474 } $ with a remainder of $ \color{red}{ 2350 } $.