Divide using synthetic division $$ ( 3x^{3}+4x^{2}-27x-36 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&3&4&-27&-36\\& & 12& 64& \color{black}{148} \\ \hline &\color{blue}{3}&\color{blue}{16}&\color{blue}{37}&\color{orangered}{112} \end{array} $$The solution is:
$$ \frac{ 3x^{3}+4x^{2}-27x-36 }{ x-4 } = \color{blue}{3x^{2}+16x+37} ~+~ \frac{ \color{red}{ 112 } }{ x-4 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&4&-27&-36\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 3 }&4&-27&-36\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 3 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&4&-27&-36\\& & \color{blue}{12} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 12 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrr}4&3&\color{orangered}{ 4 }&-27&-36\\& & \color{orangered}{12} & & \\ \hline &3&\color{orangered}{16}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 16 } = \color{blue}{ 64 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&4&-27&-36\\& & 12& \color{blue}{64} & \\ \hline &3&\color{blue}{16}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -27 } + \color{orangered}{ 64 } = \color{orangered}{ 37 } $
$$ \begin{array}{c|rrrr}4&3&4&\color{orangered}{ -27 }&-36\\& & 12& \color{orangered}{64} & \\ \hline &3&16&\color{orangered}{37}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 37 } = \color{blue}{ 148 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&4&-27&-36\\& & 12& 64& \color{blue}{148} \\ \hline &3&16&\color{blue}{37}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -36 } + \color{orangered}{ 148 } = \color{orangered}{ 112 } $
$$ \begin{array}{c|rrrr}4&3&4&-27&\color{orangered}{ -36 }\\& & 12& 64& \color{orangered}{148} \\ \hline &\color{blue}{3}&\color{blue}{16}&\color{blue}{37}&\color{orangered}{112} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}+16x+37 } $ with a remainder of $ \color{red}{ 112 } $.