Divide using synthetic division $$ ( 4x^{4}-5x^{2}-12 ) \div ( 2x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&4&0&-5&0&-12\\& & 12& 36& 93& \color{black}{279} \\ \hline &\color{blue}{4}&\color{blue}{12}&\color{blue}{31}&\color{blue}{93}&\color{orangered}{267} \end{array} $$The solution is:
$$ \frac{ 4x^{4}-5x^{2}-12 }{ x-3 } = \color{blue}{4x^{3}+12x^{2}+31x+93} ~+~ \frac{ \color{red}{ 267 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&0&-5&0&-12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 4 }&0&-5&0&-12\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&0&-5&0&-12\\& & \color{blue}{12} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}3&4&\color{orangered}{ 0 }&-5&0&-12\\& & \color{orangered}{12} & & & \\ \hline &4&\color{orangered}{12}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 12 } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&0&-5&0&-12\\& & 12& \color{blue}{36} & & \\ \hline &4&\color{blue}{12}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 36 } = \color{orangered}{ 31 } $
$$ \begin{array}{c|rrrrr}3&4&0&\color{orangered}{ -5 }&0&-12\\& & 12& \color{orangered}{36} & & \\ \hline &4&12&\color{orangered}{31}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 31 } = \color{blue}{ 93 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&0&-5&0&-12\\& & 12& 36& \color{blue}{93} & \\ \hline &4&12&\color{blue}{31}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 93 } = \color{orangered}{ 93 } $
$$ \begin{array}{c|rrrrr}3&4&0&-5&\color{orangered}{ 0 }&-12\\& & 12& 36& \color{orangered}{93} & \\ \hline &4&12&31&\color{orangered}{93}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 93 } = \color{blue}{ 279 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&0&-5&0&-12\\& & 12& 36& 93& \color{blue}{279} \\ \hline &4&12&31&\color{blue}{93}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 279 } = \color{orangered}{ 267 } $
$$ \begin{array}{c|rrrrr}3&4&0&-5&0&\color{orangered}{ -12 }\\& & 12& 36& 93& \color{orangered}{279} \\ \hline &\color{blue}{4}&\color{blue}{12}&\color{blue}{31}&\color{blue}{93}&\color{orangered}{267} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}+12x^{2}+31x+93 } $ with a remainder of $ \color{red}{ 267 } $.