Divide using synthetic division $$ ( 4x^{3}-2x-5 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&4&0&-2&-5\\& & 12& 36& \color{black}{102} \\ \hline &\color{blue}{4}&\color{blue}{12}&\color{blue}{34}&\color{orangered}{97} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-2x-5 }{ x-3 } = \color{blue}{4x^{2}+12x+34} ~+~ \frac{ \color{red}{ 97 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&0&-2&-5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 4 }&0&-2&-5\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&0&-2&-5\\& & \color{blue}{12} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrr}3&4&\color{orangered}{ 0 }&-2&-5\\& & \color{orangered}{12} & & \\ \hline &4&\color{orangered}{12}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&0&-2&-5\\& & 12& \color{blue}{36} & \\ \hline &4&\color{blue}{12}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 36 } = \color{orangered}{ 34 } $
$$ \begin{array}{c|rrrr}3&4&0&\color{orangered}{ -2 }&-5\\& & 12& \color{orangered}{36} & \\ \hline &4&12&\color{orangered}{34}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 34 } = \color{blue}{ 102 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&0&-2&-5\\& & 12& 36& \color{blue}{102} \\ \hline &4&12&\color{blue}{34}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 102 } = \color{orangered}{ 97 } $
$$ \begin{array}{c|rrrr}3&4&0&-2&\color{orangered}{ -5 }\\& & 12& 36& \color{orangered}{102} \\ \hline &\color{blue}{4}&\color{blue}{12}&\color{blue}{34}&\color{orangered}{97} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+12x+34 } $ with a remainder of $ \color{red}{ 97 } $.