Divide using synthetic division $$ ( 4x^{3}-1 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&4&0&0&-1\\& & 8& 16& \color{black}{32} \\ \hline &\color{blue}{4}&\color{blue}{8}&\color{blue}{16}&\color{orangered}{31} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-1 }{ x-2 } = \color{blue}{4x^{2}+8x+16} ~+~ \frac{ \color{red}{ 31 } }{ x-2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{2}&4&0&0&-1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 4 }&0&0&-1\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 4 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&4&0&0&-1\\& & \color{blue}{8} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 8 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}2&4&\color{orangered}{ 0 }&0&-1\\& & \color{orangered}{8} & & \\ \hline &4&\color{orangered}{8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 8 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&4&0&0&-1\\& & 8& \color{blue}{16} & \\ \hline &4&\color{blue}{8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 16 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrr}2&4&0&\color{orangered}{ 0 }&-1\\& & 8& \color{orangered}{16} & \\ \hline &4&8&\color{orangered}{16}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 16 } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&4&0&0&-1\\& & 8& 16& \color{blue}{32} \\ \hline &4&8&\color{blue}{16}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 32 } = \color{orangered}{ 31 } $
$$ \begin{array}{c|rrrr}2&4&0&0&\color{orangered}{ -1 }\\& & 8& 16& \color{orangered}{32} \\ \hline &\color{blue}{4}&\color{blue}{8}&\color{blue}{16}&\color{orangered}{31} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+8x+16 } $ with a remainder of $ \color{red}{ 31 } $.