Divide using synthetic division $$ ( 3x^{4}-48 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&3&0&0&0&-48\\& & -6& 12& -24& \color{black}{48} \\ \hline &\color{blue}{3}&\color{blue}{-6}&\color{blue}{12}&\color{blue}{-24}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 3x^{4}-48 }{ x+2 } = \color{blue}{3x^{3}-6x^{2}+12x-24} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&0&0&0&-48\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 3 }&0&0&0&-48\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 3 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&0&0&0&-48\\& & \color{blue}{-6} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}-2&3&\color{orangered}{ 0 }&0&0&-48\\& & \color{orangered}{-6} & & & \\ \hline &3&\color{orangered}{-6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&0&0&0&-48\\& & -6& \color{blue}{12} & & \\ \hline &3&\color{blue}{-6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}-2&3&0&\color{orangered}{ 0 }&0&-48\\& & -6& \color{orangered}{12} & & \\ \hline &3&-6&\color{orangered}{12}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 12 } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&0&0&0&-48\\& & -6& 12& \color{blue}{-24} & \\ \hline &3&-6&\color{blue}{12}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -24 } $
$$ \begin{array}{c|rrrrr}-2&3&0&0&\color{orangered}{ 0 }&-48\\& & -6& 12& \color{orangered}{-24} & \\ \hline &3&-6&12&\color{orangered}{-24}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -24 \right) } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&0&0&0&-48\\& & -6& 12& -24& \color{blue}{48} \\ \hline &3&-6&12&\color{blue}{-24}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -48 } + \color{orangered}{ 48 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-2&3&0&0&0&\color{orangered}{ -48 }\\& & -6& 12& -24& \color{orangered}{48} \\ \hline &\color{blue}{3}&\color{blue}{-6}&\color{blue}{12}&\color{blue}{-24}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}-6x^{2}+12x-24 } $ with a remainder of $ \color{red}{ 0 } $.