Divide using synthetic division $$ ( 2x^{4}-5x^{3}-35x^{2}+80x+48 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&2&-5&-35&80&48\\& & 6& 3& -96& \color{black}{-48} \\ \hline &\color{blue}{2}&\color{blue}{1}&\color{blue}{-32}&\color{blue}{-16}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-5x^{3}-35x^{2}+80x+48 }{ x-3 } = \color{blue}{2x^{3}+x^{2}-32x-16} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-5&-35&80&48\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 2 }&-5&-35&80&48\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-5&-35&80&48\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 6 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}3&2&\color{orangered}{ -5 }&-35&80&48\\& & \color{orangered}{6} & & & \\ \hline &2&\color{orangered}{1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-5&-35&80&48\\& & 6& \color{blue}{3} & & \\ \hline &2&\color{blue}{1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -35 } + \color{orangered}{ 3 } = \color{orangered}{ -32 } $
$$ \begin{array}{c|rrrrr}3&2&-5&\color{orangered}{ -35 }&80&48\\& & 6& \color{orangered}{3} & & \\ \hline &2&1&\color{orangered}{-32}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -32 \right) } = \color{blue}{ -96 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-5&-35&80&48\\& & 6& 3& \color{blue}{-96} & \\ \hline &2&1&\color{blue}{-32}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 80 } + \color{orangered}{ \left( -96 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrrr}3&2&-5&-35&\color{orangered}{ 80 }&48\\& & 6& 3& \color{orangered}{-96} & \\ \hline &2&1&-32&\color{orangered}{-16}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ -48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-5&-35&80&48\\& & 6& 3& -96& \color{blue}{-48} \\ \hline &2&1&-32&\color{blue}{-16}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 48 } + \color{orangered}{ \left( -48 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}3&2&-5&-35&80&\color{orangered}{ 48 }\\& & 6& 3& -96& \color{orangered}{-48} \\ \hline &\color{blue}{2}&\color{blue}{1}&\color{blue}{-32}&\color{blue}{-16}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+x^{2}-32x-16 } $ with a remainder of $ \color{red}{ 0 } $.