Divide using synthetic division $$ ( 3x^{4}+2x^{3}-15x^{2}+10 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&3&2&-15&0&10\\& & 3& 5& -10& \color{black}{-10} \\ \hline &\color{blue}{3}&\color{blue}{5}&\color{blue}{-10}&\color{blue}{-10}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 3x^{4}+2x^{3}-15x^{2}+10 }{ x-1 } = \color{blue}{3x^{3}+5x^{2}-10x-10} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&3&2&-15&0&10\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 3 }&2&-15&0&10\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 3 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&3&2&-15&0&10\\& & \color{blue}{3} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 3 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}1&3&\color{orangered}{ 2 }&-15&0&10\\& & \color{orangered}{3} & & & \\ \hline &3&\color{orangered}{5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 5 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&3&2&-15&0&10\\& & 3& \color{blue}{5} & & \\ \hline &3&\color{blue}{5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 5 } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}1&3&2&\color{orangered}{ -15 }&0&10\\& & 3& \color{orangered}{5} & & \\ \hline &3&5&\color{orangered}{-10}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&3&2&-15&0&10\\& & 3& 5& \color{blue}{-10} & \\ \hline &3&5&\color{blue}{-10}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}1&3&2&-15&\color{orangered}{ 0 }&10\\& & 3& 5& \color{orangered}{-10} & \\ \hline &3&5&-10&\color{orangered}{-10}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&3&2&-15&0&10\\& & 3& 5& -10& \color{blue}{-10} \\ \hline &3&5&-10&\color{blue}{-10}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}1&3&2&-15&0&\color{orangered}{ 10 }\\& & 3& 5& -10& \color{orangered}{-10} \\ \hline &\color{blue}{3}&\color{blue}{5}&\color{blue}{-10}&\color{blue}{-10}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}+5x^{2}-10x-10 } $ with a remainder of $ \color{red}{ 0 } $.