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