Divide using synthetic division $$ ( x^{4}+3x^{3}-7x^{2}-15x+18 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&1&3&-7&-15&18\\& & 1& 4& -3& \color{black}{-18} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{-3}&\color{blue}{-18}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{4}+3x^{3}-7x^{2}-15x+18 }{ x-1 } = \color{blue}{x^{3}+4x^{2}-3x-18} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&3&-7&-15&18\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 1 }&3&-7&-15&18\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 1 } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&3&-7&-15&18\\& & \color{blue}{1} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 1 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}1&1&\color{orangered}{ 3 }&-7&-15&18\\& & \color{orangered}{1} & & & \\ \hline &1&\color{orangered}{4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 4 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&3&-7&-15&18\\& & 1& \color{blue}{4} & & \\ \hline &1&\color{blue}{4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 4 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}1&1&3&\color{orangered}{ -7 }&-15&18\\& & 1& \color{orangered}{4} & & \\ \hline &1&4&\color{orangered}{-3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&3&-7&-15&18\\& & 1& 4& \color{blue}{-3} & \\ \hline &1&4&\color{blue}{-3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -18 } $
$$ \begin{array}{c|rrrrr}1&1&3&-7&\color{orangered}{ -15 }&18\\& & 1& 4& \color{orangered}{-3} & \\ \hline &1&4&-3&\color{orangered}{-18}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -18 \right) } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&3&-7&-15&18\\& & 1& 4& -3& \color{blue}{-18} \\ \hline &1&4&-3&\color{blue}{-18}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}1&1&3&-7&-15&\color{orangered}{ 18 }\\& & 1& 4& -3& \color{orangered}{-18} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{-3}&\color{blue}{-18}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+4x^{2}-3x-18 } $ with a remainder of $ \color{red}{ 0 } $.