Divide using synthetic division $$ ( 2x^{4}-17x^{3}+35x^{2}+9x-45 ) \div ( x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-1&2&-17&35&9&-45\\& & -2& 19& -54& \color{black}{45} \\ \hline &\color{blue}{2}&\color{blue}{-19}&\color{blue}{54}&\color{blue}{-45}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-17x^{3}+35x^{2}+9x-45 }{ x+1 } = \color{blue}{2x^{3}-19x^{2}+54x-45} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&-17&35&9&-45\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-1&\color{orangered}{ 2 }&-17&35&9&-45\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 2 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&-17&35&9&-45\\& & \color{blue}{-2} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -19 } $
$$ \begin{array}{c|rrrrr}-1&2&\color{orangered}{ -17 }&35&9&-45\\& & \color{orangered}{-2} & & & \\ \hline &2&\color{orangered}{-19}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -19 \right) } = \color{blue}{ 19 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&-17&35&9&-45\\& & -2& \color{blue}{19} & & \\ \hline &2&\color{blue}{-19}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 35 } + \color{orangered}{ 19 } = \color{orangered}{ 54 } $
$$ \begin{array}{c|rrrrr}-1&2&-17&\color{orangered}{ 35 }&9&-45\\& & -2& \color{orangered}{19} & & \\ \hline &2&-19&\color{orangered}{54}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 54 } = \color{blue}{ -54 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&-17&35&9&-45\\& & -2& 19& \color{blue}{-54} & \\ \hline &2&-19&\color{blue}{54}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -54 \right) } = \color{orangered}{ -45 } $
$$ \begin{array}{c|rrrrr}-1&2&-17&35&\color{orangered}{ 9 }&-45\\& & -2& 19& \color{orangered}{-54} & \\ \hline &2&-19&54&\color{orangered}{-45}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -45 \right) } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&-17&35&9&-45\\& & -2& 19& -54& \color{blue}{45} \\ \hline &2&-19&54&\color{blue}{-45}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -45 } + \color{orangered}{ 45 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-1&2&-17&35&9&\color{orangered}{ -45 }\\& & -2& 19& -54& \color{orangered}{45} \\ \hline &\color{blue}{2}&\color{blue}{-19}&\color{blue}{54}&\color{blue}{-45}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-19x^{2}+54x-45 } $ with a remainder of $ \color{red}{ 0 } $.