Divide using synthetic division $$ ( 4x^{4}+52x^{3}+87x^{2}-13x-22 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&4&52&87&-13&-22\\& & 4& 56& 143& \color{black}{130} \\ \hline &\color{blue}{4}&\color{blue}{56}&\color{blue}{143}&\color{blue}{130}&\color{orangered}{108} \end{array} $$The solution is:
$$ \frac{ 4x^{4}+52x^{3}+87x^{2}-13x-22 }{ x-1 } = \color{blue}{4x^{3}+56x^{2}+143x+130} ~+~ \frac{ \color{red}{ 108 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&52&87&-13&-22\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 4 }&52&87&-13&-22\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 4 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&52&87&-13&-22\\& & \color{blue}{4} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 52 } + \color{orangered}{ 4 } = \color{orangered}{ 56 } $
$$ \begin{array}{c|rrrrr}1&4&\color{orangered}{ 52 }&87&-13&-22\\& & \color{orangered}{4} & & & \\ \hline &4&\color{orangered}{56}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 56 } = \color{blue}{ 56 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&52&87&-13&-22\\& & 4& \color{blue}{56} & & \\ \hline &4&\color{blue}{56}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 87 } + \color{orangered}{ 56 } = \color{orangered}{ 143 } $
$$ \begin{array}{c|rrrrr}1&4&52&\color{orangered}{ 87 }&-13&-22\\& & 4& \color{orangered}{56} & & \\ \hline &4&56&\color{orangered}{143}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 143 } = \color{blue}{ 143 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&52&87&-13&-22\\& & 4& 56& \color{blue}{143} & \\ \hline &4&56&\color{blue}{143}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 143 } = \color{orangered}{ 130 } $
$$ \begin{array}{c|rrrrr}1&4&52&87&\color{orangered}{ -13 }&-22\\& & 4& 56& \color{orangered}{143} & \\ \hline &4&56&143&\color{orangered}{130}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 130 } = \color{blue}{ 130 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&52&87&-13&-22\\& & 4& 56& 143& \color{blue}{130} \\ \hline &4&56&143&\color{blue}{130}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -22 } + \color{orangered}{ 130 } = \color{orangered}{ 108 } $
$$ \begin{array}{c|rrrrr}1&4&52&87&-13&\color{orangered}{ -22 }\\& & 4& 56& 143& \color{orangered}{130} \\ \hline &\color{blue}{4}&\color{blue}{56}&\color{blue}{143}&\color{blue}{130}&\color{orangered}{108} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}+56x^{2}+143x+130 } $ with a remainder of $ \color{red}{ 108 } $.