Divide using synthetic division $$ ( 20x^{3}+44x^{2}-17x-5 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&20&44&-17&-5\\& & -100& 280& \color{black}{-1315} \\ \hline &\color{blue}{20}&\color{blue}{-56}&\color{blue}{263}&\color{orangered}{-1320} \end{array} $$The solution is:
$$ \frac{ 20x^{3}+44x^{2}-17x-5 }{ x+5 } = \color{blue}{20x^{2}-56x+263} \color{red}{~-~} \frac{ \color{red}{ 1320 } }{ x+5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 5 = 0 $ ( $ x = \color{blue}{ -5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&20&44&-17&-5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 20 }&44&-17&-5\\& & & & \\ \hline &\color{orangered}{20}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 20 } = \color{blue}{ -100 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&20&44&-17&-5\\& & \color{blue}{-100} & & \\ \hline &\color{blue}{20}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 44 } + \color{orangered}{ \left( -100 \right) } = \color{orangered}{ -56 } $
$$ \begin{array}{c|rrrr}-5&20&\color{orangered}{ 44 }&-17&-5\\& & \color{orangered}{-100} & & \\ \hline &20&\color{orangered}{-56}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -56 \right) } = \color{blue}{ 280 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&20&44&-17&-5\\& & -100& \color{blue}{280} & \\ \hline &20&\color{blue}{-56}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 280 } = \color{orangered}{ 263 } $
$$ \begin{array}{c|rrrr}-5&20&44&\color{orangered}{ -17 }&-5\\& & -100& \color{orangered}{280} & \\ \hline &20&-56&\color{orangered}{263}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 263 } = \color{blue}{ -1315 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&20&44&-17&-5\\& & -100& 280& \color{blue}{-1315} \\ \hline &20&-56&\color{blue}{263}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -1315 \right) } = \color{orangered}{ -1320 } $
$$ \begin{array}{c|rrrr}-5&20&44&-17&\color{orangered}{ -5 }\\& & -100& 280& \color{orangered}{-1315} \\ \hline &\color{blue}{20}&\color{blue}{-56}&\color{blue}{263}&\color{orangered}{-1320} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 20x^{2}-56x+263 } $ with a remainder of $ \color{red}{ -1320 } $.