Divide using synthetic division $$ ( 4x^{3}+9x^{2}-58x-15 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&4&9&-58&-15\\& & -20& 55& \color{black}{15} \\ \hline &\color{blue}{4}&\color{blue}{-11}&\color{blue}{-3}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 4x^{3}+9x^{2}-58x-15 }{ x+5 } = \color{blue}{4x^{2}-11x-3} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&4&9&-58&-15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 4 }&9&-58&-15\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 4 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&4&9&-58&-15\\& & \color{blue}{-20} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -11 } $
$$ \begin{array}{c|rrrr}-5&4&\color{orangered}{ 9 }&-58&-15\\& & \color{orangered}{-20} & & \\ \hline &4&\color{orangered}{-11}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -11 \right) } = \color{blue}{ 55 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&4&9&-58&-15\\& & -20& \color{blue}{55} & \\ \hline &4&\color{blue}{-11}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -58 } + \color{orangered}{ 55 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}-5&4&9&\color{orangered}{ -58 }&-15\\& & -20& \color{orangered}{55} & \\ \hline &4&-11&\color{orangered}{-3}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&4&9&-58&-15\\& & -20& 55& \color{blue}{15} \\ \hline &4&-11&\color{blue}{-3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 15 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-5&4&9&-58&\color{orangered}{ -15 }\\& & -20& 55& \color{orangered}{15} \\ \hline &\color{blue}{4}&\color{blue}{-11}&\color{blue}{-3}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-11x-3 } $ with a remainder of $ \color{red}{ 0 } $.