Divide using synthetic division $$ ( 16x^{3}+58x^{2}+39x-17 ) \div ( 2x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&16&58&39&-17\\& & -80& 110& \color{black}{-745} \\ \hline &\color{blue}{16}&\color{blue}{-22}&\color{blue}{149}&\color{orangered}{-762} \end{array} $$The solution is:
$$ \frac{ 16x^{3}+58x^{2}+39x-17 }{ x+5 } = \color{blue}{16x^{2}-22x+149} \color{red}{~-~} \frac{ \color{red}{ 762 } }{ x+5 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&16&58&39&-17\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 16 }&58&39&-17\\& & & & \\ \hline &\color{orangered}{16}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 16 } = \color{blue}{ -80 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&16&58&39&-17\\& & \color{blue}{-80} & & \\ \hline &\color{blue}{16}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 58 } + \color{orangered}{ \left( -80 \right) } = \color{orangered}{ -22 } $
$$ \begin{array}{c|rrrr}-5&16&\color{orangered}{ 58 }&39&-17\\& & \color{orangered}{-80} & & \\ \hline &16&\color{orangered}{-22}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -22 \right) } = \color{blue}{ 110 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&16&58&39&-17\\& & -80& \color{blue}{110} & \\ \hline &16&\color{blue}{-22}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 39 } + \color{orangered}{ 110 } = \color{orangered}{ 149 } $
$$ \begin{array}{c|rrrr}-5&16&58&\color{orangered}{ 39 }&-17\\& & -80& \color{orangered}{110} & \\ \hline &16&-22&\color{orangered}{149}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 149 } = \color{blue}{ -745 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&16&58&39&-17\\& & -80& 110& \color{blue}{-745} \\ \hline &16&-22&\color{blue}{149}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ \left( -745 \right) } = \color{orangered}{ -762 } $
$$ \begin{array}{c|rrrr}-5&16&58&39&\color{orangered}{ -17 }\\& & -80& 110& \color{orangered}{-745} \\ \hline &\color{blue}{16}&\color{blue}{-22}&\color{blue}{149}&\color{orangered}{-762} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 16x^{2}-22x+149 } $ with a remainder of $ \color{red}{ -762 } $.