Divide using synthetic division $$ ( -6x^{3}+35x^{2}-50x-58 ) \div ( 5x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&-6&35&-50&-58\\& & 12& -94& \color{black}{288} \\ \hline &\color{blue}{-6}&\color{blue}{47}&\color{blue}{-144}&\color{orangered}{230} \end{array} $$The solution is:
$$ \frac{ -6x^{3}+35x^{2}-50x-58 }{ x+2 } = \color{blue}{-6x^{2}+47x-144} ~+~ \frac{ \color{red}{ 230 } }{ x+2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&-6&35&-50&-58\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ -6 }&35&-50&-58\\& & & & \\ \hline &\color{orangered}{-6}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&-6&35&-50&-58\\& & \color{blue}{12} & & \\ \hline &\color{blue}{-6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 35 } + \color{orangered}{ 12 } = \color{orangered}{ 47 } $
$$ \begin{array}{c|rrrr}-2&-6&\color{orangered}{ 35 }&-50&-58\\& & \color{orangered}{12} & & \\ \hline &-6&\color{orangered}{47}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 47 } = \color{blue}{ -94 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&-6&35&-50&-58\\& & 12& \color{blue}{-94} & \\ \hline &-6&\color{blue}{47}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -50 } + \color{orangered}{ \left( -94 \right) } = \color{orangered}{ -144 } $
$$ \begin{array}{c|rrrr}-2&-6&35&\color{orangered}{ -50 }&-58\\& & 12& \color{orangered}{-94} & \\ \hline &-6&47&\color{orangered}{-144}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -144 \right) } = \color{blue}{ 288 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&-6&35&-50&-58\\& & 12& -94& \color{blue}{288} \\ \hline &-6&47&\color{blue}{-144}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -58 } + \color{orangered}{ 288 } = \color{orangered}{ 230 } $
$$ \begin{array}{c|rrrr}-2&-6&35&-50&\color{orangered}{ -58 }\\& & 12& -94& \color{orangered}{288} \\ \hline &\color{blue}{-6}&\color{blue}{47}&\color{blue}{-144}&\color{orangered}{230} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -6x^{2}+47x-144 } $ with a remainder of $ \color{red}{ 230 } $.