Question

Simplify completely:

Answer 1

\[8x^3+6x^2-4x \over -2x\]

Answer 2

\[\Large \frac{8x^3+6x^2-4x}{-2x}\]


\[\Large \frac{-2x(-4x^2-3x+2)}{-2x}\]


\[\Large \frac{\cancel{-2x}(-4x^2-3x+2)}{\cancel{-2x}}\]


\[\Large -4x^2-3x+2\]



So \[\Large \frac{8x^3+6x^2-4x}{-2x}=-4x^2-3x+2\] for all values of x where x is not zero.

Answer 3

thanks you are a lifesaver

Answer 4

yw