Answer: D
Step-by-step explanation:
To find the difference, we want to make sure both denominators are equal.
[tex]\frac{(6x)(2x-3)}{(6x)(3x^2)} -\frac{(3x^2)(x+2)}{(3x^2)(6x)}[/tex]
Now that the denominators are equal, we can distribute and multiply out.
[tex]\frac{12x^2-18x}{18x^3} -\frac{3x^3+6x^2}{18x^3}[/tex]
With the fractions multiplied out, we can actually put it into one large fraction.
[tex]\frac{12x^2-18x-3x^3-6x^2}{18x^3}[/tex]
Let's subtract the top expression.
[tex]\frac{-3x^3+6x^2-18x}{18x^3}[/tex]
This may seem like our final answer, but we can actually factor out an 3x in the numerator and denominator.
[tex]\frac{(3x)(-x^2+2x-6)}{(3x)(6x^2)}[/tex]
With the 3x factored out, they cancel out because 3x/3x=1.
Our final answer is:
[tex]\frac{-x^2+2x-6}{6x^2}[/tex]
