Simplifying $\frac{ (2x+1) (x-3) }{ 2x^3 (3-x) }$

Question

In simplifying $$\frac{2x^2-5x-3}{6x^3-2x^4}$$

I got this far $$\frac{ (2x+1) (x-3) }{ 2x^3 (3-x) }$$

but there aren't same brackets to cancel out.

Answer 1

$$\dfrac{2x^2-5x-3}{6x^3-2x^4}=\dfrac{(2x+1)(x-3)}{2x^3(3-x)}=\dfrac{(2x+1)(x-3)}{2x^3[-(x-3)]}=\dfrac{(2x+1)(x-3)}{-2x^3(x-3)}=\dfrac{2x+1}{-2x^3}$$

Answer 2

There are!

Don't forget that $(x-3)=-(3-x)$ so you can cancel to give:

$$\frac{-(2x+1)}{2x^3}$$

Just remember that $x\neq 3$ is a requirement else you're dividing by $0$ (Thanks Dr)

Answer 3

$\frac{(2x+1)(x-3)}{2x^3(3-x)}= -\frac{(2x+1)(x-3)}{2x^3(x-3)}= -\frac{(2x+1)}{2x^3} $