show that $$2x ^{2}[x \times(-1)+(3-x)] -x(3-x)(4x) \over (2x ^{2})^{2}$$ can reduce to $$-3 \over 2x ^{2}$$

Question

anonymous · Answer

$$\frac{2x^2(x (-1)+(3-x))-x(3-x)4x}{\left(2x^2\right)^2}  $$$$\frac{6 x^2-4 x^3-12 x^2+4 x^3}{\left(4 x^4\right)}  $$$$\frac{-6 x^2}{\left(4 x^4\right)}  $$$$-\frac{3}{2 x^2} $$

anonymous · Answer

how did you go from steps 1-2? like the top part?

anonymous · Answer

Work from the inside out.  Multiply x*(-1) -> -x +3 -x is -2x+3
2x^2(-2x+3)=-4x^3+6x^2
Over at the right side -x(3-x)4x is (-3x+x^2)4x = -12x^2+4x^3
Putting it all together,
-4x^3 + 6x^2 -12x^2 + 4x^3 = -12x^2    The -4x^3 and +4x^3 subtract out.  6x^2 - 12x^2 is equal to -6x^2

anonymous · Answer

thank you!!!

anonymous · Answer

Your welcome.