Question #1de6c

1 Answer
Jul 26, 2017

Answer:

No, it can't be.

Explanation:

#6x^2+6x^3=6x^2(1+x)#

#6x^2=6x^2+0x+0->ax^2+bx+c# :
By doing this, we are assuming that #f(x) = y = 6x^2 =0#.

#x=(-b+-sqrt(b^2-4ac))/(2a)->=0#

(In this case:)
#x = (-0+-sqrt(0^2-4*1*0))/(2) = = (0+- 0)/2 = 0#

So far, so good.

#6*0^2*(1+0) = 0 * (1+0) = 0 * (1) = 0#.
This is the real answer considering that what we had assumed (#f(x) = 0#) was true.
#--------------------#

A real and good simplification would just be #6x^2(1+x)#, because it does not envolve any other factors (such as #y#) in the simplification.