Question

Question #1de6c

Answer 1

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.
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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.