Question #1de6c

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Question #1de6c

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Answer 1 · 2017-07-26T15:16:09

No, it can't be.

Explanation:

$6 x^{2} + 6 x^{3} = 6 x^{2} (1 + x)$ $6x^2+6x^3=6x^2(1+x)$

$6 x^{2} = 6 x^{2} + 0 x + 0 \to a x^{2} + b x + c$ $6x^2=6x^2+0x+0->ax^2+bx+c$ :
**By doing this, we are assuming that $f (x) = y = 6 x^{2} = 0$ $f(x) = y = 6x^2 =0$ . **

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \to = 0$ $x=(-b+-sqrt(b^2-4ac))/(2a)->=0$

(In this case:)
$x = \frac{- 0 \pm \sqrt{0^{2} - 4 \cdot 1 \cdot 0}}{2} = = \frac{0 \pm 0}{2} = 0$ $x = (-0+-sqrt(0^2-4*1*0))/(2) = = (0+- 0)/2 = 0$

So far, so good.

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

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

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Question #1de6c

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