Question #1de6c

Jul 26, 2017

No, it can't be.

Explanation:

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

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

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} \to = 0$

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

So far, so good.

$6 \cdot {0}^{2} \cdot \left(1 + 0\right) = 0 \cdot \left(1 + 0\right) = 0 \cdot \left(1\right) = 0$.
This is the real answer considering that what we had assumed ($f \left(x\right) = 0$) was true.
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A real and good simplification would just be $6 {x}^{2} \left(1 + x\right)$, because it does not envolve any other factors (such as $y$) in the simplification.