Dividing by negative number in quadratics, yielding totally different result.

Question

forewords: Honestly, I've tried looking around for any answer to this question, but it's so specific i can't find any. If it's already answered i apologise.

Alright so i have this equation $$-x^{2}+4=0$$

Now usually i would remove the negative in front of the polynomial, by multiplying the whole equation by $\frac 1{-1}$ or simply said, divide by $-1$. If we do this, we get.

$$x^2-4 = 0$$

Now if we factor the first equation we get. $$-x^{2}+4=-(x+2)(x-2)$$ If we factor the other we get $$x^2-4=(x+2)(x-2)$$ Ala the same, just without the minus. Graphing these two equations, yield different result (one is a frown, and the other is smiley). Could somebody please give me a intuitive explanation of why i should not divide by the minus, and instead put it outside the parenthesis Usually when i this far down the rabbit hole, the answer is humiliatingly simple, but now i can't seem to do anything right.

Any help is greatly appreciated, thank you.

Edit

I updated with a picture to show, where my confusion lies (how to divide by -1, and still get the same graph) Calculations

Answer 1

I suppose you are plotting $$y = -x^2 + 4 \quad\text{and}\quad y = x^2 - 4,$$ which represent parabolas, not $$-x^2 + 4 = 0 \quad\text{and}\quad x^2 - 4 = 0.$$ If you divide by $-1$ both sides of one of the former equations, you get $-y$ instead of $y$, from which you should see that they are not equivalent (i.e., they do not represent the same parabola).

Answer 2

It's okay to divide by $-1$.

It's true that the graph of $y=x^2-4$ is different from the graph of $y=-x^2+4$. But we're specifically interested in when the right-hand-side ($x^2-4$ or $-x^2+4$) is zero. And the two graphs cross the $x$-axis in exactly the same places, which means that $x^2-4$ and $-x^2+4$ are zero at exactly the same values of $x$.