(x^3-3x^2)-(2x-6)=0 What is x?

Here's another approach:

(x^3-3x^2)-(2x-6)=0     factor  by grouping

x^2(x - 3) - 2(x -3)  = 0

(x^2 - 2) (x - 3) = 0        set each factor to 0  ....     and  we find that x = ±√2, 3

Here's a graph.......

$$\\(x^3-3x^2)-(2x-6)=0\\ x^3-3x^2-2x+6=0$$

Now use remainder theorem to find a root

$$\\let\\ f(x)=x^3-3x^2-2x+6\\ f(\pm1)=\pm1-3\mp2+6\;\;=\;\;3\pm1\mp2\;\;\rightarrow\;\;2\ne0\;\;or\;\;4\ne 0\\ f(3)=27-27-6+6=0\qquad therefore\;\;x=3 \;\;$is a root$$ $

 (this has been edited a little)

I used polynomial division to determine that

$$\\(x^3-3x^2-2x+6)\div(x-3)=(x^2-2)\\\\ so\\\\ x^3-3x^2-2x+6=0\\\\ (x-3)(x-\sqrt2)(x+\sqrt2)=0\\\\ x=3\quad or\quad x=\sqrt2\quad or \quad x=-\sqrt2$$

Yes, CPhill's way is better :)

LOL!!!....Melody always says that she likes to "complicate" things.....actually, both approaches are good .... the "Rational Zeros" approach is always helpful if it's not immediately obvious how to factor a polynomial [if it's actually capable of being factored, that is ]

Miss Melody is there a difference between these plus minus things?

$$\pm \;\mp$$

I not ever see the one on the right before.

That is a good question Dragonlance.

I have laid it out a little better now so hopefully it is a little clearer    

$$f(\pm1)=\pm1-3\mp2+6\;\;=\;\;3\pm1\mp2\;\;\rightarrow\;\;2\ne0\;\;or\;\;4\ne 0\\$$

The signs on the top belong with f(+1)

and the signs on the bottom belong with f(-1)