Now use remainder theorem to find a root
f(3)=27-27-6+6=0\qquad therefore\;\;x=3 \;\;$is a root$$$
(this has been edited a little)
I used polynomial division to determine that
x=3\quad or\quad x=\sqrt2\quad or \quad x=-\sqrt2$$
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.......
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?
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
The signs on the top belong with f(+1)
and the signs on the bottom belong with f(-1)