#1**+5 **

$$\\(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$$

Melody
Aug 25, 2015

#2**+10 **

Best Answer

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.......

CPhill
Aug 25, 2015

#4**0 **

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 ]

CPhill
Aug 25, 2015

#5**+5 **

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

$$\pm \;\mp$$

I not ever see the one on the right before.

Dragonlance
Aug 25, 2015

#6**+5 **

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)

Melody
Aug 25, 2015