How to find x-intercepts of f(x) = 2x^3-x-2?

I'm not really sure how to approach this question... with it being a cubic function?

Question

How to find x-intercepts of f(x) = 2x^3-x-2?

I'm not really sure how to approach this question... with it being a cubic function?

ganeshie8 · Answer

you want to solve it algebraically ?

anonymous · Answer

Yeap! @ganeshie8

ganeshie8 · Answer

lets start by dividing 2 through out so that leading coeffecient becomes 1

ganeshie8 · Answer

$2x^3-x-2 = 0$

$x^3-x/2-1 = 0$

hero · Answer

I might have an approach to this myself as well.

ganeshie8 · Answer

ive just made up my method but not quite too sure of it yet... please show your method @Hero

hero · Answer

I made mine up too bro, but I'll post something.

hero · Answer

$$2x^3 - 2 - x = 0$$$$2(x^3 - 1) - x = 0$$$$2((x - 1)(x^2 + x + 1)) - x = 0$$$$2(x -1)(x^2 + x + 1) - x = 0$$$$(x-1)\left(2(x^2 + x + 1)) - \frac{x}{(x-1)}\right) =0$$

hero · Answer

Now use zero product property

hero · Answer

@cmolina19 
@ganeshie8

hero · Answer

There's probably a mistake in there somewhere

anonymous · Answer

Wait how did you get the second set of parentheses? o.O 

I've been trying to do this on my own.. and I'm actually wondering.. is it possible to do this:
2x^3-x-2=0
2x^3-x=2
x(x^2-1)=0
x=0, x=1
???

anonymous · Answer

wait oh crap, changed the 2 to zero

hero · Answer

Your second step to third step is flawed

anonymous · Answer

ok, 
x(x^2-1)=2
x=2, x=sqrt(3)

anonymous · Answer

Yeah, that's right. I noticed it once I posted it lol

anonymous · Answer

Nevermind, plugging it into the calculator it's not correct

hero · Answer

I used difference of cubes to get the expression above

hero · Answer

$$x^3 - 1 = x^3 - 1^3$$

hero · Answer

But something is flawed with it since x = 1 is not a solution

hero · Answer

Do you get what I did (or tried to do)?

anonymous · Answer

I do, but not the last line :$ 
(x−1)(2(x2+x+1))−x(x−1))=0

hero · Answer

That's not what the last line reads.

hero · Answer

That last expression is -x/(x-1)

hero · Answer

It's what happens when you factor out x - 1

ganeshie8 · Answer

that gives back cubic again

hero · Answer

I know. You must have worked on it @ganshie8

ganeshie8 · Answer

yeah trying ur method as it looks simple but ending up wid cubic no matter wat i do

hero · Answer

x = 1 is not a solution so immediately we know something is wrong.

anonymous · Answer

Yeah, whoops. Copied and pasted it to here, so it missed some characters.

ganeshie8 · Answer

yeah that also there

ganeshie8 · Answer

try this :-
1) find one root $k$ by doing some algebra tricks (I am going to help you with this)
2) then divide $(x-k)$ turning the cubic to quadratic
3) quadratic can be easily handled

hero · Answer

Yes, of course. That's the traditional way of doing it. I was trying to use some factoring method.

ganeshie8 · Answer

traditional is painful and lengthy also, anyways i feel like giving it a try now :)

ganeshie8 · Answer

$\large 2x^3-x-2 = 0$

$\large x^3-x/2-1 = 0$ --------------(1)

say, x = u+v

$\large x^3 = u^3+v^3 + 3uv(u+v) $

$\large x^3 = u^3+v^3 + 3uvx $

$\large x^3 -3uvx - (u^3+v^3) = 0  $ -------(2)

ganeshie8 · Answer

compare x coefficient and constant terms between (1) and (2)

ganeshie8 · Answer

$3uv = 1/2 => uv = 1/6 => u^3v^3 = 1/216$ 

$u^3+v^3 = 1$

ganeshie8 · Answer

quadratic with $u^3, v^3$ as roots :-

$t^2-t+1/216 = 0$

here next steps :-
1) solve t by completing the square
2) taking cube root of solutions above, gives u value of u and v
3) you have x, since x  = u+v

ganeshie8 · Answer

give it a try

hero · Answer

Good luck with that bro.

anonymous · Answer

wait, what is t again?

ganeshie8 · Answer

t is just an intermediate variable, solve the quadratic in t. solutions will be u^3 and v^3

ganeshie8 · Answer

$t^2-t+1/216 = 0$

$216(t^2-t) + 1 = 0$

$216((t-1/2)^2)  = 53$

$(t-1/2)^2  = 53/216$

solve t

keep going