Factoring Difficult Trinomial

[wind_surfer]

Hello,

I can't figure out how to factor this trinomial --> 2x^3+3x^2-1 I tried the normal approach for trinomials with a coefficient >1 on the leading term:
- I multiplied leading coefficient and the constant = -2
- Then I was stuck because you can't multiply 1 and -2 or -1 and 2 (the factors) as well as being able to add them to get 3 to match the middle term
- What do I do?
- Is there some sort of extra step to take on this problem? As it seems to be an exception to the normal steps of factoring

Any help would be appreciated.

Thankyou

 

[Deleted member 4993]

Hello,

I can't figure out how to factor this trinomial --> 2x^3+3x^2-1 I tried the normal approach for trinomials with a coefficient >1 on the leading term:
- I multiplied leading coefficient and the constant = -2
- Then I was stuck because you can't multiply 1 and -2 or -1 and 2 (the factors) as well as being able to add them to get 3 to match the middle term
- What do I do?
- Is there some sort of extra step to take on this problem? As it seems to be an exception to the normal steps of factoring

Any help would be appreciated.

Thankyou

Use rational root theorem.

For a quick review, go to:

http://www.purplemath.com/modules/rtnlroot.htm

 

[wind_surfer]

Thanks!

 

[wind_surfer]

I still can't factor it completely.. I used the rational roots theorem and got (x+1)(x-1/2) but that isn't equal to 2x^3+3x^2-1...
Is it even possible to factor this completely? I'm confused..
Any help would be appreciated.. I've tried everything I can think of and all the ways to factor I've found online.

Thankyou

 

[HallsofIvy]

I still can't factor it completely.. I used the rational roots theorem and got (x+1)(x-1/2) but that isn't equal to 2x^3+3x^2-1...

Is it even possible to factor this completely? I'm confused..
Any help would be appreciated.. I've tried everything I can think of and all the ways to factor I've found online.

Thankyou

The rational roots theorem says that the only possible rational roots of 2x^3+ 3x^2- 1 must be among 1, -1, 1/2, and -1/2. You have determined that, of those, only -1 and 1/2 actually are roots and two linear factors are x+ 1 and x- 1/2. but 2x^3+ 3x- 1 is cubic so there must be a third linear factor. To see what that is divide by x+ 1 and x- 1/2. (Equivalently, divide by the product (x+ 1)(x- 1/2)= x^2+ (1/2)x- 1/2.)

 

[Deleted member 4993]

I still can't factor it completely.. I used the rational roots theorem and got (x+1)(x-1/2) but that isn't equal to 2x^3+3x^2-1...
Is it even possible to factor this completely? I'm confused..
Any help would be appreciated.. I've tried everything I can think of and all the ways to factor I've found online.

Thankyou

2x^3 + 3x^2 + 1 = (x+1)(2x -1)(x+1) = (x+1)²(2x-1)