Anyone know how to factor polynomials?

Hey guys, I'm talking calculus, but high school math was such a long time ago. I can do simple polynomials like:
 x^2 + 6x + 8
 
I find 2 numbers that add to 6 and multiply to  8. So I end up with
(x + 2)(x + 4)
 
But what if I have :
3x^2 -7x + 2
 
What do I do?

Ask your professor not us, colleges have tons of resources at your disposal, use them

You could always use the quadratic formula, seen on the right had side of this page:
 http://www.purplemath.com/modules/quadform.htm
 
Or, you could do it by completing the square, if you've done that method.
 

Quadratic formula.
 
EDIT: Didn't see the three. Can't you just do that in your head?
 
x = 1/3 or 2

Trying it out...

Man, you just made me realize how fucked I'm gonna be when I start my math class in a couple of weeks. I took an algebra class last semester and I already forget how to do this shit. Good thing I have an entire folder full of notes lying around.

Are you allowed to use software? I just typed into Maple:
 
factor(3*x^2-7*x+2);
 
and got:

 (3 x - 1) (x - 2)
 
Your TI-89 could easily do it too.

Pemdas. 
 
Now, it's been a while...but...
 
 3x^2 -7x + 2 --> 3x*3x-7x+2 --> 9x-7x+2 --> 2x +2 --> 2x =-2 --> x =-1

" Pemdas.   Now, it's been a while...but...   3x^2 -7x + 2 --> 3x*3x-7x+2 --> 9x-7x+2 --> 2x +2 --> 2x =-2 --> x =-1 "   

" Pemdas.   Now, it's been a while...but...   3x^2 -7x + 2 --> 3x*3x-7x+2 --> 9x-7x+2 --> 2x +2 --> 2x =-2 --> x =-1 "

As mentioned, I'd just stick to the quadratic formula.  I personally hate dealing with fractions and completing the square would just get messy. 
I also got  X = 1/3, 2.

" Hey guys, I'm talking calculus, but high school math was such a long time ago. I can do simple polynomials like:  x^2 + 6x + 8  I find 2 numbers that add to 6 and multiply to  8. So I end up with (x + 2)(x + 4)  But what if I have : 3x^2 -7x + 2  What do I do? "

ya its pretty easy actually to bad this is a gaming website

With a smile.

Pay attention in class or use excel

" As mentioned, I'd just stick to the quadratic formula.  I personally hate dealing with fractions and completing the square would just get messy. I also got  X = 1/3, 2. "

" Quadratic formula.  EDIT: Didn't see the three. Can't you just do that in your head?  x = 1/3 or 2 "

What do you call a parrot that doesn't eat?
 

" @nick_verissimo said:

" As mentioned, I'd just stick to the quadratic formula.  I personally hate dealing with fractions and completing the square would just get messy. I also got  X = 1/3, 2. "

@Muttinus_Rump said:

" Quadratic formula.  EDIT: Didn't see the three. Can't you just do that in your head?  x = 1/3 or 2 "

What the hell. He didn't want to solve them, he wanted to factor them. "

The quadratic formula works well for degree 2 polynomials, and is usually the fastest way to get complex roots. If the degree of the polynomial is higher than two, however, I always find it easier to remove all of the rational roots with synthetic division (cf. the Rational root theorem) and then worry about complex roots.

thanks guys i got it, @Cliffy's method works great
 
its calculus for scientists, they just expect us to know this stuff, nothing in the textbook or anything
 
But man its been awhile

I'm rusty as all hell (it's been about 8 years since I needed to know this), but this is how I think I'd do it:
 
3x^2 -7x + 2  
 

1. You know you'll have a (3x   ) and a (x    ) bracket (because multiplied out they'll give you the '3x^2' part).  
2. You know the two loose numbers (currently missing above) will multiply out to give you the '+2'.
3. You know the only way to get '+2' is multiplying a positive by a positive, or a negative by a negative. 
   
4. The '-7x' part tells you this needs to be a negative by a negative, so you have (3x-  )(x-  ), with a '2' to go in one bracket and a '1' to go in the other.
   
5. The '3x' will multiply out by the number in the second bracket, and the 'x' will multiply out by the number in the first bracket.  Combined they need to give you '-7x', so you obviously want the '3x' to multiply out with '-2' (giving -6x), and the 'x' to multiply out with the '-1' (giving -x).  '-6x' + '-x' = '-7x'.
   
6. Now you have all the parts:  (3x-1)(x-2).  If you multiply these out you'll (hopefully) get what you started with.
   

hey, are you in my match class or something lol? i'm doing the same thing. it's like (3x + 1(x -2) something like that.

I believe the answer is (3x-1)(x-2), although I'm a bit rusty, myself.

lol math 1200 at mount royal? no? @iam3green:

yeah, basically I did what @Jimbo did, I just didn't explain it as well.  Other problem is that you don't know that it's both negatives; you could have a -10 and a +3 and it come out to -7.  You can pair the fact that it has to be a positive number (+2) and a negative (-7x) whatever that part is called to get that it has to be two negatives though... sorry I'm confusing myself now...