1. Sep 11, 2009

### winston2020

1. The problem statement, all variables and given/known data

Factor.
$$2x^{2}+5x-12$$

I just took a semester off from school and I feel dumb. My recommendation to anyone reading is don't do that.
Anyways, back to gr. 10 math

Normally if I was going to factor this I would try to eliminate the coefficient of $$x^{2}$$ but it can't be divided evenly. IIRC you're supposed to take the factors of c and try to find 2 that add up to b, but that doesn't seem to be an option here either...

I just need a push in the right direction :)

2. Sep 11, 2009

### CFDFEAGURU

Take a look at the Binomial theorem. That should help you out.

Thanks
Matt

3. Sep 11, 2009

### Staff: Mentor

This is fairly straightforward, since there aren't many possibilities for 2x2. Start with (2x + ?)(x + ?) and by trial and error, see if there is some way to factor -12 so that you get an x-term coefficient of + 5.

4. Sep 11, 2009

### njama

Start by:

2(x2+(5/2)x-6)=2[ (x+5/4)2-25/4-6 ]

5. Sep 11, 2009

### gabbagabbahey

The question asked him to factor the quadratic, not complete the square.

6. Sep 11, 2009

### Elucidus

It is possible to factor quadratic trinomials whose lead coefficient is not 1 by multiplying the lead and tail coefficients together and finding a factor pair of that product that adds up to the middle coefficient.

Based on a set of equations by Viete:

Find u and v so that $ax^2 + bx + c = ax^2 + ux + vx + c$ and

$uv=ac$ and $u+v=b.$

Once u and v are found, then factor by grouping.

--Elucidus

7. Sep 11, 2009

### HANNONKEVINP

My Teacher taught me a method called slide and divide.

2x^2 +5x -12, you multiply the coefficient of 2x^2 times -12.

X^2 +5x-24, from there you factor normally...Once you get the factors, you divide the number of by the coefficient you used before. One little trick that you must remember is if the number turns out to be not a whole number, you bring out the denominator and put in on the x. For example (x-3/2) would become (2x-3)

8. Sep 12, 2009

### HallsofIvy

Staff Emeritus
Yes, but that's a perfectly valid way to find the factors (and works even if the coefficients in the factors are not rational).
$2((x+ 5/4)^2- 25/4- 6)= 2((x+ 5/4)^2- 49/4)$ and that is now a "difference of squares": 2(x+5/4+ 7/2)(x+5/4- 7/2)