Factoring

[gabrielh]

I'm currently taking a Math Analysis class. We started reviewing basic factoring from Algebra 2 before starting new material. I can factor a basic trinomial with the coefficient of x^2 being 1, but if the coefficient is 2 or more, I am not sure how to factor it. I am able to factor these very easily with a graphing calculator, but I'd like to know how to do it by hand.


1. The problem statement, all variables and given/known data
Factor: 2x^2 + 5x - 12


2. The attempt at a solution
I know, by using a graphing calculator, that the trinomial factors into (2x-3)(x+4). I know that if the trinomial were, for instance, x^2 - 2x - 8, one could say the product of 8 and the coefficient of x is 8, and the factors of 8 that when added together result in negative 2 are -4 and +2, therefore the factors of x^2 - 2x - 8 are (x-4)(x+2).

Can trinomials with coefficients greater than one be factored in a similar way?

Using the method described above, I come up with a product of 24. Factors of 24 that result in a sum of 5 when added together are +8 and -3, which by the method I am acquainted with would mean the factors are (x+8)(x-3), but that obviously doesn't equal 2x^2 + 5x - 12 when multiplied. I know I'm missing a step somewhere.

Thanks for the help in advance. I know this is a simple question, but it plagued me all day.

[rock.freak667]

You do it in the same manner

2x2 + 5x - 12 must be factored into something like (2x+a)(x+b)

the last times the last should give -12 or ab=-12, so what products can give -12?
There is -12 and 1, 2 and -6, 3 and -4 (and the others switch around the signs)

expanding (2x+a)(x+b) in your head, you'd see that the coefficient of x is a+2b. So looking at your choices, you want a+2b=5. Right away 12 and -1 (or 1 and -12) is eliminated.

2 and -6 is gone since 2+2(-6)≠5 or -2+2(6)≠5

so you are left with 3 and -4 or -3 and 4

3+2(-4)= -5 .So we are seeing 5 but negative, so we need to change the signs. So the choice is -3 and 4

so it is factored as (2x-3)(x+4)

as you practice, you can quickly do this and eliminate the obvious ones it can't be.

For ax2+b+c

Another way is to compute b2-4ac and find the square root of that. If it is an integer, then using the quadratic equation formula

x_1,x_2=\frac{-b \pm \sqrt{b^2-4ac}}{2a}

In your example b2-4ac=121, so √(b2-4ac)=11

x1,x2= (-5±11)/2(2)

x1=(-5+11)/4=6/4=3/2

So one root is x=3/2 or 2x+3=0

x2=(-5-11)/4=-4. So the other root is x=-4 or x+4=0

the product of these two factors will give your original quadratic as being (2x+4)(x-4)

[HallsofIvy]

First not all trinomials can be factored with integer coefficients. But trinomials with coefficients of x^2 not equal to one are only a little more tedious. To factor ax^2+ bx+ c, you need to think about what (mx+ n)(px+ q) would look like: mx(px+ q)+n(px+q)= mpx^2+ mqx+ npx+ nq= mpx^2+ (mq+np)x+ nq. So we are looking for a numbers m, n, p, q such that mp= a, nq= c, and mq+np= b. Start by thinking of factors for a and b and look at all the different ways you could form mq+ np.

With you example, 2x^2 + 5x - 12, there are two ways to factor 2: (1)(2) and (2)(1). But there are many ways to factor 12: (1)(12), (2)(6), (3)(4), (4)(3), (6)(2), and (12)(1). Since there are 2 ways to factor 2 and 6 ways to factor 12, there are (2)(6)= 12 possible combinations. Also, since the last term is negative we know the last term in the two factors must have opposite sign:
(1)(2) and (1)(12): (x+ 1)(2x- 12)= 2x^2- 10x- 12
(2)(1) and (1)(12): (2x+ 1)(x- 12)= 2x^2- 23x- 12
(1)(2) and (2)(6): (x+ 2)(2x- 6)= 2x^2- 2x- 12
(2)(1) and (2)(6): (2x+ 2)(x- 6)= 2x^2- 10x- 12
(1)(2) and (3)(4): (x+ 3)(2x- 4)= 2x^2+ 2x- 12
(2)(1) and (3)(4): (2x+ 3)(x- 4)= 2x^2- 5x- 12
(1)(2) and (4)(3): (x+ 4)(2x- 3)= 2x^2+ 5x- 12 !!!
(2)(1) and (4)(3): (2x+ 4)(x- 3)= 2x^2- 2x- 12
(1)(2) and (6)(2): (x+ 6)(2x- 2)= 2x^2+ 10x- 12
(2)(1) and (6)(2): (2x+ 6)(x- 2)= 2x^2+ 2x- 12
(1)(2) and (12)(1): (x+ 12)(2x- 1)= 2x^2+ 23x- 12
(2)(1) and (12)(1): (2x+12)(x- 1)= 2x^2+ 10x+ 12
Whew! Of course, we could have stopped when we got "(1)(2) and (4)(3): (x+ 4)(2x- 3)= 2x^2_ 5x- 12" but I wanted to show what they all looked like.

[gabrielh]

Thank you both for your in depth replies. I'm able to do this now. Again, thanks a lot :)

[Elucidus]

There is a clever factorization method (I believe developed by Viete) for non-monic quadratic trinomials with integer coefficients. (i.e. of the form ax^2 + bx + c \text{ for }a,b,c \text{ integers where } a \neq 1 \text{ or } 0).

Step 0.
Factor out any common divisors of a, b, and c and work with the cofactor.

Step 1.
Multiply a and c and call the product d. This is the key number to unlocking the factorization.

Step 2.
Determine if there is a pair of factors of d whose sum is b. If such a factor pair exists, then the polynomial is factorable, otherwise it is not factorable (and you'd stop here).

If a factor pair exists, call them m and n.

Step 3.
Rewrite ax^2 + bx + c \text{ as }(ax^2 + mx) + (nx + c) and factor the latter expression by grouping. Be careful when n is negative.

Step 4.
You should have your factorization.

The advantage to this method is that it cuts down the "try-this-then-try-this-other-way"-ness of other methods.



Example: Factor 12x^2 -31x-30.

Step 0. No common divisors exist so I'm working with it as is.

Step 1. The product of 12 and -30 gives a key number of -360.

Step 2. The factos of -360 the add to -31 are -40 and 9.

Step 3.

12x^2-31x-30=(12x^2-40x)+(9x-30).

=4x(3x-10)+3(3x-10)

=(3x-10)(4x+3).

Done.

--Elucidus