Factoring Quadratics When a Doesn't = 1

Date: 03/24/2003 at 11:51:54
From: Tracy
Subject: Factoring quadratics when a doesn't = 1

A trick to solving quadratics was presented to me. I was wondering 
if there is a proof for it.

For any trinomial ax^2+bx+c you multiply ac, and then find factors of 
the product that add to b. Once you find the match, they serve as your 
denominator and then use a as the numerator. Reduce and then the 
numerators are your firsts and the denominators are your matching 
lasts in foil.  

Thank you.
Tracy

Date: 03/24/2003 at 16:10:15
From: Doctor Peterson
Subject: Re: Factoring quadratics when a doesn't = 1

Hi, Tracy.

Your method is very closely related to what I explain here:

   Factoring Trinomials
   http://mathforum.org/library/drmath/view/56442.html 

I have adopted a related, but less 'magical', method as my preferred 
approach to these problems:

   Finding a Single Pair of Factors
   http://mathforum.org/library/drmath/view/53308.html 

I prefer this method because it involves more thinking that is 
transferable to other tasks; since factoring a trinomial has no real 
use beyond elementary algebra, I'd rather not focus too much on tricks 
that do nothing but increase your speed in factoring!

Let's try out your method on an example, to make sure I have it right:

    4x^2 + 11x + 6

    ac = 4*6 = 24; 3*8 = 24 and 3+8 = 11

    use 3 and 8 as denominators with 4 as numerator

    4/3 is in lowest terms;
    4/8 reduces to 1/2.

    Use 4/3 and 1/2 to form the factors (4x+3) and (1x+2).

    (4x+3)(x+2) = 4x^2 + 11x + 6

That worked. Your method as stated, however, has a flaw: it fails if 
a, b, and c have a common factor. (For example, try it on 4x^2 + 10x 
+ 6.) In such cases, your fractions will simplify too far. You have 
to first factor out the 2, and then factor 2x^2 + 5x + 3, whose 
coefficients are relatively prime. Or, you can use the method and 
then multiply by the necessary factor to make "a" correct.

To prove your method, let's suppose that we have found m and n so that

    mn = ac
    m+n = b

Now you are taking the fractions

    a/m, a/n

and reducing, so we have

    a/m = u/p    (so that u = ap/m)
    a/n = v/q    (so that v = aq/n)

It can be shown (I'll do this below) that

    pq = c

Now you claim that the trinomial is equivalent to

    (ux + p)(vx + q)

Let's see if this equals ax^2 + bx + c. We have:

    (ux + p)(vx + q) = uvx^2 + (vp+uq)x + pq

Now, uv = (ap/m)(aq/n) = (a^2 pq)/(mn) = (a^2 c)/(ac) = a.

And, vp+uq = apq/n + apq/m = apq(m+n)/mn = (m+n)(ac)/(ac) = m+n = b.

And, finally, pq = c.

So we've shown that our factored form is equivalent to the original.

Footnote:

To show that pq=c, let's suppose that in factoring ac as mn, we took 
the factors

    a = a1 a2
    c = c1 c2

and rearranged them as

    m = a1 c1
    n = a2 c2

Then

    a/m = (a1 a2)/(a1 c1) = a2/c1
    a/n = (a1 a2)/(a2 c2) = a1/c2

I claim that these have to be in lowest terms. To show this, suppose 
that a2 and c1 are both divisible by d. Then

    a = a1 a2 is a multiple of d
    c = c1 c2 is a multiple of d
    b = a1 c1 + a2 c2 is a multiple of d

and a, b, and c are not relatively prime as assumed. By contradiction, 
the supposition that a2/c1 was not in lowest terms is false. (The same 
reasoning shows that a1/c2 is in lowest terms.)

Thus if a, b, and c are relatively prime, p=c1 and q=c2, and pq = c 
as I claimed.

If a, b, and c are NOT relatively prime, then either p or q will be 
less than c1 or c2 respectively, and pq is a proper divisor of c. This 
makes the rest of the proof fail. In fact, the product of the factors 
you find will be the given trinomial divided by the greatest common 
factor of a, b, and c.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/