Factoring Trinomials: 9x^2 - 42x + 49

Date: 10/15/2002 at 12:56:16
From: Tiffany
Subject: Dactoring trinomials

Right now in my math course we are factoring trinomials, for instance: 

   9x^2 - 42x + 49

I am having a lot of trouble factoring this trinomial. I kind of, but 
not completely, understand x^2-21x-72. I am having the most difficulty 
factoring when there is a number in front of the x^2 (9x^2).  

Is there an easier way to do this than finding all of the common 
factors and going from there?  I really don't understand this method.  
Please help me.  

Thank you for your time.
Tiffany

Date: 10/15/2002 at 15:17:29
From: Doctor Ian
Subject: Re: Factoring trinomials

Hi Tiffany,

Is there an easier way? It depends on what you think is 'easy'. Here 
are a few different ways to go about factoring a quadratic trinomial:

   Factoring Quadratics Without Guessing
   http://mathforum.org/library/drmath/view/60700.html 

And here is a gentle guide to using the 'standard' method:

   Factoring Polynomials
   http://mathforum.org/library/drmath/view/60580.html 

In the case of 

  9x^2 - 42x + 49

the nice thing is that there aren't a lot of possibilities, because 
there aren't a lot of prime factors to deal with. The only 
possibilities for the initial terms are:

  (x + __)(9x + __)          

  (3x + __)(3x + __)

In fact, looking at the signs, we know that both of the numbers in the 
blanks will have to be negative:

  (x - __)(9x - __)          

  (3x - __)(3x - __)

And the only possibilities for the final terms are 

  (x - {1,7,49})(9x - {49,7,1})          

  (3x - {1,7,49})(3x - {49,7,1})

So that's a total of only 6 possibilities to check. 

Now, what did I mean by 'looking at the signs'? Well, assuming that 
a and b are both positive, there are a few patterns that are worth 
memorizing. The first two are:

  (x + a)(x + b) = x^2 + (a+b)x + ab         

  (x - a)(x - b) = x^2 - (a+b) + ab

These are the only cases in which the final term (ab) will be 
positive. And you can tell by looking at the sign of the middle term 
what signs will appear in the factors. 

The other pattern is

  (x + a)(x - b) = x^2 + ax - bx - ab

                 = x^2 + (a-b)x - ab

In this case, the sign of the final term will be negative, and the 
sign of the middle term can be anything at all. (But that's not a 
problem, since the sign of the final term is enough to tell you 
what's going on.) 

So back to our possibilities. To check them, we need to multiply the 
pairs by the initial coefficients, and add:

  1. (x - {1,7,49})(9x - {49,7,1})          

         (1,49):  9*1 + 1*49 = 58            (No)
         (7,7):   9*7 + 1*7  = 70            (No)
         (49,1):  9*49 + 1*1 = [too big]     (No) 

  2. (3x - {1,7,49})(3x - {49,7,1})

         (1,49):  3*1 + 3*49 = [too big]     (No)
         (7,7):   3*7 + 3*7  = 42            (Yes!)
         (49,1):  3*49 + 3*1 = [don't care] 
       
Is this a pain? Yes, it is - and the more prime factors you have, the 
more it hurts.  

However, as the second URL above points out, if you're trying to 
sketch the graph of the function (which is usually what you're trying 
to do with a function), it's a _lot_ easier than picking points at 
random. 

And, believe it or not, as you get more practice at this, you'll get 
better at instinctively avoiding the possibilities that can't make 
sense (in very much the same way that someone who is really good at 
chess doesn't even consider the kinds of dumb moves that novices have 
to work through). For example, you're not going to multiply 49 by 
_anything_ and add the result to something else to get 42. So with 
practice, you'd go right to the following, smaller set of possibilities:

  (x - {7})(9x - {7})          

  (3x - {7})(3x - {7})

So you'd have only two things to try, instead of 6.  

Try reading both of the answers from the Dr. Math archives above, and 
then try factoring some expressions. If you get stuck, write back and 
show me how far you were able to get, and we'll go from there.  Okay? 

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