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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.

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:

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/
```
Associated Topics:
High School Polynomials
Middle School Factoring Expressions

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