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### Factoring Trinomials

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Date: 02/24/2002 at 22:34:14
From: John
Subject: Explanation for this method of factoring

My teacher gave me a method for factoring trinomials (ax^2 +bx +c).
He didn't give me the name of the method but I know it's not FOIL or
the grouping method (first part is similar).

Here's how to do it:

ex. 6x^2 +19x -20
1) multiply the first coefficient by the constant
(6)(-20)= -120
2) rewrite it to x^2 +19x -120 (notice that 6 is no longer there)
3) factor the new trinomial (changes to x^2 +bx +c)
(x-5)(x+24)
4) add the first coefficient back to the equation (in front of x)
(6x-5)(6x+24)
5) find the common factor of the numbers in each bracket
ex. for the first bracket (6 and 5) there is no common factor so
it's left as is
for the second bracket (6 and 24) the common factor is 6
then divide each bracket by the common factor
6) result is (6x-5)(x+4)

Now, my question is, what's the logic/reason behind this method? How
can I prove it? Now that I know the steps, I want to know why it
works.

It may look complicated, but so far, this is by far the easiest way I
have learned to factor trinomials. If you know an easier/faster way
to do it, please feel free to comment.

Thank you.
```

```
Date: 02/25/2002 at 14:55:04
From: Doctor Peterson
Subject: Re: Explanation for this method of factoring

Hi, John.

I've never heard of this method, but it does work, and I like it,
apart from the fact that it is magic, and I don't like to do magic
without knowing why it works. Here's another example. We'll factor

6x^2 - x - 12

We take the 6 from the first term and multiply the last by it:

x^2 - x - 72

Factoring this, we see that 72 = 8*9 and 8 - 9 = -1, so

x^2 - x - 72 = (x + 8)(x - 9)

Replacing x with 6x and pulling out common factors,

(6x + 8)(6x - 9)

becomes

(3x + 4)(2x - 3)

This is the desired factorization.

Now I'll prove that it works:

Suppose you are factoring

ax^2 + bx + c

You factor

x^2 + bx + ac = (x-m)(x-n)

(In my example, m = -8, n = 9.)

Now you replace x with ax:

(ax-m)(ax-n) = (ax)^2 + b(ax) + ac

We'll have to assume that when you remove common factors, the product
of the factors you divide out is equal to a. (In my example I took
out 2 in the first factor and 3 in the second, whose product is 6.)
I think this is a consequence of the assumption that a, b, and c have
no common factors, but I'm not going to bother to prove it.

With this assumption, the resulting factorization is the same as

(ax-m)(ax-n)/a = [a^2x^2 + abx + ac]/a = ax^2 + bx + c

So if you can follow your process, you have indeed factored the
original trinomial.

Now let's take it in reverse. Suppose that you can factor your
original trinomial:

ax^2 + bx + c = a(x-p)(x-q)

so that we have

b = -a(p+q)
c = apq

(In my example, p=-4/3, q=3/2. I am only expecting them to be
rational, not integers. When we multiply by a, we know that apq is an
integer so the product of the denominators of the roots, in reduced
form, must divide a.)

Then we find that

x^2 + bx + ac = a[x^2/a + b/ax + c]
= a[a(x/a)^2 + b(x/a) + c]
= a[ay^2 + by + c]   where y=x/a
= a^2(y-p)(y-q)
= a^2(x/a-p)(x/a-q)
= (x-ap)(x-aq)

Note that our m and n from above are ap and aq, which will be
integers.

Therefore, if you can factor the given trinomial (using integers), you
can also factor x^2 + bx + ac, and will be able to follow your
procedure. So it looks as if your method is a valid method that will
apply to any factorable trinomial.

Looking in our archives to see if we are aware of this method already,
I found that the method given here is somewhat equivalent, in that it
has you factor ac into factors whose sum is b, which is just what you
do to factor x^2 + bx + ac:

http://mathforum.org/library/drmath/view/52878.html

The following pages also teach the same "AC method," which I probably
prefer to yours just because it is easier to explain why it works, and
doesn't have you doing nameless transformations on an equation for no
obvious reason:

AC Method - Math Help
http://mcraefamily.com/MathHelp/factoring3c.htm

AC Method - Southwest Texas State University
http://www.swt.edu/slac/math/ACMethod.html

Here's the only page where I found your method, also called the "AC
method":

Wikipedia
http://www.wikipedia.com/wiki.phtml?title=talk:Polynomial

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

```
Date: 02/25/2002 at 19:14:28
From: John
Subject: Explanation for this method of factoring

Is there an easier way you can prove this method? I don't quite get
your explanation (too many variables and very complicated).

Thank you.
```

```
Date: 02/25/2002 at 22:56:31
From: Doctor Peterson
Subject: Re: Explanation for this method of factoring

Hi, John.

Sorry, this is how things are proved in algebra. The parameters a, b,
and c allow us to talk about any possible quadratic trinomial, and the
roots m and n allow us to talk more easily about factoring it without
having to write them in terms of the quadratic formula, which would be
even more confusing. So all the variables are there for good reasons.
But if you read through what I wrote one step at a time and don't let
it overwhelm you, you should (at least when your algebraic
understanding is more mature) be able to follow it without much
trouble.

But until you can do that, it may help if we look at an example to see
what is happening, avoiding all the parameters. Let's take my example,
where we factor

6x^2 - x - 12

The modified trinomial is

x^2 - x - 72 = (x + 8)(x - 9)

Replacing x with 6x, this is

(6x)^2 - (6x) - 72 = (6x + 8)(6x - 9)
= 2(3x + 4) * 3(2x - 3)
= 6(3x + 4)(2x - 3)

That's your method. But we can rearrange the left side to look like
this:

6*6x^2 - 6*x - 6*12 = 6(6x^2 - x - 12)

Do you see what this shows? I'll put it all in a different order to
make it clearer. If we take our original trinomial and multiply it by
6 (the coefficient a), we get

6(6x^2 - x - 12)

Then we can distribute the 6:

6^2 x^2 - 6x - 72

Now express this in terms of 6x:

(6x)^2 - (6x) - 72

To make it easier, let's introduce a new variable y = 6x:

y^2 - y - 72

Now we can factor this:

(y + 8)(y - 9)

Having done that, we can replace y with 6x again:

(6x + 8)(6x - 9)

Now we can factor out a 2 from the first factor and a 3 from the
second:

6(3x + 4)(2x - 3)

But this is equivalent to what we started with:

6(6x^2 - x - 12) = 6(3x + 4)(2x - 3)

Divide both sides by 6, and we've factored our trinomial:

6x^2 - x - 12 = (3x + 4)(2x - 3)

That's what you are really doing, with the magic removed.

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

```

```Date: 04/23/2009 at 12:49:59
From: Li
Subject: Factoring Trinomials

Dear Dr. Math,

In refereeing for a journal article I came up with the following:
If y = ax then ax^2 + bx + c = (1/a)(y^2 + by + ac) which gives a
quick and easy method of factoring ax^2 + bx + c.  I thought that

My explanation above may be shorter and easier for students to
understand.  Also, for an experienced user, the method can be presented
in a very quick and direct form:

ax^2 + bx + c = (1/a)(ax + ?)(ax + ?) where the two ?'s should multiply
into ac and sum into b.

Best regards,

Li

```

```Date: 04/23/2009 at 23:25:20
From: Doctor Peterson
Subject: Re: Factoring Trinomials

Hi, Li.

I've never until now heard an explanation of this method that actually
justifies it and doesn't just tell the student to magically replace
one trinomial with another; both of your approaches here make a lot of
sense, and may change it from a method I avoid into one I may end up
teaching as my preferred method.  (My math department officially
forbids teaching the "slip and slide" method, as they call it, because
it does not involve transforming the expression into a series of
equivalent expressions, and so is mathematically unjustified.)  I
currently teach the "ac" method, which I learned for the first time in
the course of answering this question; that method is taught in the
texts I've used since then, along with trial and error.  Here is my
explanation of it:

Finding a Single Pair of Factors

Let me fill in the details in what you wrote, to see how I would
explain it to students.  I generally start by showing an example, and
then generalize; so I'll work with the trinomial 6x^2 - x - 12.

First, for students able to handle a change of variables: We choose
(for no obvious reason, except that it turns out to work) to rewrite
the equation in terms of a new variable u = 6x.  This means we are
replacing x with u/6:

6x^2 - x - 12 = 6(u/6)^2 - (u/6) - 12
= 1/6 u^2 - 1/6 u - 12

Now we factor out 1/6, which means we see 12 as 72/6:

u^2 - u - 72
= ------------
6

Now we just have to factor a trinomial of the easy kind; we get

(u - 9)(u + 8)
= --------------
6

We have to put the answer in terms of x, so we replace u with 6x:

(6x - 9)(6x + 8)
= ----------------
6

Now we factor out the common factor from each factor here, and cancel
those with the 6:

3(2x - 3) 2(3x + 4)
= -------------------
6

= (2x - 3)(3x + 4)

This, as you say, is the same method as on the page you refer to,
except that every step is justified and we aren't just jumping from
one trinomial to one that is not equal to it.

Your second approach doesn't require the second variable, which
beginning students would have trouble with; here's how I'd explain it:

We try putting 6x in each of the prospective factors, just as we put x
in each factor in the easy case; but we realize we have to divide that
by 6 to make it come out right, so we write

(6x + _)(6x + _)
6x^2 - x - 12 = ----------------
6

What's written so far makes sense, since we have 36x^2 / 6 = 6x^2.
Now, just as in the simple case, we look at how we can fill in the
blanks.  Call them p and q for now.  The last term of the trinomial
will be pq/6, so (pq)/6 has to equal -72, so pq must equal -12.
The middle term will be

(6x*q + p*6x) / 6 = (p + q)x

so p + q has to equal -1.  As usual we find that p and q are -9 and 8,
so we have

(6x - 9)(6x + 8)
6x^2 - x - 12 = ----------------
6

and from here we do the same as in the other method, factoring out the
common factor.

The nice thing here is that, like the ac method I usually teach, it
centers around the same process as in the first case (finding a pair
of numbers whose sum and product are known), but it doesn't require
the process of factoring by grouping, and it's easy to see why it
works (though not, perhaps, how you would think of it!).  So I'm going
to try this last approach with my next class and see if they like it.

Thanks for your contribution!  If you have any more thoughts (and
especially if I've missed a good way to explain it) please let me know.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Polynomials

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