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Factoring a Trinomial

Date: 08/22/2003 at 04:09:26
From: Jen
Subject: Factoring a trinomial

What is factoring by grouping? When factoring a trinomial, why is it 
necessary to write the trinomial in four terms?

2x^3 - 5 x^2 + 3 x
First, we factor out x such that
2x^3 - 5 x^2 + 3 x = x (2x^2 - 5 x + 3)
We obtain the form ax^2 + b x + c where a = 2, b = -5, and c = 3
Because ac = (2) (3) = 6, and b = -5, we need two integers with a 
product of 6 and a sum of -5. These integers are -2 and -3. 
Replacing -5 x by -2 x -3 x, we obtain a four-term polynomial
2x^3 - 5 x^2 + 3 x = x (2x^2 - 5 x + 3) = x (2x^2 - 2 x -3 x + 3)
= x [2x (x - 1) - 3 (x - 1)]
By grouping, we have
x [2x (x - 1) - 3 (x - 1)] = x (x - 1) (2x - 3)

Date: 08/22/2003 at 13:09:10
From: Doctor Peterson
Subject: Re: factoring a trinomial

Hi, Jen.

I don't know that I would say any of this is "necessary"; there are 
several strategies you can use to factor, and you might just be smart 
enough to see the answer immediately. (I'm not!) So the question is, 
how does the technique you are being taught work? I explained the 
method here:

   Finding a Single Pair of Factors 

and here I showed why it works (actually two variants):

   Factoring Quadratics When a Doesn't = 1 

   Factoring Trinomials 

Actually, I think we can explain the basic method as you and I present 
it more easily. We have

  ax^2 + bx + c

and we find p and q so that

  pq = ac
  p + q = b

Then we rewrite the quadratic as

  ax^2 + (p + q)x + pq/a

  ax^2 + px + qx + pq/a

Now we factor out ax from the first pair of terms and q from the 

  ax(x + p/a) + q(x + p/a)

and then factor out the "x + p/a" from both terms:

  (ax + q)(x + p/a)

or, for the sake of symmetry,

  a(x + q/a)(x + p/a)

Since it may not be true that both p and q are divisible by a, this 
may not be exactly how the answer looks using integers; but we have 
seen that writing b as p + q does always allow us to factor by 
grouping (that is, to find common factors in two pairs of consecutive 
terms, and then factor the result). And anything that allows us to 
finish off the problem so easily, while perhaps not strictly 
"necessary," is certainly welcome!

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

- Doctor Peterson, The Math Forum 
Associated Topics:
High School Basic Algebra
High School Polynomials
Middle School Algebra
Middle School Factoring Expressions

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