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### Factoring Trinomials By Grouping

```Date: 11/16/2004 at 16:29:55
From: Laura

Dear Dr. Math,

I remember seeing a grouping trick for solving quadratic equations by
factoring the trinomial.  Somehow the middle term is broken up into
two parts.  Then the first two parts are factored and the second two
parts are factored.  After that somehow the factored parts match up
and you can pull out the common factor and then you have a factored

For example, 6x^2 + 7x + 2 should factor into (2x + 1)(3x + 2).

The grouping method separates the middle term into 3x and 4x, giving

6x^2 + 3x + 4x + 2

Factor first and second parts:

3x(2x + 1) + 2(2x + 1)

Pull out the common factor:

(2x + 1)(3x + 2)

What trick would I use to figure out how to separate the middle term?
Also, is there a way to tell if this method will work before you try
it?

```

```
Date: 11/16/2004 at 16:42:18
From: Doctor Vogler
Subject: Re: Trick for factoring quadratics.

Hi Laura,

Thanks for writing to Dr. Math.  Yes, this is how I first learned to
factor quadratics.  You first multiply together the first and last
coefficients.  For example, for

6x^2 + 7x + 2

you get 12.  Next, you look for two factors of that number that add up
to the middle number.  In this case, the two factors are 3 and 4.
Then you split up the middle term into those two numbers, and I think
you can do the rest.

Be warned:  You have to be careful of signs.  For example, if the
product is negative, then one of the factors will be positive and the
other will be negative.  If the product is positive, but the middle
number is negative, then both factors must be negative.  For example, in

10x^2 + 26x - 12

the product is 10(-12) = -120, and the factors that add to 26 are +30
and -4.  In

8x^2 - 26x + 21

the product is 8*21 = 168, and the two factors that add to -26 are -12
and -14.

Yes, there is a way to determine if the polynomial is factorable.  The
way to test is by computing the discriminant of the polynomial, which is

b^2 - 4ac

ax^2 + bx + c = 0.

If the discriminant is a perfect square, then there are factors that
add up to the middle number.  If not, then there are not.  If there
are not factors, you can still solve the equation by using the
quadratic formula or completing the square.

back and show me what you have been able to do, and I will try to
offer further suggestions.

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra

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