Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Quadratic Trinomials

Date: 14 Aug 1995 13:28:05 -0400
From: Anonymous
Subject: algebra

I'm reviewing some stuff for algebra 2, and I forgot QUADRATIC TRINOMIALS.
It says "factor as the product linear binomials or write prime."  Is there
some trick to figuring this out?  Help me out, please.
* means squared

2x* + 5x + 2

25x* - 10x - 3

3x* - 10x + 3


Date: 14 Aug 1995 14:17:07 -0400
From: Dr. Ken
Subject: Re: algebra

Hello there!

Well, the foolproof way that works every time is to use the Quadratic
Formula and then multiply through by the leading coefficient.  Oh, by the
way, the standard notation for "squared" is ^2.  For instance, in the first
problem, you have 2x^2 + 5x + 2.  Using the quadratic formula, we have 
x = (-b +- Sqrt{b^2 - 4ac})/2a 
  = (-5 +- Sqrt{5^2 - 4*2*2})/2*2 
  = (-5 +- 3)/4
  = -2 or -1/2

So following the pattern, we have the following factorization:
        (leading coefficient)(x - first root)(x - second root)
   = 2(x+2)(x+1/2)
   = (x+2)(2x+1), and that's the answer.

If you use the quadratic formula and you don't get a perfect square inside
the radical (what's inside the radical is called the discriminant) or you
get a negative sign, then you can write "prime".  I mean, you could try to
show off and write 

2x^2 + 4x + 1 = 2(x + 1 + Sqrt{2}/2)(x + 1 - Sqrt{2}/2),

but somehow I don't think that's what your teacher had in mind.  Good luck
with it!


Date: 15 Aug 1995 22:59:12 -0400
From: Anonymous
Subject: Re: algebra

Is there a shorter, easier, quicker way to do this?  We didn't use the
quadratic formula in my algebra class.  We did everything else though.

Date: 16 Aug 1995 15:57:47 -0400
From: Dr. Ethan
Subject: Re: algebra

Well, I must agree with Ken that the most reliable way is to learn the
quadratic formula but if you want another way then you need to factor.
And here is how you do it.  It is a little more complicated then the
quadratic formula but it is also a little easier to understand.

Here is the idea, look at 

2x^2 + 5x + 2

We need to find two binomials of the form (ax +b)
which when multiplied together equals the desired trinomial.

Well, let's assume they are (ax+b) and (cx + d)
then (ax +b)(cx+d) = a*c x^2 + (bc + ad)x + b*d

So a*c = 2
and b*d = 2
and (bc + ad) = 5

Well, to get this we realize that we can only choose 1 or 2 for any of the

Now you just experiment till you find a combo that works.


is one such combo because

(x+2)(2x+1) = 2x^2 + 5x +2

This is just one example.  Some are harder, but all factoring works on this
principle.  Write back if this isn't enough help.

Ethan Doctor On Call
Associated Topics:
High School Basic Algebra

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994-2013 The Math Forum