The Math Forum

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

Factoring Trinomials

Date: 05/28/99 at 15:52:17
From: Becky
Subject: Factoring Trinomials

I'm having trouble factoring trinomials with a number in front of the 
first variable. For example: 6y^2-19y+10. If you could please explain 
to me how to figure this out I would appreciate it. Here's what I've 
done so far:

        (3y + ? )(2y + ? ) 

Please help. Thanks!

Date: 06/03/99 at 18:22:54
From: Doctor Kimberly
Subject: Re: Factoring Trinomials

Hello Becky,

Factoring used to give me a lot of trouble also. Here are a few things 
to keep in mind:

You have to find two numbers such that the sum of twice the first 
number and three times the second number equals -19, but when you 
multiply the two numbers, you get +10. That should give you a hint 
about the signs of the numbers. The first two numbers that you have 
obtained are right so far, but keep in mind that they could have 
been 6y and y.  

You can also plug in various numbers and use the F-O-I-L method (to 
find numbers that work). This is the easier way to do it. If this is 
the method that is not working for you right now, you can also use 
the guess-and-check strategy (but you have to know how to factor by 
grouping to use this method). Write your problem in this manner:

6y^2 + (__ + __) + 10

If you want to use the guess-and-check method, you have to find two 
numbers that give you a product of 60 (with this method, you find the 
product of the two outer numbers, 6 and 10), but they have to be the 
same numbers that give you -19 when you add them. You can make a table 
like this:

Factors of 60 | Sum of factors
 1, 60        |  60 + 1 = 61
 2, 30        |  30 + 2 = 32
 3, 20        |  20 + 3 = 23 ... and so on

Once you find the numbers, you can factor by grouping. Here is an 
example of that:

     2y^2 + 7y + 6

Since 2*6 = 12, find the factors of 12:

     Factors of 12 | Sum of factors
      1, 12        |  12 + 1 = 13
      2, 6         |   6 + 2 = 8
      3, 4         |   4 + 3 = 7

3+4 is 7, so use them: 

     2y^2 + 7y + 6
     2y^2 + (3+4)y + 6
     2y^2 + 3y + 4y + 6

Group the terms that have a common factor:

     (2y^2+4y) + (3y+6) --> Here, you could have also put 4y with 6
                             and 2y^2 with 3y.


     2y(y+2) + 3(y+2)

Notice that both of the groupings have a (y+2) term. You can use the 
distributive property to write the trinomial this way:


And this is your answer. You can multiply to check it out. (If you 
group the terms the other way, the two groups will have (2y+3) in 

If you are not comfortable with this method or if you are having 
trouble with the F-O-I-L method, please write back and let us know 
what aspect of factoring is giving you trouble. You can also get 
additional help at this location on our faq:   

Good luck with factoring!

- Doctor Kimberly, The Math Forum   
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- The Math Forum at NCTM. All rights reserved.