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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.
Factor:
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:
(2y+3)(y+2)
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
common.)
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:
http://mathforum.org/dr.math/faq/faq.learn.factor.html
Good luck with factoring!
- Doctor Kimberly, The Math Forum
http://mathforum.org/dr.math/
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