Factoring TrinomialsDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/