September 24th, 2017, 02:38 PM  #1 
Newbie Joined: Sep 2017 From: Canada Posts: 1 Thanks: 0  Factoring!
When factoring the RIGHT side of a trinomial if there are multiple correct answers then how is it decided which one is correct? For example y^24y12 The answer could either be (y6)(y+2) (y+6)(y2) (y4)(y+3) (y+4)(y3) (y1)(y+12) (y+1)(y12) I would assume the correct answer would be (y+4)(y3) but that actually factors the whole not just the right. The book states the correct answer as (y+2)(y6) But how is that decided? 
September 25th, 2017, 03:37 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,678 Thanks: 1339 
There is only one correct factorization ... $y^24y12=(y6)(y+2)$ ... all the rest will not yield the linear term $4y$ when expanded (multiplied). 
September 25th, 2017, 02:04 PM  #3  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,875 Thanks: 766  Quote:
Quote:
In general, in order to factor y^2+ ay+ b into (y+ u)(y+ v) you need to find two numbers, u and v, such that uv= b and u+ v= a. Here, a= 4 and b= 12. How can you factor 12? One way is (1)(12) but 1+ 12= 11, not 4. Another is, of course, (12)(1) but 12+ 1= 11, not 4. Another is 3(4) but 3 4= 1 not 4. (3)(4) is 12 but 3+ 4= 1, not 4. (2)(6)= 12 but 2+ 6= 4 not 4. (2)(6)= 12 and 2 6= 4! AHA!  

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