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August 2nd, 2017, 05:24 AM  #1 
Newbie Joined: Jul 2017 From: Earth Posts: 6 Thanks: 1  Need help checking/show method. Factorise a) [check] 6x^2  21xy + 8xz  28yz = 3x(2x  7y) + 4z(2x  6y) b) [method](x  1)^2  (y  2)^2 c) [check] p^2  y^4 = (p + y^2)(p  y^2) d) [method] 16a^2  (3b + 4c)^2 e) [check] 3ax + bx  3ay  by  9a  3b = 3ax  3ay  9a  3b + bx  by = 3(ax  ay  3a  3b) + b(x  y) Last edited by JoKo; August 2nd, 2017 at 05:45 AM. 
August 2nd, 2017, 05:26 AM  #2 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,453 Thanks: 489 Math Focus: Yet to find out. 
What have you tried? Let's start with just one question at a time first..

August 2nd, 2017, 05:55 AM  #3 
Newbie Joined: Jul 2017 From: Earth Posts: 6 Thanks: 1 
Well, for (d) I expanded the bracket and tried to factorise from there; $\displaystyle 16a^2  (3b +4c)^2$ $\displaystyle = 16a^2 + 9b^2 + 24bc + 16c^2$ $\displaystyle = 4(4a^2 + 6bc + 4c^2) + 3(3b^2)$ 
August 2nd, 2017, 07:07 AM  #4  
Senior Member Joined: Feb 2016 From: Australia Posts: 1,453 Thanks: 489 Math Focus: Yet to find out.  Quote:
$ (3b +4c)^2 = (3b + 4c)(3b + 4c) = (9 b^2 + 12bc + 12bc + 16c^2) = 1*(9 b^2 + 24bc + 16c^2) = ( 9 b^2  24bc  16c^2)$  
August 2nd, 2017, 07:32 AM  #5 
Newbie Joined: Jul 2017 From: Earth Posts: 6 Thanks: 1 
Thanks for pointing that out. Is the method correct ?

August 2nd, 2017, 08:19 AM  #6 
Senior Member Joined: May 2016 From: USA Posts: 855 Thanks: 345  I think you misunderstand the purpose behind factoring. (Factorizing is just a silly word.) The purpose is to get an expression into an equivalent expression that is more convenient for some purpose. There may be quite a few equivalent expressions. Which one is "correct" depends on what you want to do. So your expression for d would have been equivalent and would be therefore A "correct" answer if you had not made that sign error. In other words, your method is valid but not uniquely valid. I suspect that they want a different "correct" answer, namely an answer based on the method of difference of squares. $16a^2 = (4a)^2. \text{ And obviously } (3b + 4c)^2 \text{ is a square.}$ So $16a^2  (3b + 4c)^2 = \{4a  (3b + 4c)\}\{4a + (3b + 4c)\} =$ $(4a  3b  4c)(4a + 3b + 4c).$ Last edited by skipjack; August 4th, 2017 at 03:15 AM. 
August 2nd, 2017, 08:47 AM  #7  
Math Team Joined: Jul 2011 From: Texas Posts: 2,675 Thanks: 1337  Quote:
$3a(xy3) + b(xy3)$ $(xy3)(3a+b)$  
August 2nd, 2017, 09:04 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 18,232 Thanks: 1437 
(a) 6x²  21xy + 8xz  28yz = 3x(2x  7y) + 4z(2x  7y). That leads to (3x + 4z)(2x  7y). (c) Correct. (b) Use the difference of squares method (as per (c)). (e) 3ax + bx  3ay  by  9a  3b = (3a + b)x  (3a + b)y  3(3a + b) = (x  y  3)(3a + b) In hindsight, it can be seen that all the questions have been designed to be easy. It's a good idea to be well aware of the easiest methods and how to see whether they are immediately applicable. 

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