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Find factorials

 
 
Reply Fri 10 Dec, 2004 09:18 pm
Res.All,
I am a student of the 10th standard. I have got few problems in Mathematics. I want solution of the following questions.

Please give the factors of:-
1. a^4 - 4ab + 4b^2 -1

2. x^3 + 3x^2 - 6x -6

3. 16a^2 - 40ab + 16b^2 + 6b -1

4. x^4 - 8x^2y^2 + 4y^2


I will be grateful to you all if you can answer these questions.
Thanking you
Sorry for taking your time,
Yours
Chandani
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Type: Discussion • Score: 1 • Views: 705 • Replies: 9
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Perdition
 
  1  
Reply Sat 11 Dec, 2004 06:51 am
would be more graceful to just ask how to solve the first one if you are unfamiliar with how it is done and from that working method you could manage the rest, if there are any discrepancies in the development of the other answers then you could definetly ask on here how it can be resolved
0 Replies
 
chandani
 
  1  
Reply Sat 11 Dec, 2004 06:12 pm
Res.Sir,
I thank you very much for replying to my questions. I have solved almost all the difficulties like this in my book, but I could not solve only these 4. I know the method, but here some logic is necessary, which does not come into my mind. I am sorry if I am wrong in any way. But I am keen on getting the answers of these questions.
Please help,
Yours
Chandani
0 Replies
 
Vengoropatubus
 
  1  
Reply Sat 11 Dec, 2004 08:25 pm
I don't suppose someone could explain the method real quick anyway...
0 Replies
 
engineer
 
  1  
Reply Sun 12 Dec, 2004 08:58 pm
Tough ones...
For one, three and four, you have to try and group the terms so that you can pull common terms out of them. It is more of an art form than exact science. I didn't find any factors for these. You solve the second one via synthetic division. It has rational possible solutions of +/- 1,2,3,6. Since none of these works, you cannot factor this one further.
0 Replies
 
engineer
 
  1  
Reply Sun 12 Dec, 2004 09:31 pm
Solution for #4
I solved number four by assuming that the factors were of the form

(ca + cb + c)(ca + cb + c) where all the c's are different constants.

I chose that form based on the terms in the equation. We had to have a^2, b^2, b and a constant term.

I ended up with six equations and six unknowns. After a little simplificaiton, I got (4a - 2b - 1)(4a - 8b + 1)
0 Replies
 
engineer
 
  1  
Reply Mon 13 Dec, 2004 07:39 am
Sorry, the reply above was for number 3. For number 4, you can break up the -8x^2y^2 in half to get...

x^4 - 4x^2y^2 + 4y^2 - 4x^2y^2
(x^2 - 2y) ^2 - 4x^2y^2
(x^2 - 2y + 2xy)(x^2 - 2y - 2xy)
0 Replies
 
engineer
 
  1  
Reply Mon 13 Dec, 2004 07:47 am
Are you sure you typed the first one in correctly? It makes a lot more sense if the first term is a^2.
0 Replies
 
chandani
 
  1  
Reply Tue 14 Dec, 2004 07:46 pm
Res.All,
I thank you all for the replies. I am very sorry for taking your time,

Yours
Chandani
0 Replies
 
engineer
 
  1  
Reply Wed 15 Dec, 2004 09:20 am
No Problem
Thanks for posting some good problems!
0 Replies
 
 

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