Go Back > Fun Stuff > Puzzles

Thread Tools
Old 2003-08-27, 06:31   #1
asdf's Avatar
Sep 2002

1111002 Posts
Default Factoring Problem

I'm having trouble with this problem:

Factor: (4)(a^2)(c^2) - (a^2 - b^2 + c^2)^2

asdf is offline   Reply With Quote
Old 2003-08-27, 07:59   #2
andi314's Avatar
Nov 2002

2×37 Posts

(4)(a^2)(c^2) - (a^2 - b^2 + c^2)^2 =

- (a + b + c)·(a + b - c)·(a - b - c)·(a - b + c)

hope i could help you
andi314 is offline   Reply With Quote
Old 2003-08-29, 06:10   #3
asdf's Avatar
Sep 2002

22×3×5 Posts

I have checked and the answer is correct, and I have accidentally found a way to find the factors, but it was through major guessing. Could anyone show how to factor this out fully?
asdf is offline   Reply With Quote
Old 2003-08-29, 06:34   #4
cheesehead's Avatar
"Richard B. Woods"
Aug 2002
Wisconsin USA

22·3·641 Posts

There are online factorers. The one I frequently use is "Factoris" at It can factor an integer, a rational number, a polynomial or a rational function.

When given:


Factoris returned:

Warning. You have entered an ambiguous formula, whose interpretation by wims may differ from what you meant. Please check the formula.

Please always use ``*'' for multiplication, and always enclose function arguments in parentheses. (For experts. If
you don't want wims to interprete your expression, start it with the string ``1-1+''.)

Factorization of the polynomial 4*(a^2)*(c^2)-(a^2-b^2+c^2)^2 into irreducible components over Z :

-a4 + (2c2 + 2b2) a2 + (-c4 + 2b2 c2 - b4)= -(c-b-a)(c-b+a)(c+b-a)(c+b+a)

Remark. Factorization of multivariate polynomials is only implemented over Z for the time being.
In the next-to-last line, Factoris superscripts the exponents, but that formatting is lost in my cut-and-pasting the text here. That is, what is meant is -a^4 + (2c^2 + 2b^2) a^2 + (-c^4 + 2b^2 c^2 - b^4) = -(c-b-a)(c-b+a)(c+b-a)(c+b+a)

Factoris doesn't use "*" for multiplication in its output even though it wants you to use it in input. :)

Of course, Factoris doesn't tell you how to factor an expression. But one might get clues from its answers.
cheesehead is offline   Reply With Quote
Old 2003-08-30, 17:56   #5
axn's Avatar
Jun 2003

5×1,039 Posts
Default Re: Factoring Problem

Originally Posted by asdf
I'm having trouble with this problem:

Factor: (4)(a^2)(c^2) - (a^2 - b^2 + c^2)^2

Soln: (I am omitting the * for multiplication)

(4)(a^2)(c^2) - (a^2 - b^2 + c^2)^2

==> X^2 - Y^2, where X = 2ac and Y = (a^2 - b^2 + c^2)

==> (X+Y)(X-Y)

==> (2ac + a^2 - b^2 + c^2)(2ac - a^2 + b^2 - c^2)

==> [(a+c)^2 - b^2] [b^2 - (a-c)^2]

==>[(a+c+b)(a+c-b)] [(b+a-c)(b-a+c)]
axn is online now   Reply With Quote

Thread Tools

Similar Threads
Thread Thread Starter Forum Replies Last Post
Programming and factoring problem lavalamp Puzzles 112 2014-10-05 20:37
Factoring problem RedGolpe Factoring 9 2008-09-02 15:27
Problem with P-1 factoring... VolMike Software 5 2007-07-26 13:35
Prime95 v24.14 P-1 Factoring problem harlee Software 1 2006-12-19 22:19
Problem trial factoring + 64 bit EPF Hardware 2 2005-06-26 04:12

All times are UTC. The time now is 08:38.

Sun Dec 5 08:38:18 UTC 2021 up 135 days, 3:07, 0 users, load averages: 1.08, 1.25, 1.38

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.