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Partial factorization?
01-26-2018, 11:30 AM (This post was last modified: 01-26-2018 05:02 PM by DrD.)
Post: #10
RE: Partial factorization?
factors(16*x^8+8*x^4+1); ==> [2*x^2-2*x+1,2,2*x^2+2*x+1,2]; Unfortunately, (4*x^4+1)^2, is not included in the results vector.

This is an important factorization. A problem solving process I was validating, led to finding the square root of 16*x^8+8*x^4+1. By hand, finding the square root of an expression squared, is easy. I was using the prime to confirm, step by step, a lengthy problem, where understanding the problem solving method is just as important as the solution to the problem.

Can you think of a way to factor 16*x^8+8*x^4+1, (using the prime), that does include (4*x^4+1)^2?
I needed it to find a square root, and Didier's approach gets there, but more in general, I haven't found a command that includes a full factorization chain.

Thanks,

-Dale-
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Messages In This Thread
Partial factorization? - DrD - 01-25-2018, 10:26 AM
RE: Partial factorization? - Joe Horn - 01-25-2018, 01:15 PM
RE: Partial factorization? - DrD - 01-25-2018, 02:42 PM
RE: Partial factorization? - Arno K - 01-25-2018, 01:35 PM
RE: Partial factorization? - Arno K - 01-25-2018, 05:27 PM
RE: Partial factorization? - DrD - 01-25-2018, 06:19 PM
RE: Partial factorization? - DrD - 01-25-2018, 10:46 PM
RE: Partial factorization? - parisse - 01-26-2018, 06:32 AM
RE: Partial factorization? - DrD - 01-26-2018 11:30 AM
RE: Partial factorization? - parisse - 01-26-2018, 04:43 PM
RE: Partial factorization? - DrD - 01-26-2018, 05:07 PM
RE: Partial factorization? - Rudi - 01-26-2018, 07:13 PM
RE: Partial factorization? - parisse - 01-26-2018, 07:32 PM
RE: Partial factorization? - DrD - 01-26-2018, 08:03 PM



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