SOLUTION: I am having difficulty figuring this problem out. Could you please help me? :) 16x^4-41x^2+25=0

Question 289894: I am having difficulty figuring this problem out. Could you please help me? :)
16x^4-41x^2+25=0
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
Sure I can help. What have you tried? Where is your work?
There are four real answers two are negatives of the other two
You might guess that 16 (4^2) and 25 (5^2) has something to do with 2 of the roots.
Notice that 41=25+16
There might be some difference of squares involved since one answer is a negative of the other.
remember (a-b)*(a+b)=a^2-b^2
(4x-5)(4x+5)=16x^2-25
divide that into 16x^4-41x+25 gets
(x^2-1)
but that equals (x-1)*(x+1)
so we have (4x-5)*(4x+5)*(x-1)*(x+1)
We were very lucky

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