How do you factor #4y^(2n)-4y^(n)-3#?

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Answer 1 · 2016-01-26T00:23:49

Treat as a quadratic in #y^n# to find:

#4y^(2n)-4y^n-3=(2y^n+1)(2y^n-3)#

Without knowing anything about #n# we can treat this as a quadratic in #y^n#...

Find a pair of factors of #AC = 4*3 = 12# that differ by #B=4#

The pair #6, 2# works.

Use this pair to split the middle term and factor by grouping:

#4y^(2n)-4y^n-3#

#=4y^(2n)-6y^n+2y^n-3#

#=(4y^(2n)-6y^n)+(2y^n-3)#

#=2y^n(2y^n-3)+1(2y^n-3)#

#=(2y^n+1)(2y^n-3)#

If we knew more about #n# then further factorisation would be possible.