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.

How do you factor 4y^(2n)-4y^(n)-34y^(2n)-4y^(n)-3?