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

Jan 26, 2016

Treat as a quadratic in ${y}^{n}$ to find:

$4 {y}^{2 n} - 4 {y}^{n} - 3 = \left(2 {y}^{n} + 1\right) \left(2 {y}^{n} - 3\right)$

#### Explanation:

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

Find a pair of factors of $A C = 4 \cdot 3 = 12$ that differ by $B = 4$

The pair $6 , 2$ works.

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

$4 {y}^{2 n} - 4 {y}^{n} - 3$

$= 4 {y}^{2 n} - 6 {y}^{n} + 2 {y}^{n} - 3$

$= \left(4 {y}^{2 n} - 6 {y}^{n}\right) + \left(2 {y}^{n} - 3\right)$

$= 2 {y}^{n} \left(2 {y}^{n} - 3\right) + 1 \left(2 {y}^{n} - 3\right)$

$= \left(2 {y}^{n} + 1\right) \left(2 {y}^{n} - 3\right)$

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