how do you factor y^4-1 completely?
first, it is a difference of two squares:
(y^2-1)(y^2+1)
Next, the first term is again the difference of two squares. If you allow imaginary numbers, then the second term is a difference of two squares (y+i)(y-i)
you should recognize the difference of squares
y^4-1
= (y^2 + 1)(y^2 - 1) , ahh, once more
= (y^2 + 1)(y + 1)(y-1)
To factor the expression y^4 - 1 completely, you can use a special factoring formula known as the difference of squares.
The difference of squares formula states that a^2 - b^2 can be factored as (a + b)(a - b).
In this case, y^4 - 1 can be written as (y^2)^2 - 1^2. Now you can see that the expression fits the difference of squares pattern.
Therefore, y^4 - 1 can be factored as (y^2 + 1)(y^2 - 1).
Now let's factor further.
The expression y^2 - 1 is a difference of squares as well. It can be factored as (y + 1)(y - 1).
So, the complete factored form of y^4 - 1 is (y^2 + 1)(y + 1)(y - 1).