how do you factor y^4-1 completely?

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#1

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)

(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)

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#2

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)

y^4-1

= (y^2 + 1)(y^2 - 1) , ahh, once more

= (y^2 + 1)(y + 1)(y-1)

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