How do you factor completely #w^2 - w^2 y^4#?

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How do you factor completely #w^2 - w^2 y^4#?

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Answer 1 · 2018-04-04T19:22:51

#w^2(1+y)(1-y)(1+y^2)#

Explanation:

This is just a difference of squares. When you have #a^2 - b^2#, it's is equal to #(a+b)(a-b)#. Just simply expand it to see why. In this case, it would be

#(w)^2 - (w*y^2)^2#

#(w- wy^2)(w + wy^2)#

#w^2(1-y^2)(1+y^2)#

#w^2(1+y)(1-y)(1+y^2)#

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How do you factor completely #w^2 - w^2 y^4#?

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