How do you factor #16x^2-x^2y^4#?

1 Answer
mason m
May 12, 2016

#x^2(4+y^2)(2+y)(2-y)#

Explanation:

First, notice that both terms have a common factor of #x^2#, so we can factor it out:

#16x^2-x^2y^4=x^2(16-y^4)#

Focusing on just the #(16-y^4)#, notice that both of these are squared terms:

  • #16=4^2#
  • #y^4=(y^2)^2#

This will be useful since #(16-y^4)# is a difference of squares, which can be factored as: #a^2-b^2=(a+b)(a-b)#. Thus, we see that:

#x^2(16-y^4)=x^2(4^2-(y^2)^2)=x^2(4+y^2)(4-y^2)#

Note that #(4-y^2)# is also a difference of squares, since #y^2# is obviously squared and #4=2^2#. We see that:

#x^2(4+y^2)(4-y^2)=x^2(4+y^2)(2^2-y^2)=x^2(4+y^2)(2+y)(2-y)#

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