How do you factor #1 - 4y^2#?

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Answer 1 · 2018-03-03T09:25:45

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

#"this is a "color(blue)"difference of squares"#

#•color(white)(x)a^2-b^2=(a-b)(a+b)#

#"here "a=1" and "b=2y#

#rArr1-4y^2=(1-2y)(1+2y)#

Answer 2 · 2018-03-03T09:27:32

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

#color(blue)(1-4y^2)#

#=>color(blue)((1)^2 - (2y)^2)#

We have an identity for #color(green)(a^2 - b^2)# which is equal to #color(green)((a+b)(a-b))#.

So , #color(red)((1)^2 - (2y)^2)#

#=> color(red)((1+2y)(1-2y))#