How do you find the zeroes for #y=2x^2-2#?

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Answer 1 · 2015-06-20T16:39:04

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

So this has zeros for #x = +-1#

First separate out the obvious common scalar factor #2# to get

#y = 2x^2-2 = 2(x^2-1)#

Now #x^2-1 = x^2-1^2# is a difference of squares, so you can use the standard difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

to deduce:

#(x^2-1) = (x^2-1^2) = (x-1)(x+1)#

This will be zero when #x=1# or #x=-1#