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 $x=-1$

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