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

Jun 20, 2015

$y = 2 {x}^{2} - 2 = 2 \left({x}^{2} - 1\right) = 2 \left(x - 1\right) \left(x + 1\right)$

So this has zeros for $x = \pm 1$

#### Explanation:

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

$y = 2 {x}^{2} - 2 = 2 \left({x}^{2} - 1\right)$

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

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

to deduce:

$\left({x}^{2} - 1\right) = \left({x}^{2} - {1}^{2}\right) = \left(x - 1\right) \left(x + 1\right)$

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