Posted by **Sydney** on Saturday, July 25, 2009 at 8:59pm.

I'm working on vertical asymptotes when factoring 6x^2 - 6 (denominator). I came up with 6(x^2-1). Then I was trying to find the zeros of the denominator but 6 does not = 0 and x^2-1=0 then I got x^2=1 and x = plus or minus sqrt 1. Can you tell me what I am doing wrong?

- College Algebra -
**MathMate**, Saturday, July 25, 2009 at 11:10pm
(6x²-6)

=6(x²-1)

=6(x+1)(x-1)

If the above expression is the denominator of a given function, there will be two vertical asymptotes, namely at the zeroes of the given expression, at x=1 and x=-1.

The expression equals zero at these two places since 6*0=0.

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