College Algebra
posted by Sydney .
I'm working on vertical asymptotes when factoring 6x^2  6 (denominator). I came up with 6(x^21). Then I was trying to find the zeros of the denominator but 6 does not = 0 and x^21=0 then I got x^2=1 and x = plus or minus sqrt 1. Can you tell me what I am doing wrong?

(6x²6)
=6(x²1)
=6(x+1)(x1)
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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