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1. Explain how to solve x^2-2x-8>0
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\[x^{2}-2x-8>0\]
factor out the equation set it to zero, and solve for the x
everything good there but the answer
zeros are at \(-2\) and \(4\) and since \[x^2-2x-8\] is a parabola that faces up, it is negative between the zeros and positive outside of them
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you want to know where it is positive, outside the zeros is \(x<-2\) or \(x>4\)
in interval notation \((-\infty, -2)\cup (4,\infty)\)
ya....I just ended it at the solution of x.
Thanks for the help Both of you!
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