Question 145103: If f(x)=x^3-2x^2-7/x^6-2x^5+6x^2-4 how many vertical asymptotes does it have. I try to factor the bottom but have no luck am I missing something?
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website! Your attempt to factor the denominator shows that you have the right idea. Any real zero  of the denominator polynomial will give an asymptote with equation  .
Since there is no general solution for a 6th degree polynomial equation, you are stuck with using the Rational Zero Theorem, reproduced below courtesy of Cliff's Notes:
If a polynomial function, written in descending order, has integer coefficients, then any rational zero must be of the form ± p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.
For the given situation, your constant term factors are 1, -1, 2, -2, 4, and -4, and your lead coefficient factors are 1 and -1. Hence the possible rational zeros of the denominator are: 1, -1, 2, -2, 4, and -4.
Use either polynomial long division or synthetic division to test each of these possible zeros. The number will be a zero if the results of dividing the polynomial by  leaves a zero remainder. If you find a zero, start testing the remaining possibilities on the quotient from the division where you found the zero. If you are lucky enough to find 4 of them, your quotient will be reduced to a quadratic that you can easily solve in general. Hopefully, you will find at least two so that you can get down to a quartic. There is a general solution for the quartic (look it up in Wikipedia), but I wouldn't want to wish that horror on my worst enemy.
Good Luck.
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