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How do you quickly find the roots of a polynomial besides the factor theorem? For example, if I gave one the equation $f(x)=2x^4+3x^2-7x+8$, how could you find the roots of the polynomial in under 1 and a half minutes. Again, please don't suggest using the factor theorem.

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    Could you provide some context as to why you ask this question? Where this question originates? why you specify 90 sec to answer? – 2017-02-26
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    possible duplicate: [Is there a way to find all roots of a polynimial?](http://math.stackexchange.com/questions/133614/is-there-a-way-to-find-all-roots-of-a-polynomial-equation) – 2017-02-27
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    This task is one that not only is a computer better suited for, but there's not even a formula/ algorithm to get the exact solution for polynomials of degree greater than or equal to 5. Your choices for computational algorithms include Newton's method, Regula Falsi, Brent's method, [etc](https://en.wikipedia.org/wiki/Root-finding_algorithm). – 2017-02-27
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    $f(x)$ has no real roots. It can't have negative roots by the rule of signs, it can't have roots in $[0,1]$ by inspection ($7x \lt 8$), neither $\gt 1$ because it's strictly increasing up from $6$ in that range. – 2017-02-27
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    @BobbieD: you might find it instructive to learn something about [real algebraic geometry](https://en.wikipedia.org/wiki/Real_algebraic_geometry). Exact computation in the field of real algebraic numbers is one that computers are better suited to than we are, but neither we nor the computers are much hindered by not always having solutions by radicals (which is what the degree 5 limitation is about). – 2017-02-27
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    @RobArthan I'll take a look at that. Thanks. :) – 2017-02-27
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    I'm guessing that the time limit of "under 1 and a half minutes" comes from a test taking regime similar to ACT or GRE math exams. One should certainly have familiarity with the Rational Roots Thm. for polynomials with integer coefficients, but the OP seems to have picked this integer quartic mostly for the sake of illustrating their frustration. – 2017-02-27

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Your example is irreducible over the rationals, and has no real roots. Since it's a quartic, there are expressions for the roots in terms of radicals, but they are not pleasant.