Finding zeros of a polynomial

12x^4 - 9x^3 - 58x^2 + 4x + 36

Is the only way to find the zeros of any given polynomial to guess and check or is there a more mechanical method like the quadratic formula that works on every single polynomial?

Please be advised this is for an beginner Algebra class so more advanced techniques cannot be used since I won't know what any of it means.

Re: Finding zeros of a polynomial

there is a "general formula" for polynomials of degree 2,3, and 4. be advised that this general formula is quite complex, and produces very nasty-looking solutions, in many cases. for polynomials of degree 5 or higher there is no "general formula" (although for "some" polynomials, we can still factor them easily).

your polynomial has integer coefficients. so the first thing one might check for is rational roots. these will be of the form ąp/q where q is an integer factor of 12, and p is an integer factor of 36.

that still gives a lot to check for, p might be 1,2,3,4,6,9,12,18 or 36, and q might be 1,2,3,4,6 or 12.

i come up with 1,2,3,4,6,9,12,18,36,1/2,1/3,1/4,1/6,1/12,2/3,3/4,3/2,9/2,9/4 and their negatives as rational roots to check. that 38 "easy checks".

entering this polynomial at Wolfram|Alpha yields: 12x^4 - 9x^3 - 58x^2 + 4x + 36 - Wolfram|Alpha

which indicates that the 4 roots of your polynomial are not rational. explaining how to get the "general solution" of a quartic is rather an advanced topic, if you feel up to it, see here:

Quartic function - Wikipedia, the free encyclopedia

Re: Finding zeros of a polynomial

So pretty much the only "reasonable" and sure way to solve these is with a calculator of some sort? I'm just afraid when I take Calculus soon and I don't know how to factor these polynomials I might be behind in class and fail.

Re: Finding zeros of a polynomial

calculus is not about "factoring" polynomials. it is about studying how certain "nice" functions behave. polynomial functions fall into the "nice" category, so they are used as non-trivial examples. here is typically what you would do in calculus to study p(x) = 12x^{4} - 9x^{3} - 58x^{2} + 4x + 36:

you would observe that for large x, the x^{4} term dominates, so that p(-x) and p(x) are both very large, for |x| large ("roughly-speaking, U-shaped, with some "wiggles" at the bottom").

you would also observe that p(-1) = 12 + 9 - 58 - 4 + 36 = -5 < 0, which tells you p crosses the x-axis at least twice (at least 2 real roots), once going down, and once coming back up.

since p(0) = 36 > 0, we know that we have at least 2 negative roots.

since p(1) = 12 - 9 - 58 + 4 + 36 = -15 < 0, which tells us p crosses the x-axis 2 more times (and since a polynomial of degree 4 can have at most 4 roots, this must be it), at some positive values, that is, we have 2 positive roots.

this is "qualitative" information, we did not actually FIND the roots, we just get "an idea of what p looks like" (sort of a curvy W shape). by looking at the "slope" of p, we can find the 2 places where p "bottoms out", and "the top of the hump" in-between. this can be used to "narrow down where the roots lie".

we can even use these techniques to get approximations of these roots. what we do is approximate the curve p by a tangent line to p, the tangent line has a linear equation, and we can solve those for x-intercepts. we then "feed that solution back into where we take the tangent" (i am playing fast and loose with what is known as Newton's method). this allows us to get "reasonable approximations" to the roots (which can be used for practical applications).

in terms of differentiating (finding slopes) and integrating (finding areas), the nice thing about polynomials is that we can go "term by term", breaking a polynomial into a sum of functions cx^{k}, and then "add the results". so polynomials are "nice to work with" in calculus, because they behave themselves (unlike, say, f(x) = 1/x, which gets "very naughty" around 0).

Re: Finding zeros of a polynomial

Good thing to "practice"; set one up for yourself; like:

(x + 1)(x + 2)(x - 4)(2x - 1) = 0

Multiply it out; you'll get:

2x^4 - 3x^3 - 19x^2 - 6x + 8 = 0

So solution to that one is x = -1, x = -2, x = 4 and x = 1/2 ; got that?

And watch what's happening when you're multiplying out.

This'll somewhat "familiarize" you with the mechanics...

Hope that helps you out...