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Math Help - Proving a polynomial has 3 roots, what does it means?

  1. #1
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    Question Proving a polynomial has 3 roots, what does it means?

    Hi,

    I am asked to prove a 4th degree polynomial:
    p(x) = x^4+x^3-3

    Has at least 2 roots...


    And that some other equation doesn't roots in interval (-oo,-2)
    and that it has 3 roots in (-oo,+oo)

    what does it means that an equation has roots? or to find it?
    like with the quadratic formula that you find the +- x's ? (is has either 1 or 2 right?)

    Is there a simple way to find roots of such equations (with x^3 or higher degree)?

    Thanks (again).
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  2. #2
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    Re: Proving a polynomial has 3 roots, what does it means?

    Hey ryu1.

    Hint: Consider the second derivative being 0 and how that correlates with the information about the roots.

    Also do the roots have to be real or can they be complex as well?
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  3. #3
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    Re: Proving a polynomial has 3 roots, what does it means?

    what does it means that an equation has roots? or to find it?
    A root of a polynomial f(x) is a number a such that f(a)=0. For example the roots of a quadratic equation is given by the quadratic formula. For a more concrete example, the number a=1 is a root for the polynomial f(x)=x^3-1 since f(a)=a^3-1=1-1=0.

    Is there a simple way to find roots of such equations (with x^3 or higher degree)?
    Not in general. There are some truly monstrous equations for general polynomials of degree 3 and 4. However, it has been proven that such an formula is impossible for degrees 5 and higher (not just unknown, but cannot possibly exist). So your best bet in these questions is to look for properties for your specific polynomial, not a general one. I cannot guarantee I can give you an answer which you can understand without knowing how much you know, but here is an attempt.

    I am asked to prove a 4th degree polynomial:
    p(x) = x^4+x^3-3
    Has at least 2 roots...
    This approach is different from chiro's, so you can choose what you like. First, we note that p(x)\to \infty as  x\to \pm \infty, so there is some number N>0 such that f(-N)>0 and f(N)>0. Next, we see that p(x) is negative at some point, for example p(0)=-3. Applying the intermediate value theorem on [-N, 0] and [0,N] gives the result.
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