I do not accept this problem as being true at all.
n= 3 is an odd number but I do not believe that a cubic equation of this form can have at most 2 roots. Just graph the cubic equation y = y = x^3-5x-1 and see how many roots it has. Also a polynomial of odd degree can never have two real roots as complex roots comes in pairs. So suppose the degree of a polynomial is 7. The the number of complex roots can be 0 (and then there are 7 real roots) or the number complex roots are 2 (so there are 5 real roots) or the number of complex roots are 4 (and there are 3 real roots) or the number of complex roots are 6 (and there is 1 real root)
So (in theory) it is not wrong to say that there are at most 2 real roots in a polynomial of odd degree of the form x^n +ax+b it is more exact to say that there is 1 real roots in a polynomial of odd degree of the form x^n +ax+b. The problem is that many odd degree polynomials of the form y = x^n +ax+b have more than 1 root.
Is there a problem as well with the statement of even degree polynomials?
Have you omitted some conditions? For example, must x be real (so you are talking about real roots), and are a and b assumed to be positive? Possibly it may be true under some such conditions. But we can't help you prove it without (a) being sure of the real problem, and (b) seeing where you need help in your work.
You put this under calculus; is there a reason to suppose that calculus methods will be of use? What methods have you learned that you think are applicable?
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