If P(x) is a polynomial of odd degree, show that the equation p(x)=0 has at least one solution.
It Doesnt because its x=0 and 0 doesnt have a soulution because its nothing but just a plain old 0.and it has no function either.:)
0 is like not even a numbe or whatever u wanna call it.but its like its just there for the heck of it.
and because no number will ever be less than 0 unless your using a negative number.
Your answer is not clear to me, I cannot translate my thoughts into a mathematical solution. If the polynomial is defined and continuous on some closed interval [a,b] and f(a) and f(b) have opposite signs then f(c)=0 for at least one point c in (a,b). Odd integers: 1, 3, 5, 7... 2n-1 is the formula describing this arithmetic sequence.
@zepdrix
Remember what makes an odd function special? It satisfies this property: \(\large\rm f(-x)=-f(x)\)
Woops, I mean intermediate value theorem. That's the one, right?
Zep I think it has nothing to do with that. It's about odd degree polynomial like this F(x) = x^3 + x^2 + x
Oh oh odd `degree`, woopssss :((
According to the fundamental theorem of algebra we will have At least 3 solutions for an odd degree polynomial like the example I listed
For large values of x will the function approximately equal to the first term of the polynomial and how can I show mathematically that the function is sometimes positive and sometimes negative in general not for a specific odd degree polynomial.
Okay so there are 2 things that effect the nature of the solutions of a polynomial Consider the following polynomial x^3 + x^2 + 9 Now 1st you will need to find the factors of 9 and then the factors of (1) the leading coefficient Q 9: +-1, +-3, +-9 P 1: +-1 And then you will have to divide Q/P factors to define the possible solutions (so in general those Q and P) are what can determine the nature of the solutions whether complex of positive or negative solutions.
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