Thread: Odd Degree Polynomials and the Fundamental Theorem of Algebra

This is my problem:

__________________________________________-

1. How can the Fundamental Theorem of Algebra be used to show that any polynomial of
odd degree has at least one root?

Make sure that:

1. every claim is justified

2. a classmate reading your answer would understand

3. vocabulary is used correctly

4. the solution is vivid (there are no missing details)

5. your answer is concise (verbose answers are virtually impossible to understand)


__________________________________________

I don't even know where to start. Can someone please give me a step by step on the proof, as well as any tips? Thanks in advance.

Originally Posted by Freshmanmath

This is my problem:

__________________________________________-

1. How can the Fundamental Theorem of Algebra be used to show that any polynomial of
odd degree has at least one root?

Make sure that:

1. every claim is justified

2. a classmate reading your answer would understand

3. vocabulary is used correctly

4. the solution is vivid (there are no missing details)

5. your answer is concise (verbose answers are virtually impossible to understand)


__________________________________________

I don't even know where to start. Can someone please give me a step by step on the proof, as well as any tips? Thanks in advance.

By the Fundamental Theorem of Algebra, every polynomial with real coefficients has exactly as many complex roots as its degree (some of which may be repeated). Non-real roots always appear as complex conjugates, which means there is always an even number of non-real roots. So an odd degree polynomial must have an extra root which does not have a complex conjugate, so therefore must be real. Therefore, every polynomial with real coefficients of odd degree has at least one real root.

Thanks, that's very informative. Would it also be possible for you to say, give me a simple example and apply what you said step by step? I'm more of a visual learner, and I really want to get this down solid. Excellent post, nonetheless.

Still require some help, please.

let's suppose that p(x) is in R[x], of odd degree. there are two cases:

1. p(x) has a real root
2. p(x) does not have a real root

we want to show that (1) always happens, or equivalently, that (2) never happens.

p(x) is a complex polynomial as well, so by the fundamental theorem of algebra, p(x) has n linear complex factors (some of these factors may be repeated, that is correspond to multiple roots).

now first (and we don't need to use FTA to prove this), let's show that if z is a complex number with z* it's complex conjugate, p(z) = 0 implies p(z*) = 0:

recall that for two complex numbers z and w, z = a+ib, w = x+iy, that:

(z+w)* = (a+x + i(b+y))* = a+x - i(b+y) = a - ib + x - iy = (a+ib)* + (x+iy)* = z* + w*
(zw)* = [(a+ib)(x+iy)]* = [(ax - by) + i(ay + bx)]* = ax - by - i(ay + bx) = ax - (-b)(-y) + i(a(-y) + b(-x)) = (a - ib)(x - iy) = z*w*

in particular, the last formula means that (zⁿ)* = (z*)ⁿ, for all natural numbers n.

also recall that z* = z if and only if b = 0, that is: z is real.

suppose that p(x) = a₀+a₁x+a₂x²+.....+a_nxⁿ.

then p(z*) = a₀+a₁z*+a₂(z*)²+.....+a_n(z*)ⁿ

= a₀*+a₁*z*+a₂*(z²)*+.....+a_n*(zⁿ)*

= a₀*+(a₁z)*+(a₂z²)*+...+(a_nzⁿ)*

= (a₀+a₁z+a₂z²+...+a_nzⁿ)* = [p(z)]* = 0* = 0.

so if p(x) splits over C (which is what FTA states), then the root of p(x) (for REAL polynomials only, not complex ones) occur in conjugate pairs.

note that (x - z)(x - z*) = x² - (z+z*) + zz*, and z+z* and zz* are both real:

(a+ib) + (a-ib) = 2a, (a+ib)(a-ib) = a²-(-b²) + i(a(-b) + ab) = a²+b².

so the conjugate pairs produce real quadratic factors of p(x), unless z = z*. but z = z* iff z is real, and case (2) asserts this never happens.

so, in that case, we get that p(x) is a product of quadratic polynomials, in particular, p(x) is of even degree.

since p(x) is assumed to be of odd degree, case (2) leads to a contradiction.

Another great answer! I'm going to try to put that in idiot's terms as best as possible.