Do polynomials $ P(t)$ of an odd degree have at least one real root belong to $(t-a)Q(t)$?

Question

This is a continuation of a question where ker(T) = (t-a)Q(t) = P(t). Show that {P(t) ∈ R[t] | deg(P(t)) = 3} ⊂ $∪_{a∈R}$ker(T).

So the mark scheme says that all polynomials in R[t] of an odd degree have at least one real root a, therefore it belongs to ker(T). Why does it have to be an odd degree that possesses a reel root?

A polynomial doesn't belong to $∪_{a∈R}$ker(T) if it doesn't have a reel root $(t^2+1)^k$ is an example of one for degrees n = 2k. I don't understand this part either.... can someone explain to me, i feel like my understanding of polynomials is dreadful

Answer 1

An odd polynomial has at least one real root is due to the fact that complex roots come in conjugate that is if $\alpha$ is a complex root of a polynomial then $\overline{\alpha}$ is a root too.

Answer 2

If you have an odd degree polynomial, the highest degree term eventually "dominates", so it takes over for sufficiently large values. If you take a large enough positive number as the polynomial's input, the highest term $a_n x^n$ is the sign of the leading coefficient, and the other terms are negligible. Likewise, if you take a large enough negative number, $a_n x^n$ is the opposite of the leading coefficient's sign if it is odd degree. So an odd degree polynomial has both a positive and negative value "somewhere". Since $0$ is between a negative and positive number, and the polynomial is continuous on the entire real line, by the Intermediate Value Theorem, there exists a point such that the polynomial equals $0$ somewhere on the real line.