Is there a polynomial function $P(x)$ with real coefficients that has an infinite number of roots? What about if $P(x)$ is the null polynomial, $P(x)=0$ for all x?
Polynomials – Is There a Polynomial with Infinitely Many Roots?
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Related Solutions
First case: If the number of real roots $r$ of $f(x)$ is greater than one, then $f'(x)$ has at least $r-1$ real roots. (The limitation "greater than one" is not necessary but the statement is trivial if $r\le 1$.) Given any two roots $a<b$ of $f(x)$, $f$ is continuous and differentiable on $[a,b]$, so by Rolle's theorem $f'(c)=0$ for some $a<c<b$.
There may be more roots of $f'(x)$ than those between roots of $f(x)$, so the only upper bound is the obvious one of $n-1$. Ask if you need examples. It seems to me that if multiplicity is taken into account that the number of real roots of $f'(x)$ has the same parity (even/odd) as the number of real roots of $f(x)$, but I haven't proven it yet. If multiplicity is not taken into account, the parity can be anything.
Second case: If $f(x)$ has degree $n$ and has $n$ real roots, then each consecutive pair of roots of $f(x)$ defines a root of $f'(x)$, which makes $n-1$ roots of $f'(x)$. Since $f'(x)$ is a polynomial of degree $n-1$, this is all possible roots. This continues for all later derivatives, so you are correct: all its derivatives will have all real roots.
Third case: The contrapositive of the second case tells us that if any of its derivatives have any non-real roots, then $f(x)$ also has some non-real roots.
Fourth case: The converse of the third case is not true. For example, $f(x)=x^2+1$ has two non-real roots, but its derivative $f'(x)=2x$ has one real root.
Comment case: You asked, "Suppose $f'(x)$ is a $5$ degree polynomial with $3$ real roots. What are the possible no. of roots that $f(x)$ can have($3,4,5$ etc.?)."
The formulas for $f(x)$ and $f'(x)$ are given in the diagram, where $C$ is a real constant, zero in the graph. You can see that $f'(x)$ is a degree $5$ polynomial with $3$ real roots.
The dashed horizontal lines show the possible number of real roots of $f(x)$ for varying values of $C$. There are $0$ real roots for $C=3$, $1$ real root for $C\approx. 2.638$, $2$ real roots for $C=1$, $3$ real roots for $C\approx -1.757$, and $4$ real roots for $C=-1.9$. My discussion for the first case shows that there cannot be more than $4$ real roots since $f'(x)$ has $3$ real roots.
Best Answer
The only polynomial with infinitely many roots is $$P(x)=0.$$ You can prove this without appealing to the fundamental theorem of algebra. In particular, let us prove the following:
We prove this by induction. In the linear case, we obviously have that $P(x)=mx+b$ has exactly one root at $\frac{-b}m$. Next, suppose $P(x)$ is a polynomial of degree $n$. If it has no roots, it satisfies the hypothesis trivially. Otherwise, let $r$ be a root. One can, by using polynomial long division, determine that there is a degree $n-1$ polynomial $P_2$ such that $$P_2(x)\cdot (x-r)=P(x).$$ You can check that this condition is actually equivalent to saying that $r$ is a root, since, when doing polynomial long division, you'll find that the remainder is exactly $P(r)$.
However, by the zero-product law, this means that $P$ has a root exactly when either $x-r$ or $P_2(x)$ is $0$. By inductive hypothesis, $P_2(x)$ is $0$ for at most $(n-1)$ distinct values of $x$ and clearly $x-r$ is zero only at $r$. Thus, $P(x)$ can have at most $n$ zeros. This completes the proof.
Notice that this, unlike the fundamental theorem of algebra, holds in any field - that is, we only need multiplication, addition, and their inverses to be suitably well behaved.