#### Fundamental Theorem of Algebra

This theorem can be stated as: a non-zero, n-degree polynomial has exactly n roots which can be either real or complex.

An implication of this theorem is that if an (seemingly) n-degree has more than n roots, then it must be an identity equaling $0$, i.e. all its coefficients are zero. This also means that this polynomial has infinitely many roots. For example, the equation $2x+1=0$ has exactly one root. But the equation $0x=0$ is an identity which can be satisfied by any real or complex number $x$.

The problem appeared on 2017/10/12 is a typical one that can be solved by this theorem.

 (3673) Show that $$\frac{(x+a)(x+b)}{(c-a)(c-b)}+\frac{(x+b)(x+c)}{(a-b)(a-c)}+\frac{(x+c)(x+a)}{(b-c)(b-a)}=1$$ without expanding the left side of the equation.

This problem essentially asks us to prove this is an identity regardless of what values $x$, $a$, $b$, and $c$ take. To show this, let's first construct a polynomial

$$f(x)=\frac{(x+a)(x+b)}{(c-a)(c-b)}+\frac{(x+b)(x+c)}{(a-b)(a-c)}+\frac{(x+c)(x+a)}{(b-c)(b-a)}-1$$

Then, it is easy to show that $f(−a)=f(−b)=f(−c)=0$. Clearly, $a$, $b$, and $c$ are all distinct otherwise the original expression becomes undefined. Hence, we find $3$ distinct roots of $f(x)$. Meanwhile, it is obvious that $f(x)$ is at most $2$-degree. Therefore, we can assert $f(x)=0$ is an identity.

The fundamental theorem of algebra is an important one even though it is not well covered in classroom. The book has one section dedicated to this topic. Here is a more challenging example from this book.

 (2074) Let polynomials $P(x)$, $Q(x)$, $R(x)$, and $S(x)$ satisfy: $$P(x^5) + xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x)$$Prove: $(x-1) | P(x)$

In fact, it can be shown that both $(x−1)|Q(x)$ and $(x−1)|R(x)$ hold as well.