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**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

Competition Algebra 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.

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