## Precalculus (6th Edition) Blitzer

According to the linear factorization theorem, if $f\left( x \right)={{a}_{n}}{{x}^{n}}+{{a}_{n-1}}{{x}^{n-1}}+...+{{a}_{1}}x+{{a}_{0}}$, such that $n\ge 1$ and ${{a}_{n}}\ne 0$, then $f\left( x \right)={{a}_{n}}\left( x-{{c}_{1}} \right)\left( x-{{c}_{2}} \right)...\left( x-{{c}_{n}} \right)$. Here, ${{c}_{1}}, {{c}_{2}}\text{,} ... \text{,} {{c}_{n}}$ are complex numbers, which can possibly be real as well as non-distinct. Also, $\left( x-{{c}_{1}} \right), \left( x-{{c}_{2}} \right), ..., \left( x-{{c}_{n}} \right)$ are called linear factors because their degree is 1. Hence, the given statement is the same as above, but expressed in words.