Can every polynomial be factored into constant and linear complex factors?

That is, can any polynomial, $a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x^1+a_0$, be expressed $b_0\left(x + b_1\right)\left(x + b_2\right)\ldots \left(x + b_n\right)$ where $b_i \in \mathbb{C}$?

• do you want $b_i \in \mathbb{Z}$ or $\mathbb{C}$? (the first one is usually for integers, the latter for complex numbers) Jan 27 '15 at 17:13
• isn't it $b_i\in \mathbb{C}$?
– Zero
Jan 27 '15 at 17:13
• The fundamental theorem of algebra states that every non-constant polynomial in complex numbers splits into linear factors in the field of complex numbers. Jan 27 '15 at 17:15
• You two are right, that was a typo. Jan 27 '15 at 17:15
• @Peter, did you mean the FT of Algebra? Jan 27 '15 at 17:17

Yes. Normally you would see $-$'s in the factors though, because then the $b_n$'s are the roots of the polynomial.

Every polynomial over $\mathbb{R}$ can be factored into linear or quadratic terms. If you have a real quadratic, it can be factored into linear terms over $\mathbb{C}$. This is what is meant by the fact that $\mathbb{C}$ is an algebraic completion of $\mathbb{R}$: $\mathbb{C}$ has the roots of all polynomials over $\mathbb{R}$.

The only condition of the fundamental theorem of algebra (FTA) is that the polynomial must be non-constant, i.e. some $a_i \ne 0$ for an $i > 0$. Then it states that there is a unique factorisation $$\sum_{i=0}^k a_i x^k = b_0 \prod_{i=1}^k (x-b_i)$$