Are all quadratics factorable into a product of two binomials?

I'm learning algebra in school, and my teacher said that all quadratics are factorable into a product of two binomials. I then realized however that some quadratics would have imaginary roots, and therefore wouldn't be able to be put into factored form. Who's wrong here, my teacher or me? For example, can $x^2 + 4x + 1$ even be expressed in factored form? Thanks in advance.

• That example of yours has two real roots --- is that intended, or did you mean to write a polynomial with complex roots? Nov 21, 2017 at 18:35
• Factored over what? $\mathbb{Z}$? $\mathbb{R}$? $\mathbb{C}$? Nov 21, 2017 at 20:12
• @anomaly - "I'm learning algebra in school" pretty much precludes your question from having any useful context for him. Nov 22, 2017 at 3:38
• @FedericoPoloni No, sorry about that. My mistake, I meant to include an example with imaginary roots. Nov 22, 2017 at 11:46
• @PaulSinclair: If you're aware of imaginary roots, you know about complex numbers. And while the concepts ℕ,ℤ, and ℚ are introduced only after their contents, usually ℝ is introduced up front. Certainly anyone knowing about complex numbers will be familiar with ℂ. Perhaps you might not be already familiar with the concept of factoring over ... but the conceptual difference between factoring over ℂ vs. factoring over ℝ is pretty intuitive. (The actual techniques, less so) Nov 22, 2017 at 13:36

It really depends on whether you want to have complex factors or not. If you can have complex factors, every expression can be. If not, then only if $b^2\ge4ac$ would they be factorable.

Take $x^2+1$, it can be factored into $(x-i)(x+i)$ but none of the factors are real.

• If we want to be totally literal with respect to the wording of the question, I think we have to expand the provisos to include "trivially" binomial terms, like $(x+0)$ as a factor for $x^2+4x$. Nov 21, 2017 at 21:32
• And to be even more pendantic, to have imaginary factors, you would need $b= 0$ and $ac < 0$ (for a quadratic with real coefficients). It is complex factors that allow everything to be factored. That said, the use of "imaginary" to mean "complex" has a very long history and has been done by mathematical giants. But I thought I'd point out what is considered the correct terminology. Nov 22, 2017 at 19:32
• @PaulSinclair edited In. Nov 23, 2017 at 0:57

You're either right or wrong depending on context, and so is your teacher.

One of the beautiful properties of the complex numbers is that when you allow polynomials to have complex coefficients, every polynomial is the product of linear factors in exactly one way (not counting different orders of multiplication of the factors as different ways to express the product). This is known as the Fundamental Theorem of Algebra. (You can tell it's an important theorem, because it's called "fundamental.")

If you allow only real coefficients of polynomials, the Fundamental Theorem has a slightly different form: every polynomial is the product of linear factors and/or irreducible quadratic factors with real coefficients in exactly one way (not counting different orders of multiplying the factors). The irreducible quadratic factors are precisely the ones that have complex roots.

So when we work with polynomials, we may ask ourselves, "Are we working with real polynomials today or with complex polynomials?" Depending on the answer to that question, you can factor every polynomial into linear terms, or there are some that you cannot factor in that way.

Each polynomial of degree $n\ge 1$ has exactly $n$ complex roots and hence it can be decomposed into a product of $n$ binomials (with complex coeffitients of course). (See Fundamental Theorem of Algebra)

In your example there are real roots: \begin{align} x^2+4x+1&=x^2+4x+4-3 \\ &=(x+2)^2-3 \\ &=(x+2-\sqrt{3})(x+2+\sqrt{3}) \end{align}

• If you want to factor numerically, you can use a tool such as Factoris. Type x^2+4x+1 and note that you can choose under "Options" what ring to factor over. Nov 22, 2017 at 8:45

Let $P(z) = \sum_{i=0}^n a_i z^i$, so that $P$ is a polynomial of degree $n$. Suppose that $z,a_1,\dots,a_n \in \mathbb{C}$. Then, there exists $r_1,\dots,r_n \in \mathbb{C}$ such that $P(z) = a_n\prod_{i=1}^n (z-r_i)$. This is more or less what the Fundamental Theorem of Algebra states (or implies, depending on who you ask). Thus, under these circumstances, every polynomial of degree $n$ can be written as a product of exactly $n$ binomial factors, not necessarily distinct.

However, if we suppose that $r_1,\dots,r_n \in \mathbb{R}$, then this is no longer the case, even if we also have $z,a_1,\dots,a_n \in \mathbb{R}$. For example, the polynomial $P_0(z) = z^2+z+1$ has no such representation, and equivalently, there is no $z^* (\in \mathbb{R})$ such that $P_0(z^*) = 0$. Since $\mathbb{Q}, \mathbb{Z},$ and $\mathbb{N}$ are subsets of $\mathbb{R}$, they have similar properties.