Basic algebra: Prove that an n-degree polynomial is expressible as the product of n binomials.

Question

Using basic algebra, how does one prove that an n-degree polynomial is expressible as the product of $n$ binomials? Here I am allowing for binomials of the form $xb_i+\alpha_i=x0+1.$

This is something I should know like the back of my hand, but I don't. The answer I am seeking is a high school-level demonstration. I am motivated by the following statement in Hermann Weyl's The Classical Groups:

A poly­nomial of degree $n$ has at most $n$ different zeros; this follows in the well-known way by proving that $f\left(x\right)$ contains the factors $\left(x — \alpha_1\right)\left(x — \alpha_2\right)\dots$ if $\alpha_1,\alpha_2,\dots$ are distinct zeros.

I have sources which provide "advanced" proofs. I am seeking an intuitively satisfying proof that could be presented to a reasonably intelligent high school student with the expectation of understanding. I am more interested in the method of proof, than the proof itself.

Answer 1

I don't think you will have a basic algebraic proof, at a high school level, because you require complex numbers to even deal with $x^2+1$. Even this supposedly simple proof requires calculus.

Answer 2

A poly­nomial of degree $n$ has at most $n$ different zeros; this follows in the well-known way by proving that $f\left(x\right)$ contains the factors $\left(x — \alpha_1\right)\left(x — \alpha_2\right)\dots$ if $\alpha_1,\alpha_2,\dots$ are distinct zeros.

This statement does not claim that a polynomial of degree $n$ can be expressed as the product of $n$ binomials. In fact, the claim in the title of your question is false in real algebra, although it is true in complex algebra.

What the claim in the book actually asserts is that if the numbers $\alpha_1,$ $\alpha_2, \ldots,$ $\alpha_k$ are $k$ distinct zeros of a polynomial $f(x),$ then $$f(x) = (x - \alpha_1)(x - \alpha_2)\cdots(x - \alpha_k)g(x) \tag1$$ where $g(x)$ is a polynomial. It never says that so many zeros necessarily exist.

A real polynomial of any even degree may have no real zeros at all. A real polynomial of any odd degree may have just one real zero and no others. In general, a polynomial $f(x)$ of degree $n$ can have fewer than $n$ zeros, but it cannot have more than $n$ zeros. If it had more than $n$ zeros, the factorization in Equation $(1)$ would show that $f(x)$ was a polynomial of degree greater than $n,$ a contradiction.