A student asked me about an exercise which said something like

Factorize $p$ over $\Bbb R[x]$, where $p$ was a given polynomial of degree 4.

The idea was to use the rational root theorem to find two roots, and the quadratic formula to find the last two, and then just write $$p=k\prod_{i=1}^4(x-x_i).$$

When explaining this to the student, he said "but those $x_i$ are the roots, that does not mean that the polynomial can be written that way".

Embarrassingly enough, I couldn't convince the student.

So my question is,

given $p$ a polynomial over a field $K$, with all its roots $x_1,\dots,x_n$, why can I write $p=k\prod_{i\geq 0} (X-x_i)$, where $k$ is the principal coefficient?

  • $\begingroup$ Does the student doubt that $p$ can be factored as a product of linear factors, or just that the linear factors are $x - x_i$? $\endgroup$ May 11, 2016 at 19:13

1 Answer 1


What you are looking for is a weak form of the Remainder theorem which states that over a field, $a$ is a root of $p(x)$ if and only if $(x-a)$ divides $p(x)$. Now, over an algebraically closed field (like $\mathbb C$) repeatedly apply this to get $p(x)$ as a product of linear factors.

That is, if $x_1$ is a root of $p$, we have $p(x)=(x-x_1)p_1(x)$ and the degree of $p_1=\deg(p)-1$. Next if $x_2$ is a root of $p_1$, we have $p_1(x)=(x-x_2)p_2(x)$ and the degree of $p_2=\deg(p_1)-1$ and keep doing this until you are left with a $p_n$ whose degree is $0$ and this gives your constant $k$ in front.


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