Why are all polynomials with real roots factorizable?

Question

$$f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0$$

I'm trying to understand why all polynomials with real roots are factorizable. The explanation relies on the fact that all polynomials which are divided by a first degree polynomial $(x-b)$ where $b$ is a root will be of the form$$(x-b)(a_{n-1}x^{n-1}+\cdots+a_0)+R,$$where $R$ is a real number.

I dont understand why this is so. Why is there no possibility of remainders which are not real numbers?

On a separate note, does this mean that polynomials with no real roots are unfactorizable?

Answer 1

This is the euclidian algorithm.

Let $f \in \mathbb{R}[X]$ be polynomial and $z$ be a root. Then $f(z) = 0$, so take $g = (X - z)$. Since $\mathbb{R}[X]$ is euclidian, you find polynomials $q, r \in \mathbb{R}[X]$ such that $f = qg + r$ with $r = 0$ or $\mathrm{deg}(r) < \mathrm{deg}(g)$. But $r(z)= (f-qg)(z) = f(z) - q(z)g(z) = 0$, so $r(z) = 0$. Eitherway, if $r$ wasn't $0$, then $\mathrm{deg} (r) < \mathrm{deg} (g) = 1$, so $\mathrm{deg} (r) = 0$, and $r$ must be constant. Since $r(z)=0$, you have $r=0$ and $f = gq$.

Normally $\mathrm{deg} (0)$ wouldn't be defined to be $0$, but this doesn't make any difference.

Maybe you are looking for a more elementary reasoning. Then you can argue more visually: Let $f_n = f = a_n X^n + a_{n-1}X^{n-1} \ldots + a_0$ be polynomial such that $f(z) = 0$. Then you can take $f_{n-1} = f_n - a_n (X - z)^n$, which has degree one less then $f$ and still satisfies $f_{n-1}(z) = 0$. Continue setting $f_{k-1} = f_k - e(f_k)(X-z)^k$ (where $e(f_k)$ denotes the leading coefficient of $f_k$), noting $\deg f_k = k$ and $f_k(z) = 0$, down to $f_0$ which is constant and by construction satisfies $f_0 (z) = 0$, so $f_0 = 0$. You always subtracted multiples of $X-z$ so $f - q (X-z) = 0$ for some subtractions subsumed in a polynomial $q$. So the result follows.

This is essentially the same as above.

Answer 2

$R$ is $0$ actually when $b$ is a root of the polynomial.

Since, polynomials over a field form EUCLIDEAN Domain, hence $p(x)=q(x)(x-b)+r(x)$ where $p(x),q(x)$ and $r(x)$ are polynomials with $deg(q(x))=deg(p(x))-1$ and $deg(r(x))\lt deg(x-b))$ or $0$

When $b$ is a root $\implies p(b)=0\implies r(b)=p(b)-q(b)(b-b)=0\implies p(x)=q(x)(x-b)$ only with $0$ remainder.

And you can always factorize a polynomial not necessarily in the field over which the polynomial but some other field called splitting field for that polynomial

Answer 3

Suppose $f(a)=0$ then $f(x)=f(x)-f(a)=(x-a)g(x)$ where deg $g(x)=n-1$ and $g(x)$ has real coefficients.

[factorising $x^r-a^r$ for each $r$]

Now given $b \in \mathbb R$ let $h(x)=f(x)-f(b) \text{ with } f(b) \in \mathbb R$.

Then $h(b)=0$ so that there is some polynomial with real coefficients such that $h(x)=(x-b)p(x)$ [using the first remark].

So that $f(x)=(x-b)p(b)+f(b)$

This gives an easy way of calculating the remainder and showing it is real without doing the long division.

Answer 4

Here is an argument that works for children (if there are any reading this site :)

We have $f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0$, i.e. $f(x)=x(x^{n-1}+a_{n-1}x^{n-2}+\cdots+a_1)+a_0$. So we are done for $b=0$. For general $b$, use the substitution $x'=x-b$.