# Why are all polynomials with real roots factorizable?

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

I'm trying to understand why do 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 have

$(x-b)(a_{n-1}x^{n-1}+\cdots+a_0)+R$ 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?

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No to the second question, e.g., $(x^2+1)^2$ is factorizable. –  Gerry Myerson Oct 3 '12 at 7:39
@GerryMyerson, but not into linear binomials –  akkkk Oct 3 '12 at 7:42
When you do long division of the polynomial $f$ (which I assume has all real coefficients, though you didn't explicitly say this) by $x-b$ (which also has all real coefficients), all the computations involve real numbers only. So there's no possibility of getting a non-real remainder. –  Ted Oct 3 '12 at 7:44
Hmm, but that doesn't explain why you don't get a real remainder that can't be reduced to a polynomial. –  acjohnson55 Oct 3 '12 at 8:17
7 questions and 0% accept rate? –  mrf Nov 1 '12 at 11:38
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## 4 Answers

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.

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Thanks for replying. You mention 'such that f=qg+r with r=0 or deg(r)<deg(g). '. Why must r = 0 or otherwise deg(r)<deg(g)? Sorry if the question is very basic. –  John Oct 3 '12 at 7:57
This is the property (or definition) of Euclidian Rings. If you don't know this yet, just check the second part of my answer. It's basically the proof that this is true for the special case of a linear $g$. –  k.stm Oct 3 '12 at 8:01
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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 $x'=x-b$.

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$R$ 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

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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.

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