So I understand how to do integration on rational functions with a linear and a quadratic denominator, and I understand how to do a partial fraction decomposition, but I was wondering what happens if the polynomial is higher degree and can't be factored. Later in the page, it says this:

However, it can be shown that any polynomial with real coefficients is a product of linear and/or irreducible quadratic factors with real coefficients.

And I was wondering, how do we know this and is this definitely true?

  • $\begingroup$ Do you know complex number? $\endgroup$
    – user99914
    Commented Nov 28, 2014 at 4:02
  • $\begingroup$ Yes, I do know complex numbers $\endgroup$
    – ASKASK
    Commented Nov 28, 2014 at 4:03
  • 2
    $\begingroup$ It's a consequence of all $n$th order polynomials having exactly $n$ roots. $\endgroup$
    – user541686
    Commented Nov 29, 2014 at 2:42
  • $\begingroup$ If anyone was curious, the original page is https://people.clarkson.edu/~sfulton/ma132/parfrac.pdf. $\endgroup$
    – Math2718
    Commented Feb 9, 2020 at 3:21

6 Answers 6


Yes, this is true. In particular, we have to use a theorem called the Fundamental Theorem of Algebra which states the following:

Any polynomial of degree $n$ with complex coefficients is the product of $n$ linear factors.

For instance, $ax^2+bx+c$ can always be written as $a(x-k_1)(x-k_2)$ for potentially complex $k_1$ and $k_2$.

Then, we need another piece of machinery to apply this theorem to the real numbers. In particular, we need to know about the complex conjugate defined as follows: $$\overline{x+iy}=x-iy.$$ You can check that $\overline{a}\cdot\overline{b}=\overline{ab}$ and $\overline{a}+\overline{b}=\overline{a+b}$. Basically, this reflects the complex plane over the real axis, and, in doing so, preserves multiplication and addition. However, for any real number $\overline{x}=x$. We can use this in the following way:

Suppose $f(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots a_1x + a_0$ is a polynomial with real valued coefficient. Then, we can calculate, for any complex number $z$ that: $$\overline{f(z)}=f(\overline z)$$ which is proven since, for instance $\overline{a_n z^n}=\overline{a_n} \cdot \overline{z^n}=a_n \overline{z}\,\! ^n$ since $a_n$ is real. In particular, this means that, suppose that $a+bi$ is a root of $f$. Then so must $a-bi$ be a root of $f$ - that is, complex roots come in so-called "conjugate pairs".

We can exploit this in the following way: A polynomial $f(x)$ is the product of its leading coefficient, $a_n$, and of the terms $(x-z_i)$ where $z_i$ is a sequence indexing its roots. Now, if $z_i$ is real, then we're fine, because it is a linear factor. The quadratic factors arise because, instead of writing, for a complex root $a+bi$ and its conjugate $a-bi$ the product $(x-a-bi)(x-a+bi)$, involving complex terms, we can multiply the expression out to receive $(x^2 - 2ax + a^2 + b^2)$ which is a quadratic in $x$ with real coefficients. Doing this with all the factors, we see the polynomial is writable as a product of linear and quadratic factors.

  • 1
    $\begingroup$ Wow thank you, this answer is incredible $\endgroup$
    – ASKASK
    Commented Nov 28, 2014 at 4:17
  • $\begingroup$ In the last paragraph, it should be "A monic polynomial $f(x)$...." $\endgroup$ Commented Nov 28, 2014 at 4:20
  • 5
    $\begingroup$ There's a glitch in this answer: one has to verify that a nonreal root and its conjugate have the same multiplicity. Not really hard, however, but it involves induction in some way: divide the polynomial by the linear or quadratic factor and you're left with a polynomial with lower degree. $\endgroup$
    – egreg
    Commented Nov 29, 2014 at 14:18
  • 3
    $\begingroup$ @egreg I am aware of that; I figured anyone who cared about that could also resolve it - though, if you want to avoid induction, notice that a root $x$ of multiplicity $n$ has, for any $i<n$ that $f^{(i)}(x)=0$ and the derivatives of $f$ have real coefficients, so $f^{(i)}(\overline{x})=\overline{f^{(i)}(x)}$. You can also argue by continuity, since the set of polynomials with no multiple roots is dense. $\endgroup$ Commented Nov 29, 2014 at 15:54
  • 1
    $\begingroup$ Can this be a valid proof ? Suppose we were to factor out all real roots of Q(x). Then we would be left with a polynomial P(x) having only real roots. Now, P(x) cannot have an odd degree as any polynomial of odd degree has at least one real root. Thus, the degree of P(x) must be even. Because complex roots always occur in conjugate pairs, and if the degree of P(x) were to be n then n/2 roots would be z1 and other n/2 roots would be conjugate of Z1. Thus they have the same multiplicity. Is this proof correct ? $\endgroup$ Commented Aug 13, 2021 at 10:51

In general, any polynomial of degree $n$ has a factorization into linear complex factors. This is a consequence of the fundamental theorem of algebra.

If $p(x)$ is a real polynomial with factor $x-w$ for $w$ complex, then $x-\bar w$ is also a factor (because $p(\bar w)=\overline{p(w)} = 0$.) When $w$ is not real ($w\neq \bar w$) we then know that $(x-w)(x-\bar w) = x^2 -(w+\bar w)x+w\bar w$ is a factor of $p(x)$, and $w+\bar w$ and $w\bar w$ are both real.

So, by induction, we can always factor $p$ as a product of linear and quadratic polynomials.

  • $\begingroup$ Thanks, this is what I was looking for. $\endgroup$
    – ASKASK
    Commented Nov 28, 2014 at 4:14
  • $\begingroup$ Each pair of conjugate non-real roots provides a quadratic factor, each real root a linear factor. $\endgroup$
    – orangeskid
    Commented Jan 13, 2015 at 1:47
  • $\begingroup$ Can you explain the last line: "So, by induction, we can always factor p as a product of linear and quadratic polynomials."? $\endgroup$
    – MrAP
    Commented Mar 25, 2018 at 15:48

It's true because the algebraic closure of $\Bbb R$ is $\Bbb C$ which is a field extension of degree 2.

Any irreducible polynomial $p$ would create a finite field extension of $\Bbb R$, namely $\Bbb R[x]/(p)$. But this means the degree of $p$ divides 2.

  • 2
    $\begingroup$ Sorry I'm unfamiliar with the concept of fields, can you explain this in a more basic sense? $\endgroup$
    – ASKASK
    Commented Nov 28, 2014 at 4:09

Yes, because if $z$ is a complex root of $P(x)$ with real coefficients, you can readily show that $\bar{z}$ is also a root since $P(\bar{z}) = \overline{P(z)} $. Therefore $u(x) =(x-z)(x-\bar{z}) $ divides $P(x)$.

But $u(x)$ is a quadratic polynomial with real coefficients.


This is what Carl Friedrich Gauss proved in his 25-page doctoral dissertation at the age of 22. The title of his dissertation is 'New Proof of the Theorem That Every Algebraic Rational Integral Function In One Variable can be Resolved into Real Factors of the First or the Second Degree." In the foreword to the translation, it is mentioned that this was the first proof of the statement as all other earlier proofs were flawed.

See http://archive.larouchepac.com/files/pdf/gauss_fundamental_1.pdf

  • 1
    $\begingroup$ The link that you are sharing is not working. $\endgroup$
    – mike
    Commented Sep 18, 2021 at 6:20
  • $\begingroup$ Thanks for letting me know. Here is a link to the Latin edition. google.com/books/edition/… $\endgroup$
    – AYO
    Commented Sep 19, 2021 at 9:41

For example, take $P(x) = x^n-1$. The factorization into complex linear factors is

$$(x^n-1) = \prod_{k=0}^{n-1}( x - (\cos \frac{2 k \pi}{n} + i \sin \frac {2 k \pi}{n}) )$$

From the pairs of non-real conjugate roots $\{ (\cos \frac{2 k \pi}{n} + i \sin \frac {2 k \pi}{n}) ,(\cos \frac{2 (n-k) \pi}{n} + i \sin \frac {2 (n-k) \pi}{n}) \}$ for $0< k < \frac{n}{2}$ we get the quadratic factors $$( x - (\cos \frac{2 k \pi}{n} + i \sin \frac {2 k \pi}{n}) )(x-(\cos \frac{2 (n-k) \pi}{n} + i \sin \frac {2 (n-k) \pi}{n}))=\\=x^2 - 2\cos\frac{2 k \pi}{n}\, x + 1 $$

Thus we get the real factorization of $x^n-1$:

$$x^n-1 =(x-1)\prod_{k=1}^{\frac{n-1}{2}} (x^2 - 2\cos\frac{2 k \pi}{n}\, x + 1 )$$ if $n$ odd and

$$x^n-1 =(x-1)(x+1)\prod_{k=1}^{\frac{n}{2}-1} (x^2 - 2\cos\frac{2 k \pi}{n}\, x + 1 )$$

if $n$ is even.


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