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I know that you can't factor a real polynomial into $\Pi_{i=1}^N(x-a_i)$ in general. But is it possible to factor every real finite polynomial into this form:

$(\Pi_{i=1}^N a_ix^2 + b_ix + c_i) (\Pi_{i=1}^Mx-d_i)$ where the second term is possibly empty?

This is at least true for polynomials with odd powers because they must have at least one real root. (My reasoning was false, edit see below) So I wonder if it is also true for even powers?

Edit: Berci has pointed out a lapse in my reasoning. For an odd power, once you factor out the first root, it is conceivable that you may hit an even-degree polynomial that is irreducible.

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8  
Did you ever heard about the fundamental theorem of algebra? –  user26857 Feb 15 '13 at 1:20
    
Well, your argument for odd powers works only for $3$. –  Berci Feb 15 '13 at 1:21
    
@Berci Oh right. I'll make the correction. –  Mark Feb 15 '13 at 1:26
    
@YACP So because there is a complex root, there must be the conjugate root as well in order to keep the coefficients real. Then that proves the assertion? –  Mark Feb 15 '13 at 1:30
    
@Mark Of course. –  user26857 Feb 15 '13 at 1:57

3 Answers 3

up vote 7 down vote accepted

The answer is yes, here is a proof sketch.

1) Let $P(x)$ be a polynomial with real coefficients. By the fundamental theorem of algebra it has a root, call it $a$.

2) If $a\in \mathbb R$ then $(X-a)$ divides $P(x)$.

3) If $a\notin \mathbb R$ then, since the coefficients are all real, it follows that $\bar a$ is also a root of $P(x)$.

4) It follows that $(x-a)(x-\bar a)$ divides $P(x)$.

5) Verify directly that $(x-a)(x-\bar a)$ is a polynomial with real coefficients.

6) Repeat steps above as many times as needed to obtain $P(x)$ as the product of linear factors and quadratic factors, all with coefficients in $\mathbb R$.

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Yes I believe this is what YACP was hinting at. Thanks! –  Mark Feb 15 '13 at 1:37

Yes, this is true. It is equivalent to the fundamental theorem of algebra which says that $\mathbb{C}=\mathbb{R}(i)$ is algebraically closed.

If $p \in \mathbb{R}[x]$ is a monic polynomial, then it is a product of linear factors $x-a$ with $a \in \mathbb{C}$. The complex conjugation leaves $p$ invariant, hence acts on these roots $a$. If it leaves $a$ invariant, this means $a \in \mathbb{R}$ and we keep $x-a$. If not, we expand $(x-a)(x-\overline{a})=x^2-(a + \overline{a})x + a\overline{a} \in \mathbb{R}[x]$.

For example, $p=x^3+3x^2+x+3$ has the roots $i,-i,-3$, so that it factors as $(x^2+1)(x+3)$.

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http://en.wikipedia.org/wiki/Complex_conjugate_root_theorem

$\left(x-a-bi\right)\left(x-a+bi\right)=x^2-2ax+a^2+b$

So, if polynomial with real coefficients has any complex root, that it has quadratic "polynomial factor". If it doesn't have any complex root, then it also has such "factor" (or it's polynomial of first degree). Factor it out and repeat.

P.S. http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra (from it follows if it has at least one complex root $x_0$, than divide the polynomial by $x-x_0$ and repeat until all polynomial is gone - in other words, every polynomial of degree $n$ has $n$ (complex or real) roots).

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