Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove that the polynomial below has at least one nonreal complex root

$$x^5+\frac{x^4}2+ \frac{x^3}3+\frac{x^2}4+\frac x{24}+\frac 1{120}$$

I have tried to prove that there exist $k\in \Bbb R$, such that the above polynomial can be factored into $(x^2+k)P(x)$, where deg$(P)=3$. But somehow I couldn't work it out.

share|improve this question

2 Answers 2

up vote 35 down vote accepted

Between any two real roots of a polynomial there should be at least one root of its derivative. So the maximum possible of roots in the polynomial is the number of roots of the derivative plus one.

In this case, we have $f(x)=x^5+x^4/2+x^3/3+x^2/4+x/24+1/120$, and $$ f'''=60x^2+12x+2, $$ which has no real roots. So $f''$ has at most one real root; so $f'$ has at most two real roots, and finally $f$ has at most three real roots. We conclude that $f$ has at least two complex roots.

share|improve this answer
7  
Wow - what a beautiful answer! +1 :) –  AWertheim Mar 12 at 0:39
2  
I see, application of Rolle's theorem! Very Nice! I have been thinking about the graph of the polynomial, but I couldn't give a rigorous proof. –  Bohan Lu Mar 12 at 0:51

Here is a way to do it without resorting to calculus:

Eliminate the quartic term (making it into a "depressed quintic") by making the substitution $x=z-1/10$ (as $1/10$ is one fifth of the coefficient of $x^4$), which turns the polynomial into $$z^5+\frac{7 z^3}{30}+\frac{17 z^2}{100}+\frac{z}{6000}+\frac{239}{37500}\text{.}$$ By Descartes' Rule of Signs, this has no positive roots and either three or one negative roots. $0$ certainly isn't a root, so it must have at least 2 non-real roots.


As an aside, you could use the discriminant to show further that it must have exactly one real root, but the discriminant is $2258539/17915904000$ and is not something you could calculate by hand, given the formula for the discriminant of a monic quintic is pretty terrible (which you can verify with Wolfram|Alpha).

share|improve this answer
    
Interesting method! I have never heard of nor seen something like "depressed quintic". Does this manipulation or method have a general application toward a particular type of problem? What are some backgrounds of this method? –  Bohan Lu Mar 13 at 3:24
1  
@BohanLu depressing (making a linear substitution to eliminate the term of second-highest order) a polynomial is a key first step in solving polynomials. Depressing a quadratic is basically "completing the square", depressing a cubic/quartic is the first step to solving those equations, and even though quintics don't often have solutions in radicals, depressing them makes them simpler for solving/characterization. (Continued below: –  Mark S. Mar 13 at 5:17
1  
For this problem, the technique is to give Descartes' Rule of Signs a better chance, but there is also a generalization of the discriminant method. If your quintic is $x^5+px^3+qx^2+rx+s$ and the discriminant is positive, then it has one real root precisely if at least one of the following is nonpositive: $-p$, $40rp-12p^3-45q^2$, and $12p^4r-4p^3q^2-40p^2qs-88p^2r^2+117pq^2r+125ps^2-27q^4-300qrs+160r^3$. These formulas can be found in demonstrations.wolfram.com/…, which also cites "A Complete Discrimination System for Polynomials". –  Mark S. Mar 13 at 5:25

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.