Prove that the roots of a polynomial cannot be all real

[anemone]

Let $a,\,b,\,c$ and $d$ be any four real numbers but not all equal to zero.

Prove that the roots of the polynomial $f(x)=x^6+ax^3+bx^2+cx+d$ cannot all be real.

 

[Mayhem]

Since I don't know how to solve something like this, would it be valid to show that there is a complex root? Since the fundamental theorem of algebra states that for an n-th degree polynomial, there are n solutions (real or complex), thus if one of those roots are complex, by the nature of the question, not all roots can be real.

 

[PAllen]

That would be ok if you could show there is a complex root for every choice of a,b,c,d.

 

[pasmith]

Solution:

Spoiler

We have $$x_1^2 + \dots + x_6^2 = I_1^2 - 2I_2$$ where the elementary symmetric polynomials $I_1 = x_1 + \dots + x_6$ and $I_2 = x_1x_2 + x_1x_3 + \dots + x_5x_6$.

If $x_1, \dots, x_6$ are the roots of $x^6 + ax^3 + bx^2 + cx + d$ then $I_1$ is minus the coefficient of $x^5$ and $I_2$ is the coefficient of $x^4$, both of which are zero. Thus $$x_1^2 + \dots + x_6^2 = 0.$$ Now if the roots are all real they are all zero, which contradicts the requirement that at least one of $a, b, c, d$ is non-zero.

 

[mathwonk]

hint #1: cogito, ergo sum.

hint #2: If a real polynomial has n real roots, then its derivative has ≥ n-1 real roots.

.......

[SPOILER #1: Descartes rule of signs implies at most 4 real roots.]

[SPOILER #2: Assume a ≠0. If there are 6 real roots then the 3rd derivative has ≥ 3 real roots, but it has form 120x^3 + 6a = 0, where a ≠ 0. The other cases are similar or easier. The hint follows from the product rule for derivatives and Rolle's theorem, and if all roots are distinct, it's obvious from looking at the graph, i.e. just Rolle suffices.]

Remark: As to the likelihood of a high schooler solving this, I believe the Descartes rule of signs was traditionally taught in high school even decades ago, and of course many now teach derivatives and graphing. Ironically, the most elementary approach, via symmetric functions used by pasmith, may not be taught much in high schools. I think I did not learn it there, but when I finally did learn it while studying field theory in college, it made polynomials seem so much more understandable that I wondered why it was not taught earlier. Of course an intelligent and curious student could easily discover it on her own.

 

[bob012345]

Solution:

Spoiler

We have $$x_1^2 + \dots + x_6^2 = I_1^2 - 2I_2$$ where the elementary symmetric polynomials $I_1 = x_1 + \dots + x_6$ and $I_2 = x_1x_2 + x_1x_3 + \dots + x_5x_6$.

If $x_1, \dots, x_6$ are the roots of $x^6 + ax^3 + bx^2 + cx + d$ then $I_1$ is minus the coefficient of $x^5$ and $I_2$ is the coefficient of $x^4$, both of which are zero. Thus $$x_1^2 + \dots + x_6^2 = 0.$$ Now if the roots are all real they are all zero, which contradicts the requirement that at least one of $a, b, c, d$ is non-zero.

Very clever solution!