Prove that the roots of a polynomial cannot be all real

In summary, a polynomial is a mathematical expression consisting of variables and coefficients, combined using addition, subtraction, and multiplication. Real roots are the values that satisfy a polynomial equation and make it equal to zero when substituted. However, a polynomial cannot have all real roots according to the Fundamental Theorem of Algebra, which states that a polynomial can have at most n complex roots. This can also be proven using the Intermediate Value Theorem, which states that a polynomial must have at least n+1 intervals with opposite signs in order to have n real roots. Even if all the coefficients of a polynomial are real, it does not guarantee that the polynomial will have all real roots, as complex numbers are also included in the set of real numbers.
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anemone
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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.
 
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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.
 
  • #3
That would be ok if you could show there is a complex root for every choice of a,b,c,d.
 
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Solution:
We have [tex]x_1^2 + \dots + x_6^2 = I_1^2 - 2I_2[/tex] where the elementary symmetric polynomials [itex]I_1 = x_1 + \dots + x_6[/itex] and [itex]I_2 = x_1x_2 + x_1x_3 + \dots + x_5x_6[/itex].

If [itex]x_1, \dots, x_6[/itex] are the roots of [itex]x^6 + ax^3 + bx^2 + cx + d[/itex] then [itex]I_1[/itex] is minus the coefficient of [itex]x^5[/itex] and [itex]I_2[/itex] is the coefficient of [itex]x^4[/itex], both of which are zero. Thus [tex]
x_1^2 + \dots + x_6^2 = 0.[/tex] Now if the roots are all real they are all zero, which contradicts the requirement that at least one of [itex]a, b, c, d[/itex] is non-zero.
 
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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.
 
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  • #6
pasmith said:
Solution:
We have [tex]x_1^2 + \dots + x_6^2 = I_1^2 - 2I_2[/tex] where the elementary symmetric polynomials [itex]I_1 = x_1 + \dots + x_6[/itex] and [itex]I_2 = x_1x_2 + x_1x_3 + \dots + x_5x_6[/itex].

If [itex]x_1, \dots, x_6[/itex] are the roots of [itex]x^6 + ax^3 + bx^2 + cx + d[/itex] then [itex]I_1[/itex] is minus the coefficient of [itex]x^5[/itex] and [itex]I_2[/itex] is the coefficient of [itex]x^4[/itex], both of which are zero. Thus [tex]
x_1^2 + \dots + x_6^2 = 0.[/tex] Now if the roots are all real they are all zero, which contradicts the requirement that at least one of [itex]a, b, c, d[/itex] is non-zero.
Very clever solution!
 
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Related to Prove that the roots of a polynomial cannot be all real

What is a polynomial?

A polynomial is a mathematical expression that consists of variables and coefficients, combined using operations such as addition, subtraction, and multiplication. It can have one or more terms, and the highest power of the variable in the expression is called the degree of the polynomial.

What are real roots of a polynomial?

Real roots of a polynomial are values of the variable that make the polynomial equal to zero when substituted into the expression. In other words, they are the solutions to the polynomial equation.

Can a polynomial have all real roots?

No, a polynomial cannot have all real roots. This is because the Fundamental Theorem of Algebra states that a polynomial of degree n can have at most n distinct complex roots. Since real numbers are a subset of complex numbers, a polynomial cannot have more real roots than complex roots.

How can you prove that the roots of a polynomial cannot be all real?

One way to prove this is by contradiction. Assume that the polynomial has all real roots. This means that the polynomial can be factored into linear factors, with each factor representing a root. However, according to the Fundamental Theorem of Algebra, the polynomial can have at most n distinct complex roots. Since real numbers are a subset of complex numbers, this contradicts our assumption that the polynomial has all real roots. Therefore, the roots of a polynomial cannot be all real.

Can a polynomial with all real coefficients have all real roots?

Yes, a polynomial with all real coefficients can have all real roots. This is because the coefficients of a polynomial do not affect the nature of its roots. The only restriction is that the degree of the polynomial must be even, as odd-degree polynomials always have at least one real root.

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