How to show that a 5-th degree polynomial has a root?

  • #1
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Homework Statement


"Show that for some ##x\in ℝ##, that ##x^5+2x^4+3x^3+2x^2+x=1##."

Homework Equations




The Attempt at a Solution


Okay, so I know from Descartes' rule of sign that the function ##f(x)=x^5+2x^4+3x^3+2x^2+x-1## has exactly one positive root, since the sign of the coefficients change exactly once in the entire function. But I am asked to show that it has a positive root, and I am not fully sure that this explanation would cut it.

I also have that ##x^5+2x^4+3x^3+2x^2+x=\frac{1}{2}x^5+2x^4+3x^3+2x^2+\frac{1}{2}x+\frac{1}{2}x(1+x^4)=\frac{1}{2}x\sum_{k=0}^4 \binom 4 k x^k+\frac{1}{2}x(1+x^4)=\frac{1}{2}x[(1+x)^4+(1+x^4)]##. But I have to show that this expression is identically one, and I haven't an idea of how to do this.
 

Answers and Replies

  • #2
mathwonk
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intermediate value theorem. i.e. show it is somewhere less than 1 and somewhere greater than 1, and then use the theorem that the set of values is an interval.

Notice they do not ask you to actually find the value x where f(x) = 1, merely to prove there is one.

(I missed this problem freshman year in college. i didn't realize you are supposed to actually read and internalize the theorems that are presented in the course. I thought you were just supposed to brainstorm every problem on your own. the moral is that the theory presented in the course has uses.)

Similarly every odd degree polynomial has a root.
 
  • #3
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Gotcha. Thanks for the tip.
 
  • #4
LCKurtz
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Homework Statement


"Show that for some ##x\in ℝ##, that ##x^5+2x^4+3x^3+2x^2+x=1##."

Homework Equations




The Attempt at a Solution


Okay, so I know from Descartes' rule of sign that the function ##f(x)=x^5+2x^4+3x^3+2x^2+x-1## has exactly one positive root, since the sign of the coefficients change exactly once in the entire function. But I am asked to show that it has a positive root, and I am not fully sure that this explanation would cut it.
Yes, it would "cut it", presuming you have had Descartes' rule of signs.
 
  • #5
Show that for some ##x\in ℝ##, that ##x^5+2x^4+3x^3+2x^2+x=1##."

Odd degree functions looks kind of like a “S” laying on its side,so there must be an x-intercept
 
  • #6
Ray Vickson
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Homework Statement


"Show that for some ##x\in ℝ##, that ##x^5+2x^4+3x^3+2x^2+x=1##."

Homework Equations




The Attempt at a Solution


Okay, so I know from Descartes' rule of sign that the function ##f(x)=x^5+2x^4+3x^3+2x^2+x-1## has exactly one positive root, since the sign of the coefficients change exactly once in the entire function. But I am asked to show that it has a positive root, and I am not fully sure that this explanation would cut it.

I also have that ##x^5+2x^4+3x^3+2x^2+x=\frac{1}{2}x^5+2x^4+3x^3+2x^2+\frac{1}{2}x+\frac{1}{2}x(1+x^4)=\frac{1}{2}x\sum_{k=0}^4 \binom 4 k x^k+\frac{1}{2}x(1+x^4)=\frac{1}{2}x[(1+x)^4+(1+x^4)]##. But I have to show that this expression is identically one, and I haven't an idea of how to do this.
If quoting and using a theorem does not "cut it", what would cut it for you?
 

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