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Prove that a cubic has no rational roots

  • Thread starter kingwinner
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1) Prove that the acute angle whose cosine is 1/10 cannot be trisected with straightedge and compass.

...
I worked it out and at the end found out that , if I can prove that the cubic polynomial 40x3 - 30x -1 has no rational roots, then I am done.

Now, is there any way to prove (e.g. divisibility, elementary number theory) that 40x3 - 30x -1 has no rational roots without using the rational root test? (since it is very long and tedious and no calculators are allowed)

I was trying to prove this by contradiction.
Suppose x=m/n is a rational root in lowest terms.
But I am stuck at arriving at a contradiction...how can I possibly do so?


Any insights?

Thanks for any help!
 

Answers and Replies

tiny-tim
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I worked it out and at the end found out that , if I can prove that the cubic polynomial 40x3 - 30x -1 has no rational roots, then I am done.
Hi kingwinner! :smile:

The rational root theorem says that any rational root of a_nx^n + … + a_0, can be written as p/q, where p is a factor of a_0 and q is a factor of a_n.

So any rational root of your equation must be of the form ±1/b, where b is a factor of 40.

b can only be 1 2 4 8 5 10 20 or 40.

Now, the turning-points of the function are at 120x^2 = 30, or x = ±1/2.

At x = 1/2, the function is -11. At x = 0, it is -1.

So between x = 0 and 1/2 it is negative.

So that immediately eliminates all possible positive roots except 1.

At x = -1/2, it is 9. So there is one root between -1/2 and 0.

If I were you, I'd start by checking one of the highest possible negative roots! :smile:
 
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Hi,

Thanks very much! But is there any way to prove it using divisibility and elementary number theory? (since this question is from a non-calculus based course)

I have a rough idea, but just can't finish it:
We want to prove that 40x3 - 30x -1 has no rational roots
40x3 - 30x -1 = 0
Suppose (proof by contradiction) x=m/n is a rational root in lowest terms
=> 40m3 - 30mn2 - n3 =0
=> 40m3 - 30mn2 = n3

If p is a prime and p|m, then p|n3 => p|n since p is a prime which contradicts that x=m/n is in lowest terms => no such p => m=+/-1

Similarly, if q is a prime and q|n, then q|40m3 , and now I am stuck...since this argument above does not seem to work here due to the "40" appearing here...
 
tiny-tim
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If p is a prime and p|m, then p|n3 => p|n since p is a prime which contradicts that x=m/n is in lowest terms => no such p => m=+/-1
Yes, very good! You have just proved half of the rational root theorem! :smile:

In the letters of my previous post, your m is my p, which has to be ±1.

And your n is my q = b, which has to be 1 2 4 8 5 10 20 or 40.
 
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...Similarly, if q is a prime and q|n, then q|40m3
=>q|m or q|40 (since q is prime)

q|m => contradicts gcd(m,n)=1 => no such q => n=+/-1 => x = m/n = +/-1
q|40=> ?????

What does q|40 imply?
 
tiny-tim
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Nothing more than ±q = 1 2 4 8 5 10 20 or 40, I'm afraid. :redface:
 
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Oh, that's right! So I think the rational root test is the only way to go...after all
 

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