# Prove that a cubic has no rational roots

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

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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.

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!

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tiny-tim
Homework Helper
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! 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! 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
Homework Helper
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! 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.

...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
Nothing more than ±q = 1 2 4 8 5 10 20 or 40, I'm afraid. 