Does $f(x)=x^3+x^2-8x+1$ have rational roots?

Question

Problem

Does $f(x)=x^3+x^2-8x+1$ have rational roots ?

Attempt to solve

A citation from our lecturer

Possible rational roots are in form of:

$$ \frac{\text{factor of constant}}{\text{factor of highest degree}} $$

now if i would rewrite equation so that factor of every degree is visible:

$$ f(x)=(1) \cdot x^3+(1) \cdot x^2 + (-8) \cdot x + (1)\cdot 1 $$

$$ \text{factor of constant} = 1 $$ $$ \text{factor of highest degree} = 1 $$ $$ \implies \frac{1}{1} = 1 \neq 0 \implies \text{"no rational roots"}$$

However the example solution for this problem suggests that $1$ and $-1$ are factors but i cannot see how. If you would use $-1$ as factor it would change the polynomial in to another one. It claims that constant and highest degree factor have $-1$ as factor in common but i cannot see how.

Another thing is i don't quite get how this implies this polynomial doesn't have any rational roots ? If someone could provide an explanation on what's going on, that would be great.

This follows from Vieta's theorem. The product of all roots is equal to constant divided by coefficient of the highest degree. — Vasili, Oct 16 '18 at 17:30
@Vanya Not true, the proofs requires more structure than that, e.g. the existence of gcds, e.g. see this proof. — Bill Dubuque, Oct 16 '18 at 18:12

score 2 · Accepted Answer · answered Oct 16 '18 at 17:22

2

The method described by your lecturer says that if the polynomial has rational roots, then these roots can be obtained by a certain process. Since that process provides no rational roots, the conclusion is that $f(x)$ has no rational roots.

And what's so strange about the fact that $f(x)$ has no rational roots. The roots of $x^2-2$ are $\pm\sqrt2$, which are irrational. Therefore, $x^2-2$ has no rational root. What's supposed to be the problem?

answered Oct 16 '18 at 17:22

José Carlos Santos

427,504

The problem is i don't think i quite understand on how this process for obtaining the roots works. Or more specifically why does it work ? – Tuki Oct 16 '18 at 17:25
See the proof. – José Carlos Santos Oct 16 '18 at 17:26

score 1 · Answer 2 · answered Oct 16 '18 at 19:43

1

Alternative approach: $x^3+x^2-8x+1$ is irreducible over $\mathbb{F}_2$.
In particular it is irreducible over $\mathbb{Q}$, so it has no rational roots.

answered Oct 16 '18 at 19:43

Jack D'Aurizio

353,855

Does $f(x)=x^3+x^2-8x+1$ have rational roots?

Problem

Attempt to solve

2 Answers2