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.
Best Answer
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?