How do I put in words that $2\cos^2 x +\sin x-1$ cannot be factored? I know that $2\cos^2 x +\sin x-1$ can't be factored, but I don't really know how to explain it. The best explanation I can come up with is "it would be like factoring $2x^2 + y -1$; it just doesn't work", but that hardly explains exactly why it wouldn't work. Does anyone know a better way of putting it?
EDIT:
I'm not asking if it's impossible to factor at all, I know the equation can be changed to only having sines and from there factored like a regular quadratic. That's not the question. The question is how do I put into words that the equation, exactly as written above, cannot be factored unless you change the equation into $-2 \sin^2⁡x+\sin⁡x+1$ and from there factor it as $(-2 \sin⁡x-1)(\sin⁡x-1)=0$. I've already done that latter part, I just need a way to explain why I have to do it, why it's necessary in the first place to change $2\cos^2 x +\sin x-1$ in order to factor it. I'm sorry for being unclear.
EDIT 2:
To clear up some confusion, here's what the assignment asks for:
Given the equation $2 \cos^2⁡x+\sin⁡x-1=0$,


*

*Explain why this equation cannot be factored.

*Use a trigonometric identity to change the equation into one that
can be factored.

*Factor the equation.

*Determine all solutions in the interval 0≤x≤2π.
I've had no problems with steps 2, 3 and 4, so I did those first. However, I'm not sure how to word the answer to 1 right, because I can't think of anything else than "it just doesn't work".
 A: you know that $\cos x^2+\sin x^2=1$. Because of this, the polynomial can be written as
$$
2(1-\sin^2 x)+\sin x-1=-2\sin ^2 x+\sin x +1=(2\sin x +1)(-\sin x+1) 
$$
EDIT
When you do not want to use any relationships between $\sin $ and $\cos$, you have two independent variables indeed, so $2x^2+y-1$. This can't be factored in general. To show that, you can try to find some values of $x$ and $y$ (do you want to 'assume' $\sin$ and $\cos $ have range $[-1,1]$) for which the expression is prime. For example $x=1$ and $y=1$.
EDIT 2
A number $p$ is prime if its only factors (positive divisors) are $1$ and $p$ itself. $2$ is prime, but $6=2\cdot 3$ is not for example. When $2x^2+y-1$ is prime, it cannot have any factors except for $1$ and itself.
I now realize this is not a conclusive proof, because it can occur that a factor is $1$ sometimes. (See some comments on this post)
EDIT 3
The question only says 'explain', so a full proof doesn't seem to be necessary. The only term that can be factored is the $2\cos^2x$. A factorization will look like $(\cos x+a)(2\cos x+b)$ (It is possible that the two factors in front of the $\cos$ are both $\sqrt 2$ etc, but that won't work. Lets assume they have to be integers.) Now, $2a+b=0$ and $ab=\sin x-1$. It follows that $b=-2a$, thus $-2a^2=\sin x-1$. This results in the (not so nice) factorization
$$
2\cos^2x+\sin x-1=2\left(\cos x + \sqrt{\frac{1-\sin x}2}\right)\left(\cos x - \sqrt{\frac{1-\sin x}2}\right)
$$
