Find K for negative roots Given
$$m^3+m^2+2m+K=0 $$
how do I find the $K$ values that allow just negative roots?
Edit: 
Even though some answers are already posted, I´ll clarify - as suggested in the comments - that this question arise from my earlier one: Find K values that make a differential equation solution stable.
As far as I understand, the answer should be the same for both questions, as complex roots guarantee stable solutions, and then we´d just need the real roots to be negative.
 A: The function $f(x)=x^3+x^2+2x+K$ is increasing, for the derivative
$$f'(x)=3x^2+2x+2$$
is positive.
Then the polynomial has only one real root. It is clear that if $K=0$, this root is $0$. Since $f(0)=K$ and $f$ is increasing, the root is negative if $K>0$ and positive if $K<0$.
A: Hint: If all 3 roots are negative and real, then the product of roots must be negative.
If 2 roots are complex, and there is only 1 real negative root, what must the product of roots be?
You may like to read up on Vieta's formulae.
A: Some algebra knowledge makes this problem a lot easier.  Let's say the roots of the cubic are $a_1, a_2$, and $a_3$.  Then notice that:
$$f(x) = (x-a_1)(x-a_2)(x-a_3)$$
Expanding this out, we get:
$$f(x) = x^3 - (a_1 + a_2 + a_3)x^2 + (a_1a_2 + a_2a_3 + a_1a_3)x - a_1a_2a_3$$ 
Now it is a fact that, if $a+bi$ is a complex root of a polynomial, then so too is its complex conjugate $a-bi$.  A direct consequence of this is that a cubic has either $3$ real roots or a single real root and two complex roots.  
The first thing to notice is that we cannot have three real negative roots, since the sum of the roots, which is the coefficient on $x^2$, is positive.  Therefore, the polynomial in question can have only one real root.
Next, the constant term is the signed product of the roots. Given that there must be two complex roots, then note that their product $(a+bi)(a-bi) = a^2 + b^2$ is positive, and so it would be necessary for $K$ to be ____ if the real root is to be negative. 
