Number of Real roots of cubic I just have a quick question about a polynomial and its roots,
For example, I was solving the differential equation $$\frac{dy}{dx}=\frac{3x^2+4x+2}{2y-2}$$
I solved it using the basic methods of separation and solved the solution implicitly as $y^2-2y=(x^3+2x^2+2x+D)$,  $D \in \mathbb{R}$
so when I want to solve for $y$ explicitly, I solve the quadratic and get the form,
$$y=1\pm\sqrt{1+(x^3+2x^2+2x+D)}$$
Now, my question is in regard to the cubic $\mathbb{P}=x^3+2x^2+2x+1+D$
My book noted without explanation that , $\mathbb{P}$ is such that for all $D \in \mathbb{R}$, only one real root exists, call it $r(D)$.
My question is, how was this seen so easily? I think there might be something very simple I am missing potentially. I hope someone can help with the explanation, thanks!
 A: The derivative of $\mathbb{P}(x)$ is $3x^2 + 4x + 2$ (doesn't depend on $D$!). By the quadratic formula the discriminant of this derivative is $16-24 = -8$, so the derivative has no real roots. This means that $\mathbb{P}(x)$ has no critical points (ie, no points where the derivative is 0). Thus, $\mathbb{P}(x)$ is either monotonically increasing, or monotonically decreasing. Either way, it will cross the real axis exactly once.
A: Let $\Bbb{P}(x) = x^3 + 2x^2 + 2x + 1 + D$ and calculate it's derivative:
$$
\Bbb{P}'(x) = 3x^2 + 4x + 2.
$$
Since the discriminant of this quadratic polynomial is
$$
\Delta = (4)^2 - 4(3)(2) = -8 < 0,
$$
it has no real roots.  Why is this a problem?
Suppose that $\Bbb{P}(x)$ had more than one real root.  Pick two of them and call them $a$ and $b$ (with $a < b$, say).  By the Mean Value Theorem (really, its special case called Rolle's Theorem), there is a number $c$ such that $a < c < b$ and
$$
\Bbb{P}'(x) = 0.
$$
But, we just saw that $\Bbb{P}'(x)$ is never zero.  Contradiction.
