How to show the equation $x^4 + 2cx^3 + 6x^2 + 60x =-11$ has exactly two real solutions? How can we show that $x^4 + 2cx^3 + 6x^2 + 60x =-11$ has exactly two real roots? $c$ is any element in the interval $(-2,2)$.
 A: Let $p(x)=x^4+2cx^3+6x^2+60x+11$ and now let's do a little calculus. Differentiating:
$$p'(x)=4x^3+6cx^2+12x+60.$$
We want to look at the zeroes of $q:=p'$. Again, we differentiate:
$$q'(x)=12x^2+12cx+12=12(x^2+cx+1)$$
which has zeroes $x=\frac{-2c\pm\sqrt{c^2-4}}2\in\mathbb{C}\setminus\mathbb{R}$ since $c\in(-2,2)$. This shows $q$ is strictly increasing. Since $q(x)\to\pm\infty$ as $x\to\pm\infty$, $q$ takes both positive and negative values and hence has a zero; since $q$ is increasing, this zero is unique. Hence $p$ has one stationary point. Since $p(x)\to\infty$ as $x\to\pm\infty$, it suffices to show that $p(x)<0$ for some value of $x$. By considering the graph of $p$, this will give us exactly two zeroes. Now
$$p(-1)=1-2c+6-60+11<0$$
so we are done.
A: Let $f(x)=x^4+2cx^3+6x^2+60x+11$. Its second derivative is:
$$f''(x)=12x^2+12cx+12$$
But for $x\ge 0$,
$$0\le 12(x-1)^2<f''(x)$$
and for $x<0$,
$$0\le 12(x+1)^2<f''(x)$$
that is, $f'$ is increasing and therefore (since $f'$ is a polynomial with odd degree) $f'$ has exactly one zero. Rolle's theorem implies that $f$ has at most two zeros.
Now, $f(-10)>0$, $f(-1)=1-2c+6-60+11=-42-2c<0$ and $f(0)=11>0$. Bolzano's theorem implies that $f$ must have at least two zeros.
Thus, $f$ has exactly two zeros.
