How do you find the turning points of a polynomial without using calculus?
You want to know for which $c$ it is the case that $P(x)+c$ has a double root. We could mess around with the discriminant of the cubic, but that's probably too much work. Instead, suppose $P(x) + c = -(x-a)^2(x-b)$, so that $$ -x^3 + 12 x + 3 + c = - x^3 + (2a+b)x^2 -(a^2 + 2ab)x +a^2 b $$ From this, we read off $2a+b = 0$, $a^2 + 2ab = -12$, and $3+c = a^2 b$. From the first two, solutions $(a,b)$ are $(-2,4)$ and $(2,-4)$. We don't even need to solve for $c$ because the double root (the turning point) occurs at $x=a$, so the turning points are $(-2,P(-2)) = (-2, -13)$ and $(2,P(2)) = (2,19)$.
For a cubic, any turning point is a double root of a suitable translation on the Y-axis. i.e. we seek double roots of $P(x)+k = -x^3+12x+3+k$ for some value of $k$. So we must have for some $a, b$, $$-x^3+12x+3+k = (x-a)(x-b)^2$$
Equating coefficients, we get $$a+2b=0,\quad 2ab+b^2=-12, \quad ab^2 = 3+k$$ The first two gives you $b=\pm2$, which are the turning points!
Here is another way with AM-GM. Note that from symmetry it is enough to find the positive turning point of $f(x) = -x^3+12x$. As $f(0) = 0, f(1) = 11, f(2\sqrt3) = 3$, we must have a local maximum in $[1, 2\sqrt3]$.
So we maximize $2f^2 = (2x^2)(12-x^2)(12-x^2)$ which is a product of three positive terms with a constant sum, which means the maximum is when the terms are all equal, viz. $2x^2=12-x^2 \implies x^2=4$.
. Do this: let $Q(x)=P(x)-3=-x^3+12x=-x(x+\sqrt{12})(x-\sqrt{12})$. This has $2$ turning points, because there are three roots, hence there are two turning points between the two pairs of consecutive roots. $P$ is just a translation of $Q$, hence has $2$ turning points as well.
To find the turning points, I thought I would make this little manipulation on the negative of the function: $x^3-12x-3 = (x^3+3x^2+3x+1) -(3x^2+15x+4)=(x+1)^3 - (3x^2+15x+4)$. Now $(x+1)^3$ changes sign only at $-1$, so it is down to us to see the behaviour of the function $3x^2+15x+4$. This factorizes as $3(x+\frac{5}{2})^2 - 14.75$. Finally, $x^3-12x-3 = (x+1)^3 - 3(x+\frac{5}{2})^2 + 14.75$.
This inspection tells us that the turning points on both sides are between $1$ and $2.5$ by plugging these values in. I'm not sure more can be done analytically other than "imitate calculus".