TITLE: How do you find the turning points of a polynomial without using calculus?
QUESTION [11 upvotes]: I have a polynomial $P(x) = -x^3+12x+3$, and I am asked to find the turning points of it, and hence state how many zeroes it has. Since this chapter is separate from calculus, we are expected to solve it without differentiation.
However, I'm not sure how I could solve this. I'm having trouble factorising it as well since the zeroes seem to be irrational.

REPLY [22 votes]: 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 $(2,P(2)) = (2,19)$.

REPLY [10 votes]: 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. $2x^2=12-x^2 \implies x^2=4$.