How do you find the turning points of a polynomial without using calculus? 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.
 A: 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 $(2,P(2)) = (2,19)$.
A: .
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".
A: 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. $2x^2=12-x^2 \implies x^2=4$.
A: If you know
how to differentiate,
just find the roots of
$P'(x) 
= -3x^2+12
$,
which are
$x = \pm 2$.
If you don't,
$P(x) 
= -x^3+12x+3
$,
so,
imitating the invention of calculus,
$\begin{array}\\
P(x+h) 
&= -x^3+12x+3\\
&= -(x+h)^3+12(x+h)+3\\
&= -x^3-3x^2h-3xh^2-h^3+12x+12h+3\\
&= (-x^3+12x+3)-h(3x^2+3hx+h^2-12)\\
&= P(x)-h(3x^2+3hx+h^2-12)\\
\end{array}
$
so,
as expected
$\dfrac{P(x+h)-P(x)}{h}
=-(3x^2+3hx+h^2-12)
$.
Letting
$h \to 0$,
we find that
the turning point of $P$
are
when
$3x^2 = 12$
or
$x = \pm 2$.
There is probably a clever way to do this,
and this ain't it.
