$f(x),$
such that all of the following hold: (a) it has at least one inflexion point, (b) all inflexion points have $y$
coordinate equal to $0,$
(c) all its roots are real, and (d) it is symmetric with respect to the $y$
axis.

 Evaluate
$\frac{d}{dx}{(x^23)^3}.$
 Find the roots of the polynomial
$x^27x+12.$
 Find the inflexion point of the curve
$y=x^33x^2+5x1.$
 Evaluate

What does
$f$
being symmetrical about the$y$
axis imply about its degree$n?$

Does the fact that
$f$
has a point of inflexion restrict the values that$n$
can take? 
How can one express a polynomial in terms of its roots?

If you arbitrarily choose one root of the polynomial, what conditions does that set on other roots?

Once you have arbitrarily selected a pair of roots, how can you select the other pair to satisfy the inflexion point condition?

Since
$f$
is symmetrical about the$y$
axis, it must be of even order$n.$
At an inflexion point, the second derivative changes sign. For$n \leq 2,\,$
$f''(x)=0$
and so we deduce$n > 2.$
Hence, the smallest possible degree is$n=4.$
If the$y$
coordinate equals$0$
at all the inflexion points, then every inflexion point is a root. If all roots are real, then the polynomial may be expressed as a product of real monomials, i.e.$f(x)=(xa)(xb)\cdots.$
The symmetry of the function also means that the roots must be symmetrical, so$f(x)=(xa)(x+a)(xb)(x+b)$
$=(x^2a^2)(x^2b^2).$
We can arbitrarily choose a pair of symmetric roots, so let
$b=1.$
We then find the second derivative of$f,$
which is$f''(x)=12x^22a^22.$
The roots of this derivative,$\pm\sqrt{(a^2+1)/6},$
are points of inflexion. In order to satisfy the condition that all inflexion points are roots of$f,$
we equate these to$\pm a$
and solve to get$a=\frac{1}{\sqrt5}.$
Therefore$f(x)=(x^21)(x^21/5)$
in this case.