Setting up a polynomial that has what I am referring to as a "wiggle root" at $x=0$ EDIT: I am loathe to delete/restructure this post, so I will include the following: A function has a "wiggle root" at $c$ if $f(c)=0$ and $f(x)$ crosses the $x$-axis.
I am trying to set up a polynomial such that it has a wiggle root at $x=0$. I have gotten it such that there are wiggle roots at $x=-1$ and $x=1$. A wiggle root is defined as the behavior of what you see at $x = 1$ and $x = -1$ in the image below. I do not know how to describe it conceptually other than the derivative at that point is such that it approaches from the positive $y$-axis from the left and negative $y$-axis from the right.


My questions are:
  
  
*
  
*What part of the polynomial determines whether the root is a wiggle root or not?
  
*How do I create a polynomial that has a wiggle root and also satisfies these conditions: the polynomial should have 2 imaginary roots, 2 real roots, and double roots. It should be fully factorable.
  

Here are some other polynomials that I have tried:
$$P(x) = (x+1)(x+1)(x-1)(x-1)(x+i)(x-i)$$
or
$$P(x) = x^6+2x^5+x^4−x^2−2x−1$$
 A: What you describe as a "wiggle root" is a stationary point of inflection of the curve $y=f(x)$ that is also a root of the equation $f(x)=0$.
You therefore require:
1) $f(a)=0$ : to make $a$ a root of the equation
2) $f'(a)=0$ : to make $(a,0)$ a stationary point of the curve
3) $f''(a)=0$ : to make $(a,0)$ a point of inflection
      As egreg noted in a comment this last condition is
      necessary but not sufficient. The polynomial must
      have the factor (x-a) raised to an odd power greater
      than or equal to 3 to fully achieve a "wiggle root".

You have tried to set up your polynomial as $f(x)=(x-a)^2(x-b)^2(x-ci)(x+ci)$, but this will not have stationary points of inflection at $x=a$ and $x=b$. To achieve that you will need something like:
$$f(x)=(x-a)^3(x-b)^3(x-ci)(x+ci)$$
To have a stationary point of inflection at $x=0$ you will have to include the factor $x^3$, giving you:
$$f(x)=x^3(x-a)^3(x-b)^3(x-ci)(x+ci)$$
I wasn't sure if you wanted there to be two real roots and a third root at $x=0$, but if you just want two real roots (one of which is $x=0$), then you want:
$$f(x)=x^3(x-a)^3(x-ci)(x+ci)$$
A: Definition: A polynomial, $f(x)$, has a "wiggle root" at $a$ if and only if $f(a)=0$ and the degree of the root is an odd number greater than $1$.
Claim: If you are interested in finding polynomials that have wiggle roots at $a$, you just have to look at $(x-a)^3(x-a)^{2k}p(x)$. As $p(x)$ ranges over all polynomials that are not zero at $a$, this will actually give you every polynomial with a wiggle root at $a$.
Proof of the above claim:
1) It's clear that this form always has a wiggle root by the definition I described.
2) Let $p(x)$ have a wiggle root at $a$. Then $\exists$ a maximum $k\in\mathbb{N}$ such that $(x-a)^{2k+1}|p(x)$ and so $p(x)=q(x)(x-a)^{2k+1}$. Then we also have that $p(x)=q(x)(x-a)^3(x-a)^{2(k-1)}$ and so we are done, since $k$ being maximum implies that $(x-a)\not|q(x)$
