Multitangent to a polynomial function I'm trying to build some exercises on tangents of functions for beginner students in mathematical analysis. In particular I would like to suggest the study of polynomial functions  $ y = p (x) $ of which is possible to determine the graph with elementary methods and also determine (if it exists) the  $n$-tangent, i.e. a straight line $ y = mx + q $ (with $ m \ne 0$  to avoid trivial solutions) which has $ n $ distinct points of tangency with the graph and no other intersection points with it, so that the system
$$
\begin{cases}
y=p(x)\\
y=mx+q
\end{cases}
$$ 
has $n$ double solutions.
For bi-tangents I find, for example, functions of the form:
$$
y= a(x^4-3k^2x^2+2k^3x)
$$
that have as bi-tangents the straight lines 
$$
y=2ak^3x-\dfrac{9}{4}ak^3
$$
or:
$$
y=a\left( \dfrac{1}{4}x^4 -\dfrac{3k}{2}x^3+\dfrac{9k^2}{4}x^2-k^3x\right)
$$
with bi-tangents  $y=ak^3x$.
I cannot, however, find an example with a $ 3 $ -tangent, i.e. a polynomial of sixth degree $ y = p (x) $ such that $ p (x) $ and $ p '(x) $ are decomposable (more or less easily) in factors of degree  $n \le 2$, and that at the same time has a $3$-tangent such that hits points of tangency can be determined without using  the general formula to solve a cubic equation.
Someone knows any function of this type, or may suggest an efficient way to find it? 
In other words: find a function:
$$
y=a_6x^6+a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0
$$
such that $ y $ and $y'$ are factorizable with factors of degree $n \le 2$ and there exist $m,q \in \mathbb{R}$ (or better $\in \mathbb{Q}$) such that
$$
a_6x^6+a_5x^5+a_4x^4+a_3x^3+a_2x^2+(a_1-m)x+a_0-q
 =a_6\left( x^3+Bx^2+Cx+D\right)^2
$$ 
and the latter $3^{rd}$ degree polynomial is also factorizable.

Added after the Answer.
The answer of Michael Burr don't fit the request  that $f(x)$ and $f′(x)$ has roots that we can find solving equations of degree $\le 2$.
Here I sum up and a bit generalize the problem:
Let $f(x) \in \mathbb{R}[x]$ be a polynomial of degree $2n > 4$ and $f'(x) $ its derivative. I want determine the coefficients of $f(x)$ in such a way that:
all the real roots of $f(x)$ and $f'(x)$ can be found solving equations of degree $\le 2$, and there are $m,q \in \mathbb{R}$ such that the polynomial $g(x)=f(x)+mx+q$ has $n$ double roots.
Or proof that such a polynomial can not exists.

After three solutions, I found the easy way!
Start with $g(x)=(x-p)^2(x-r)^2(x-s)^2$.
For any $m$ and $q$, consider $y=m(x-p)+q$.  Add this to $g$, to get $f$, then $f$ passes through $(p,q)$ and the line $y=m(x-p)+q$ is tangent to the curve of $f$ at $p$, $r$, and $s$. 
