Graph of $x^n+nx^{n-1}y+nx^{n-2}y^2+...+nxy^{n-1}+y^n=0$ Today, I was taught about the 'homogenous' equation of 2-degree that represents a pair of straight lines intersecting at the origin. The equation was $ax^2+2hxy+y^2=0.$
Back at home, I was 'playing' with Desmos(the online graphing calculator) and plotting the graphs of functions of the form (let me generalise it),

$$x^n+nx^{n-1}y+nx^{n-2}y^2+...+nxy^{n-1}+y^n=0,\space \space \space n\in \mathbb{N}$$

I tried(with the help of Desmos) with 5 values for $n$, i.e. for $n=1,2,3,4,5$ and saw that

It 'always' represents straight lines.

So, am I correct? Why or why not. How does this depend on $n$?.
Thanks, in advance for any help.
Edit-Based on a valuable comment by @Carl Schildkraut I made some changes.
 A: By plugging $y=mx$ into
$$x^n+nx^{n-1}y+nx^{n-2}y^2+\dots+nxy^{n-1}+y^n=0$$
we get
$$x^n(1+n(m+m^2+\dots+m^{n-1})+m^n)=0$$
which means that the locus is made by a number $d$ of lines through the origin $y=mx$ where $d$ is the number of distinct real solutions of the polynomial equation 
$$P(m):=1+n(m+m^2+\dots+m^{n-1})+m^n=0$$
Note that if $m$ is a solution then it has to be negative and also $1/m$ is a solution.
If $n$ is odd then $P(-1)=0$, that is one of the lines is $y=-x$. If $n$ is even then $P(-1)=1-n+1<0$ for $n>2$, and $P(0)=1>0$, hence by continuity there is a solution in the interval $(-1,0)$ and by the previous remark there is a solutions in $(-\infty,-1)$.
From dxiv's comment below, by multiplying $P$ by $m−1\not=0$ (note that $P(1)\not=0$) we obtain 
$$f(m):=m^{n+1}+(n−1)m^n−(n−1)m−1=0$$
then
$$f'(m)=(n+1)m^{n}+n(n−1)m^{n-1}−(n−1)$$
and 
$$f''(m)=(n+1)nm^{n-1}+n(n−1)^2m^{n-2}=nm^{n-2}((n+1)m+(n−1)^2)$$
Since for $n>2$, $f''$ has exactly two real solutions namely $0$ and $-(n-1)^2/(n+1)$. Then, by Rolle's theorem, $f'$ has at most three real zeros and $f$ has at most four real zeroes. Since $f(m)=(m-1)P(m)$ and $P(1)\not=0$, it follows that $P$ has at most THREE distinct real solutions.
