hypersurface intersected with generic line Let $f \in \mathbb{R}[x_1,\cdots,x_m]$ be a homogeneous multivariate polynomial of degree $n$. Now, for $u,v \in \mathbb{R}^n$ the form $f(\lambda u + \mu v)$ can be written as $\mu^n h(\lambda/\mu)$, where $h$ is a polynomial in one variable of degree $n$. 
Question: Why is it true that if $f$ has no multiple factor in $\mathbb{R}[x_1,\cdots,x_m]$, then there exist $u,v$ such that $h$ has precisely $n$ distinct roots? Is there a simple algebraic or geometric explanation?
 A: You mean $h$ has $n$ (degree of $f$) distinct roots in $\mathbb C$ (not in $\mathbb R$), equivalently $h$ is separable over $\mathbb R$. 
A geometric explanation is Bertini's theorem. Consider the closed subvariety $X$ defined by $f$. You want a line in the projective which intersects $X$ along a smooth subvariety (of dimension $0$). Then the intersection has degree $n$ by Bézout. 
The usual Bertini's theorem is for smooth quasi-projective subvarieties. Here you need the following variant: if $X$ is a reduced closed subvariety of a projective space over a perfect infinite field $k$, then there exists a hyperplane $H$ such that the singular locus  of $H\cap X$ has positive codimension in the singular locus of $X$. 
Using this result repeatedly you ended with a line intersecting $X$ along a smooth zero-dimensional subvariety. 
I don't have a reference for the above variant. But the proof is just a variant of the usual proof  (see Hartshorne, II.8.1.18; fix a closed point in each irreducible component of the singular locus, and add the open condition that the hyperplanes we are looking for do not pass through any of these points). 
