How prove this $f(x)$ and $g_{t}(x)$ be relatively prime. 
Let the real  coefficient polynomials
  $$f(x)=a_{n}x^n+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$$
  $$g(x)=b_{m}x^m+b_{m-1}x^{m-1}+\cdots+b_{1}x+b_{0}$$
  where $a_{n}b_{m}\neq 0,n\ge 1,m\ge 1$, and let
  $$g_{t}(x)=b_{m}x^m+(b_{m-1}+t)x^{m-1}+\cdots+(b_{1}+t^{m-1})x+(b_{0}+t^m).$$
  Show that
there exist positive $\delta$, such for any $t$ such that $0<|t|<\delta$, and such
  $f(x)$ and $g_{t}(x)$ be relatively prime.

I fell this result is very well, because although two polynomial are not relatively prime, we can do it to one of the polynomial tiny perturbation makes relatively prime.
But I can't prove my problem.
Thank you very much
 A: Here is a series of steps, which seem mostly true to me.
Fact: $f(x)$ has at most $n$ distinct roots.
Claim: There exists a map  $G: [0,1] \rightarrow (\alpha_1, \alpha_2, \ldots, \alpha_m)$ which is differentiable in each coordinate, and $ \alpha_i$ are roots of $ g_t(x)$ (with multiplicity).
Possible Proof: Use Inverse/Implicit function theorem till 2 roots meet. Then be very careful?
Claim: On any open interval, $\frac{dG}{d\alpha_i}$ is not identically 0.
Proof: If it does, then there is a number $ \alpha$ such that $ g_t(\alpha) = 0 $ on that open interval. But this implies that $ \sum t^i \alpha^{m-i} = 0 $ infinitely often, which is a contradiction (since there is at most $m$ values of $t$ that can satisfy the polynomial).
Hence, the result follows, where we take a small enough $t$, such that the roots of $g_t$ all avoid the roots of $f$. The final claim allows us to move off of common roots of $f$ and $g_0$ .
A: let $x_i$ be the roots of $f$.
$g_t$ and $f$ are not primes iif $g_t(x_i) \neq 0$ for every $x_i$.
So $g_t$ and $f$ are primes iif $t$ no in the finite set of roots of $$b_{m}x_i^m+(b_{m-1}+t)x_i^{m-1}+\cdots+(b_{1}+t^{m-1})x_i+(b_{0}+t^m)$$
A: You just need to check that the resultant $r(t)=Res_x(f(x), g_t(x))$, which is a polynomial in $t$, is not the zero polynomial. Any root of $r(t)$ marks a value of $t$ where both polynomials have a common factor. If $r(t)\ne 0$, the polynomials $f(x)$ and $g_t(X)$ are relatively prime.
And as in the other answers, as $r(t)$ only has a finite number of roots...
