Show that a polynomial of degree 4 is birational equivalent to a polynomial of degree 3

Suppose that $f_{4}(x)$ is a polynomial of degree 4 with no multiple roots, $C$ is the curve defined by $y^{2}=f_{4}(x)$, I want to show that there is a polynomial $f_{3}(x)$ of degree 3 with no multiple root such that $C$ is birational equivalent to the curve defined by $y^{2}=f_{3}(x)$, I completely don't know how to do it, can anyone help me

This is a cross-site duplicate of this MO post. I present the highest-scoring answer from there as a community-wiki answer here for the benefit of future MSE users.

Myshkin:

The method explained in Husemöller's book on elliptic curves is as follows:

Take a general quartic $v^2=f_4(u)=a_ou^4+a_1u^3+a_2u^2+a_3u+a_4$, and let

$$u=\frac{ax+b}{cx+d}\qquad v=\frac{ad-bc}{(cx+d)^2} y$$

Substituting you get:

$$v^2=\frac{(ad-bc)^2}{(cx+d)^4}y^2=f_4\bigg(\frac{ax+b}{cx+d}\bigg)$$

which implies

$$(ad-bc)^2y^2=f_4\bigg(\frac{ax+b}{cx+d}\bigg)(cx+d)^4=\sum_{i=0}^4a_i(ax+b)^{4-i}(cx+d)^i=$$ $$=c^4f_4\bigg(\frac{a}{c}\bigg)x^4+f_3(x)$$

where $f_3(x)$ is a cubic polynomial whose coefficient of $x^3$ is $c^3f'_4(a/c)$. For $a/c$ a simple root of $f_4$ and $ad-bc=1$, this leaves the cubic equation $y^2=f_3(x)$.

From here you can use the tools you mention in the question to take care of the cubic.