Solve $x^4+3x^3+6x+4=0$... easier way?

Hint:

First check that $0$ is not a solution, hence $x\neq0\,$, so it is legal to divide by $x^2$. We get

$$ x^2+3x+\frac6x+\frac4{x^2}=0. \tag1 $$

Now note that

$$ \left(x+\frac2x\right)^2=x^2+4+\dfrac{4}{x^2}\iff x^2+\dfrac{4}{x^2}=\left(x+\dfrac2x\right)^2-4. $$

So $(1)$ can be written as :

$$ \left(x^2+\dfrac4{x^2}\right)+\left(3x+\dfrac6x\right)=0\iff \left(x+\dfrac2x\right)^2-4+3\left(x+\dfrac2x\right)=0. $$

Now use the substitution $u=x+\frac2x$ and you get the quadratic :

$$ u^2+3u-4=0. $$

Answer to a comment:

Awesome, but how did you see that? And is this just then a specific case...?

I remarked that the coefficients of the equation were symmetric in the following sense :

$$ x^4+3x^3+6x+4=0\iff (x^4+\color{#C00}2^2)+3(x^3+\color{#C00}2x)=0. $$

So I tried to divide by $x^2$, then to find a relation between $x^2+\tfrac4{x^2}$ and $\left(x+\tfrac2x\right)^2$, so that I can convert it into a quadratic.

Reorganize the term like $$(x^4+4+4x^2)+(3x^3+6x)-4x^2=(x^2+2)^2+3x(x^2+2)-4x^2$$ than it is easy to come up with $$(x^2+4x+2)(x^2-x+2)$$

In general any quartic equation can be solved. If the polynomial happens to be the product of two irreducible polynomials over $\mathbb Z$, the following method allows to find the coefficients easily. Reducing the coefficients mod $2$, one has $$f(x):=x^4+3x^3+6x+4=x^4+x^3=x^2(x^2+x).$$ Lifting back to ${\mathbb Z}$-coefficients, one may assume that $$f(x)=(x^2+2ax\pm 2)(x^2+bx\pm 2),$$ where $a,b\in {\mathbb Z}$ and $b$ is odd. By comparison of coefficients, one has $$2a+b=3,2ab\pm 4=0,\pm 4a\pm 2b=6,$$ which shows immediately that one needs to take the ‘plus’ sign from $\pm$. From the first two equations, by eliminating $a$, one has $b=4$ or $b=-1$, so $b=-1$ since it is odd. It follows that $a=2,b=-1$ and $$f(x)=(x^2+4x+2)(x^2-x+2).$$