Can all cubic/quartic polynomials be expressed in a form with only one x term? Quadratic expressions $ax^2+bx+c$ can all be expressed in a form with only one x term: 
$$a(x+\frac{b}{2a})^2+c-\frac {b^2}{4a}$$
Is the same true for all cubic or quartic expressions?
Is there a name for this property (expressable in a form with only one x term)?
 A: Your only choice to reach third degree would be $a(x+b)^3+c$, which has only three degrees of freedom when four are neede for the general cubic. Specifically, $x^3+x$ cannot be obtained this way.
A: Depending on what you mean by "in a form with only one $x$ term", probably not.
If you mean to ask whether cubic polynomials can (analogously to the vertex form you've written for the quadratic) all be written in the form
$$A(x - B)^3 + C, \qquad (\ast)$$
the answer is definitely no: The expression only has three real parameters, but the general cubic depends on four (the coefficients of the four powers of $x$ that occur). One can, however, write a cubic polynomial
$$a x^3 + b x^2 + c + d,$$
as a cubic polynomial
$$a \left(x + \frac{b}{3a} x\right)^3 + c' \left(x + \frac{b}{3a}\right) + d',$$
in $\left(x + \frac{b}{3a}\right)$ and which in particular has no quadratic term in that expression.
The statements for quartic (and higher-degree) polynomials are analogous.
I don't know of any special name for polynomials of the form $(\ast)$, but observe that a cubic polynomial has this form exactly if its roots form an equilateral triangle in the complex plane (centered at $B$ in fact): If $B + z$ is one solution to $(x - B)^3 = -\frac{C}{A}$, then the three roots are
$$B + z, \qquad \omega(B + z), \qquad \omega^2(B + z),$$
where $\omega$ is a nontrivial third root of unity.
Likewise, a polynomial of degree $n > 3$ can be written in the form $(\ast)$, where we replace the exponent $3$ with $n$, if its roots are the vertices of some regular $n$-gon in the complex plane. In particular, any such polynomial can have at most two roots.
