How to solve this function $f(x)$ for degree $3$ or $4$ 
If $f(x)=0$ is a polynomial whose coefficients are either $1$ or $-1$ and whose roots are all real, then the degree of $f(x)$ can be equal to$:$


$A$. $1$


$B$. $2$


$C$. $3$


$D$. $4$

My work$:$
For linear only four polynomials are possible which are $x+1$ and $-x+1$ and $x-1$ and $-x-1$. All of which have real roots. So the answer is $A$.
For quadratic we have $8$ polynomials which are $x^2+x+1$ and $x^2+x-1$ and $x^2-x+1$ and $x^2-x-1$ and $-x^2+x+1$ and $-x^2-x+1$ and $-x^2+x-1$ and $-x^2-x-1$ Here four of the polynomials have real roots which are $x^2+x-1$ and $-x^2-x+1$ and $-x^2+x+1$ and $x^2-x-1$. So we can say that $f(x)$ can be of degree two. So $B$ is also the answer.
For cubic and quartic I don't know how to proceed. Any hints or solutions are appreciated.

By the Rational Root Theorem, if such a polynomial has a rational root, it must be $x = \pm 1$.  It turns out that 8 of the 16 possible cubics (the ones with even numbers of + and - coefficients) have such a root.
For a quartic, we get $f(x) = \pm x^4 \pm x^3 \pm x^2 \pm x \pm 1 = \pm 1 \pm 1 \pm 1 \pm 1 \pm 1$, which always works out to an odd number, never zero.  So $f$ has no rational roots.  Proving that there are no real solutions can be done using Descartes' Rule of Signs, as in the other answers.
