From reading about algebraic numbers, the definition is that an algebraic number is a number that is a root of a non-zero polynomial in one variable with integer (or, equivalently, rational) coefficients.
All rational numbers are considered algebraic because any rational number is the root of a non-zero polynomial, namely $ax + b$
If I understand correctly though, $ax + b$ is a special case of the binomial equation $ax^n + b = 0$ and for $n = 1$ we have the polynomial $ax + b$ which we consider that has roots in the rational domain.
But I don't understand why we consider that the rational numbers are algebraic numbers if they are not solutions to $ax^n + b$ as well. E.g. they are not for $x^2 - 2 = 0$
Is it not required that all equations of higher degree be reducible to equations of the type $ax + b$ in order to consider the rationals as algebraic numbers?
The roots of $x^2-2$ are algebraic, but also the roots of $x^7-23x^4+37x-1$ (a polynomial I wrote with random coefficients).
There's nothing in the definition of algebraic number that requires to consider only polynomials of the form $ax^n+b$. The roots of such a polynomial are algebraic numbers, but the set of algebraic numbers is larger.
Of course general polynomials with integer coefficients are not reducible to factors of the form $ax+b$, otherwise every algebraic number would be rational.
We do want that rational numbers are algebraic, don't we? If you're worried about the degree, then consider that the rational number $r$ is a root of $(x-r)^n$, which has the degree you want and rational coefficients.
And if you want a polynomial of the form $ax^n+b$ with $n>1$, this can be accommodated as well. Say you have $r=p/q$, with $p$ and $q$ integers. Then $r$ is a root of $q^nx^n-p^n$, for every $n$.