I was taught that finding a factor (and hence a solution) of a cubic equation may be easier if I try if the factors of the free term are roots of the equation. For example, if one has an equation $x^3-4x^2+9x-10=0$, then they should try substituting the factors of 10 (1, 2, 5, 10) and see if it makes the polynomial equal to zero. If it does, then (x - this factor) will be a factor of the polynomial. Then the remaining factor (quadratic in the case of a cubic equation) can be easily found by trial and error or by long division. My questions are:
1) Why does substituting factors of the free term work? What's the intuition and/or proof behind this?
2) What are the limitations of this method? Will it always work, if no, what are the limitations/assumptions?
3) Will this method work for higher order polynomials too (for example quartic)?
This method is known as the Rational Roots Theorem, and more specifically one can test all factors of the form $\frac{s}{t}$, where $s$ is a divisor of the free term (the constant) and $t$ is a divisor of the coefficient of the highest powered term.
This method only gives rational roots, so it does not work for a polynomial such as $x^2-2=0$, which has irrational roots.
Yes, this method extends to any polynomial.