swampwiz
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The easiest proof uses topology, which has a nice intuitive feel, allowing folks to use aspects of it without going into the nitty-gritty.
First presume that these solutions exist in ℂ, then presume a path that x takes in which it is a circle of very large radius in ℂ, and thus which encircles all the solutions. The corresponding path of the polynomial output will also be in ℂ, and in the form large, wavy circle-ish share - i.e., the ultimate term will be a large circle, with the lesser terms generating the waviness - and such that the waviness is too small for the path to miss winding around the origin.
Now decrease the radius of the path of x, which will cause the polynomial path to become a smaller, relatively more wavier circle, and at some point, the waves will be relatively strong enough to touch the origin. Since the origin represents the value 0, this must mean that there is some value in the circle of x that is a solution of the polynomial, and so the polynomial can be factored into a pair of factors in which one is a linear term with the negative of this just found solution and a residual factor that has degree -1 from the original.
Now simply repeat the exercise using the residual factor; each iteration will result in another solution being found, leaving as the residual a corresponding lower degree polynomial. At some point the residual polynomial will be of degree 1, and thus the solution for it is trivially the free term. The net effect of all this is that the number of solutions that are found will correspond to the degree of the original polynomial.
QED
First presume that these solutions exist in ℂ, then presume a path that x takes in which it is a circle of very large radius in ℂ, and thus which encircles all the solutions. The corresponding path of the polynomial output will also be in ℂ, and in the form large, wavy circle-ish share - i.e., the ultimate term will be a large circle, with the lesser terms generating the waviness - and such that the waviness is too small for the path to miss winding around the origin.
Now decrease the radius of the path of x, which will cause the polynomial path to become a smaller, relatively more wavier circle, and at some point, the waves will be relatively strong enough to touch the origin. Since the origin represents the value 0, this must mean that there is some value in the circle of x that is a solution of the polynomial, and so the polynomial can be factored into a pair of factors in which one is a linear term with the negative of this just found solution and a residual factor that has degree -1 from the original.
Now simply repeat the exercise using the residual factor; each iteration will result in another solution being found, leaving as the residual a corresponding lower degree polynomial. At some point the residual polynomial will be of degree 1, and thus the solution for it is trivially the free term. The net effect of all this is that the number of solutions that are found will correspond to the degree of the original polynomial.
QED