 Euler's general approach to proving the fundamental theorem of algebra may be described as follows. Consider a depressed polynomial of degree 2 to the n plus 1. Assume it can be factored as a product of 2 degree 2 to the n polynomials. The coefficients of the factor satisfy some equation. If the equation has a real solution, then, using that real solution, two real factors can be found. In practice, the equations are nonlinear, so actually solving them is difficult, and in fact it turns out to be impossible beyond a certain point. So instead, Euler reexamined the quartic, which E assumed factored as... Assuming the quartics had four roots a, b, c, and d... We'll get to that in a moment. We must have the factorization. And since the x cubed coefficient is zero, the sum of a, b, c, and d must be zero. Now, this quadratic factor had to be expressible as the product of two of the linear factors. For example, we might have u squared minus ux plus beta be the product of x minus a and x minus c, and this means that u is equal to a plus c. And since there are four chews to six different choices of the two factors, it follows that there are six different ways to form a quadratic factor, and so this coefficient u has six possible values. And in fact, we can list those explicitly. Now, remember a plus b plus c plus d is zero, and this means half of these values are the negatives of the others. So all six values can be expressed in terms of three, which we can call p, q, and r, say a plus b is p, and so c plus d is negative p, a plus c is q, so b plus d is negative q, and a plus d is r, so b plus c is negative r. To find u, we'll let the six possible values be the solutions to an equation, and because our solutions occur in pairs, we can multiply the pairs together to get, which will be a six-degree equation. But notice that the constant term will be negative p squared, q squared, r squared, and since this is negative, it follows there's a real value u that solves the equation. Now let's consider the octic and assume that it has factors of the form. Then, arguing as before, the eight roots of the octic, added four at a time, gives the possible values of u, and again as before, if the roots are then the factorization is, and so we must have the sum equal to zero. There's eight choose for 70 possible values, so u will be the solution to a 70th-degree equation. Since the sum of the roots is zero, the sum of half of these roots will be the negative of the sum of the other half. So the 70 roots occur in pairs, if p is a root, so is negative p. So the values of u could be found by solving the 70th-degree polynomial. Since there are 35 factors, our constant term when we expand this will be, so our equation has at least two real solutions. Euler then repeats the argument for the depressed 16th-degree equation. As before, assuming a factorization into two 8th-degree polynomials, the coefficient u will have 16 choose 8, 12,870 different values. These will be the roots to a 12,870th-degree equation, but the roots will be a form plus or minus p, so this 12,870th-degree equation will be the part of Euler's work is the first actual proof of the fundamental theorem of algebra if you overlook a rather significant gap. Completing that gap would be the work of Lagrange a little bit later on. In the meantime, we'll leave the reader to figure out where that gap is, and here's a hint, it's not his assumption that the roots exist.