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*Polynomials*by Barbeau. I think I may have provided a decent proof for one of the exercises, but I'd like a second opinion. Here's the exercise:

*Show that every quadratic equation with complex coefficients has at least one complex root, and therefore can be written as the product of two linear factors with complex coefficients.*

My proof:

Assume for contradiction that [tex] p \left( t \right) = qt^2 +wt+z [/tex] is a complex polynomial with [tex] q \neq 0 [/tex]. Further, suppose that [tex]p[/tex] has two real roots. Since the discriminant must vanish we have [tex] w^2 = 4qz [/tex]. Furthermore, if the discriminant vanishes we are left with [tex] -w/2q [/tex] where [tex]q[/tex] and [tex]w[/tex] must cancel. Hence, [tex] w = -2q [/tex]. We now have two simultaneous equations; substitution yields [tex] \left( -2q \right)^2 = 4qz \Rightarrow q=z [/tex]. Then, [tex] q=z \Rightarrow w = -2z [/tex]. Therefore, [tex] p \left( t \right) =qt^2 +wt+z = zt^2 -2zt + z \Rightarrow t^2-2t+1 [/tex], contradicting the assumption that [tex] p[/tex] is a complex polynomial.

It then follows that any complex polynomial may be written in the form [tex] q \left( t- r_1 \right)\left( t-r_2 \right) [/tex], where [tex] q \in \mathbb{C} [/tex] and [tex] r_1,r_2 [/tex] are roots.

The critical assumption is that the discriminant must vanish. I'm nearly sure that it does, I'm having difficulty rigorously justifying it though. If the discriminant did

*not*vanish then we'd be left with a complex number and its conjugate in the numerator of the quadratic formula and nothing would cancel. So, for the polynomial to have two real roots then it must vanish, right?

The author implies actually evaluating the square root of the discriminant in the quadratic formula, but if I define [tex] q = q_1 + q_2i [/tex], [tex] w = w_1 + w_2i [/tex], and [tex] z = z_1 + z_2i [/tex], that's going to lead to a huge mess...or am I missing something?

Any help would be appreciated. Thanks!