Anonymous at Sat, 27 Jul 2024 08:05:07 UTC No. 16298632

I will break it into simpler steps and hopefully it will be easier to follow.

Our hypotheses are:

- A polynomial [math]P(x) = x^2 + px + q[/math].
- Two different roots [math]\alpha[/math] and [math]\beta[/math] of [math]P(x)[/math].

We will shot that [math]P(x) = (x-\alpha)(x-\beta)[/math].

Step 1. Divide [math]P(x)[/math] by [math]x-\alpha[/math].

Since [math]\alpha[/math] is a root of [math]P(x)[/math], [math](x-\alpha)[/math] divides [math]P(x)[/math] (this is called Bezout's theorem in Gelfand's book). Therefore,
[eqn]P(x) = (x-\alpha)Q(x)[/eqn] for some polynomial [math]Q(x)[/math].

Step 2. Divide [math]Q(x)[/math] by [math]x - \beta[/math].

It is also the case that [math]\beta[/math] is a root of [math]P(x)[/math], so [math]P(\beta) = 0[/math]. If we make this substitution in the equation of Step 1, we obtain that
[eqn]0 = (\beta - \alpha)Q(\beta).[/eqn] If a product of two things equals zero, it must be that one of them is zero. If [math](\beta - \alpha) = 0[/math], it would immediately follow that [math]\alpha = \beta[/math], which does not happen since the roots are different (by our second hypothesis). Therefore, [math]Q(\beta) = 0[/math] or, in other words, [math]\beta[/math] is a root of [math]Q(x)[/math]. By applying the same reasoning from Step 1 to [math]Q(x)[/math] and [math]\beta[/math], we obtain that
[eqn]Q(x) = (x-\beta)R(x)[/eqn] for some polynomial [math]R(x)[/math]. By substituting [math]Q(x)[/math] in the equation of Step 1, we get:
[eqn]P(x) = (x-\alpha)(x-\beta)R(x).[/eqn]

Anonymous at Sat, 27 Jul 2024 08:06:21 UTC No. 16298634

>>16298632
(cont.)

Step 3. [math]R(x) = 1[/math].

Multiplying two polynomials adds their degrees. The degree of [math](x-\alpha)(x-\beta)[/math] is [math]2[/math], so the degree of [math](x-\alpha)(x-\beta)R(x)[/math] is [math]2[/math] plus the degree of [math]R(x)[/math]. But the degree of [math]P(x)[/math] is [math]2[/math] (our first hypothesis), so it must be that the degree of [math]R(x)[/math] is zero. In other words, [math]R(x) [/math] is constant.

If [math]R(x) = c[/math] a constant, then
[eqn]P(x) = c(x-\alpha)(x-\beta).[/eqn] By carrying out the product, you can see that the leading coefficient of [math]P(x)[/math] is [math]c[/math]. But the leading coefficient of [math]P(x)[/math] is [math]1[/math]. In conclusion, [math]R(x) = c = 1[/math] and
[eqn]P(x) = (x-\alpha)(x-\beta).[/eqn]


- How does this prove Vieta's theorem?

Suppose that [math]x^2 + px + q[/math] has two different roots [math]\alpha[/math] and [math]\beta[/math]. We have just proven that
[eqn]x^2 + px + q = (x-\alpha)(x-\beta).[/eqn]
Carrying out the product [math](x-\alpha)(x-\beta)[/math], we get that
[eqn]x^2 + px + q = x^2 - (\alpha + \beta) + \alpha\beta .[/eqn] Two polynomials are equal if their coefficents are equal, thus, [math]p = -(\alpha + \beta)[/math] and [math]q = \alpha\beta[/math], which proves Vieta's theorem.

Anonymous at Sat, 27 Jul 2024 11:04:19 UTC No. 16298754

>>16297560
Unrelated: how do you find Gelfand's precalculus series? functions, trigonometry, algebra and method of coordinates