The sum and product of an nth degree polynomial

phyguy321
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Homework Statement


Suppose f(x) \in Complex[x] is a monic polynomial of degree n with roots c1,c2,...cn. Prove that the sum of the roots is -a_{n-1} and their product is (-1)^{n}a_{0}

Homework Equations





The Attempt at a Solution


(x-c1)(x-c2)...(x-cn) = x^{n} + (c1+c2+...+cn)x^{n-1}...(c1*c2*...*cn)

I just need a realistic proof this assumes too much
 
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In what way do you think that's assuming too much? Do you know the Fundamental Theorem of Algebra?
 
but how do i know that (x-c1)(x-c2)...(x-cn) = xLaTeX Code: ^{n} + (c1+c2+...+cn)xLaTeX Code: ^{n-1} ...(c1*c2*...*cn)?
 
Count powers of x. There's only one way to make x^n and x^0. There are n ways to make x^1. You just imagine multiplying it out.
 
phyguy321 said:
but how do i know that (x-c1)(x-c2)...(x-cn) = xLaTeX Code: ^{n} + (c1+c2+...+cn)xLaTeX Code: ^{n-1} ...(c1*c2*...*cn)?

Because you know how to multiply polynomials?
 
so that's a legit proof then?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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