Solving an equation of order $x$

Question

Two questions

1)

Looking at Vieta's formula for cubics and quadratics I think I noticed something. Is it just my imagination or are the following statements true?

a) There will be a number of roots equal to the highest power n of the equation.

b) The sum of the roots will be the coefficient of the $x^{n-1}$ term

c) the product of the roots will be the constant term.

2)

Can information, deduced from vieta's formulae, be used to set up a system of equations to find the roots of any equation of order $x$?

Answer 1

1.a. True

1.b. Actually, the negative of the sum of the roots …

1.c. Actually, the positive or negative of the product of the roots, according to whether the degree of the equation is even or odd, respectively.

NOTA BENE! All pf the above assumes that the coëfficient of the term of highest degree has been altered, through division or multiplication, to unity, with the other coëfficients altered accordingly.

$2.$ Certainly, setting the various functions of the presumptive roots equal to the coëfficients can lead to the set of solutions, but whatever difficulties are involved in the good, old-fashioned solution will be carried over into the solution of this new system. There is no escape!

Answer 2

If you are looking at polynomial equations over $\mathbb C$, you are right. The fundamental theorem of algebra tells you that every polynomial $p$ of order $n$ has exactly $n$ (perhaps non-distinct) zeros (the zeros may be complex, though). This means every polynomial can be rewritten as $$p(x) = a(x-x_1)(x-x_2)\cdots(x-x_n)$$ where $x_i$ are some complex numbers. If you expand this polynomial, you get that $$p(x) = ax^n + a(x_1+\dots+x_n)x^{n-1} + \dots + (ax_1x_2\dots x_n).$$

If the original polynomial was $x^n + a_{n-1} x^{n-1} + \dots + a_0$, then $a=1$ and your conclusions hold.