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Vieta's Formulas

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Vieta's Formulas are a set of equations relating the roots and the coefficients of polynomials.

Introduction

Vieta's Formulas can be used to relate the sum and product of the roots of a polynomial to its coefficients. The simplest application of this is with quadratics. If we have a quadratic x^2+ax+b=0 with solutions p and q, then we know that we can factor it as

x^2+ax+b=(x-p)(x-q)

(Note that the first term is x^2, not ax^2.) Using FOIL to expand the right side we get

x^2+ax+b=x^2-(p+q)x+pq

We know that two polynomials are equal if and only if their coefficients are equal, so x^2+ax+b=x^2-(p+q)x+pq means that a=-(p+q) and b=pq. In other words, the product of the roots is equal to the constant term, and the sum of the roots is the opposite of the coefficient of the x term.

A similar set of relations for cubics can be found by expanding x^3+ax^2+bx+c=(x-p)(x-q)(x-r).

We can state Vieta's formula's more rigorously and generally. Let P(x) be a polynomial of degree n, so P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0, where the coefficient of x^{i} is {a}_i and a_n \neq 0. As a consequence of the Fundamental Theorem of Algebra, we can also write P(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n), where {r}_i are the roots of P(x). We thus have that

a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 = a_n(x-r_1)(x-r_2)\cdots(x-r_n).

Expanding out the right hand side gives us

a_nx^n - a_n(r_1+r_2+\!\cdots\!+r_n)x^{n-1} + a_n(r_1r_2 + r_1r_3 +\! \cdots\! + r_{n-1}r_n)x^{n-2} +\! \cdots\! + (-1)^na_n ...

The coefficient of x^k in this expression will be the kth symmetric sum of the r_i.

We now have two different expressions for P(x). These must be equal. However, the only way for two polynomials to be equal for all values of x is for each of their corresponding coefficients to be equal. So, starting with the coefficient of x^n, we see that

a_n = a_n
a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)
a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)
\vdots
a_0 = (-1)^n a_n r_1r_2\cdots r_n

More commonly, these are written with the roots on one side and the a_i on the other (this can be arrived at by dividing both sides of all the equations by a_n).

If we denote \sigma_k as the kth symmetric sum, then we can write those formulas more compactly as \sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}, for 1\le k\le {n}.

See also

External Links

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