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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 ) with solutions and , then we know that we can factor it as

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

We know that two polynomials are equal if and only if their coefficients are equal, so means that and . 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 term.

A similar set of relations for cubics can be found by expanding .

We can state Vieta's formula's more rigorously and generally. Let be a polynomial of degree , so P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0, where the coefficient of is and . 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 are the roots of . 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 r_1r_2\cdots r_n.

The coefficient of in this expression will be the th symmetric sum of the .

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

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)
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 on the other (this can be arrived at by dividing both sides of all the equations by ).

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

See also

External Links

Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.
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