Vieta's formulas: Difference between revisions
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Theorem 14.1.4 (Vieta’s Formula For Higher Degree Polynomials) | Theorem 14.1.4 (Vieta’s Formula For Higher Degree Polynomials) | ||
In a polynomial <math>a_n x^n + a_{ | In a polynomial <math>a_n x^n + a_{n-1} x ^ {n-1} ..... a_1 x^{1} + a_0</math> with roots <math>r_1 r_2 r_3 ... r_n </math> | ||
the following holds: | |||
\begin{align*} | \begin{align*} | ||
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Note that the negative and positive signs alternate. When summing the products for odd number of terms, we will have a negative sign otherwise we will have a positive sign. | Note that the negative and positive signs alternate. When summing the products for odd number of terms, we will have a negative sign otherwise we will have a positive sign. | ||
This can be used in a variety of problems, such as : | |||
== Intermediate Level == | |||
AIME I 2001/3 | |||
AIME I 2014/5 | |||
AIME 1996/5 | |||
AIME I 2005/8 | |||
AIME 1993/5 | |||
AIME II 2008/7 | |||
Try these out! | |||
This theorem relates to polynomials | |||
{{stub}} | {{stub}} | ||
Revision as of 09:56, 15 October 2025
Theorem 14.1.4 (Vieta’s Formula For Higher Degree Polynomials)
In a polynomial 
with roots 
the following holds:
\begin{align*} r_1 + r_2 + r_3 + \cdots + r_n (the sum of all terms) &= −\frac{a_{n−1}}{a_n} \\ r_1r_2 + r_1r_3 + \cdots + r_{n−1}r_n (the sum of all products of 2 terms) &= \frac{a_{n−2}}{a_n} \\ r_1r_2r_3 + r_1r_2r_4 + \cdots + r_{n−2}r_{n−1}r_n (the sum of all products of 3 terms) &= −\frac{a_{n−3}{a_n} \\ &\vdots \\ r_1r_2r_3 \cdots r_n (the sum of all products of n terms) &= (−1)^n \frac{a_0}{a_n} \end{align*}
Note that the negative and positive signs alternate. When summing the products for odd number of terms, we will have a negative sign otherwise we will have a positive sign.
This can be used in a variety of problems, such as :
Intermediate Level
AIME I 2001/3 AIME I 2014/5 AIME 1996/5 AIME I 2005/8 AIME 1993/5 AIME II 2008/7
Try these out!
This theorem relates to polynomials This article is a stub. Help us out by expanding it.
