Difference between revisions of "Vieta's formulas"
(→Intermediate Level) |
|||
| Line 16: | Line 16: | ||
This can be used in a variety of problems, such as : | This can be used in a variety of problems, such as : | ||
| − | == Intermediate Level == | + | == Intermediate Level ==\\ |
| − | AIME I 2001/3 | + | AIME I 2001/3\\ |
| − | AIME I 2014/5 | + | AIME I 2014/5\\ |
| − | AIME 1996/5 | + | AIME 1996/5\\ |
| − | AIME I 2005/8 | + | AIME I 2005/8\\ |
| − | AIME 1993/5 | + | AIME 1993/5\\ |
| − | AIME II 2008/7 | + | AIME II 2008/7\\ |
| − | + | \\ | |
| − | Try these out! | + | Try these out!\\ |
| − | + | \\ | |
This theorem relates to polynomials | This theorem relates to polynomials | ||
{{stub}} | {{stub}} | ||
Revision as of 10:57, 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.