Difference between revisions of "Vieta's formulas"
(Created page with "Theorem 14.1.4 (Vieta’s Formula For Higher Degree Polynomials) In a polynomial anx n + an−1x n−1 + ... + a1x + a0 = 0 with roots r1, r2, r3, ...rn the following hold...") |
|||
| Line 1: | Line 1: | ||
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 | + | In a polynomial <math>a_n x^n + a_{n−1} x^{n−1} + \cdots + a_1 x^1 + a_0 = 0</math> with roots <math>r_1, r_2, r_3, \ldots, r_n</math> 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} \\ | |
| − | n−1 + | + | 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. | |
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | 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. | ||
Revision as of 09:13, 25 September 2025
Theorem 14.1.4 (Vieta’s Formula For Higher Degree Polynomials)
In a polynomial $a_n x^n + a_{n−1} x^{n−1} + \cdots + a_1 x^1 + a_0 = 0$ (Error compiling LaTeX. Unknown error_msg) 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.