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 | 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 08:13, 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.
