Someone knows smth about viets formulas

Vieta's Formulas were discovered by the French mathematician François Viète. 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 $x^2+ax+b=0$ with solutions $p$ and $q$, then we know that we can factor it as:

$x^2+ax+b=(x-p)(x-q)$

(Note that the first term is $x^2$, not $ax^2$.) Using the distributive property to expand the right side we now have

$x^2+ax+b=x^2-(p+q)x+pq$

We know that two polynomials are equal if and only if their coefficients are equal, so $x^2+ax+b=x^2-(p+q)x+pq$ means that $a=-(p+q)$ and $b=pq$. 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 $x$ term.

A similar set of relations for cubics can be found by expanding $x^3+ax^2+bx+c=(x-p)(x-q)(x-r)$.

We can state Vieta's formulas more rigorously and generally. Let $P(x)$ be a polynomial of degree $n$, so $P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0$, where the coefficient of $x^{i}$ is ${a}_i$ and $a_n \neq 0$. 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 ${r}_i$ are the roots of $P(x)$. 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 $x^k$ in this expression will be the $(n-k)$-th elementary symmetric sum of the $r_i$.

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

$a_n = a_n$

$a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)$

$a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)$

$\vdots$

$a_0 = (-1)^n a_n r_1r_2\cdots r_n$

More commonly, these are written with the roots on one side and the $a_i$ on the other (this can be arrived at by dividing both sides of all the equations by $a_n$).

If we denote $\sigma_k$ as the $k$-th elementary symmetric sum, then we can write those formulas more compactly as $\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}$, for $1\le k\le {n}$. Also, $-b/a = p + q, c/a = p \cdot q$.

Step-by-step explanation:

Step-by-step explanation:

Since each town he has allows him to create 1.13 times as many villagers as he had in the one before.

The population of the next village will be a multiplication of the population of the previous village by 1.13.

This forms a sequence in which the next term is obtained by multiplication of the previous term by a constant. This type of sequence us called a Geometric Sequence and the constant is called the Common ratio.

For any number of terms, the nth term of a Geometric Progression is determined using the formula:

Where a= First Term

r= common ratio

n= number of terms

The game gave Chase 4 villagers to start with.

Therefore, his first term a=4

The common ratio, r= 1.13

To predict the number of villagers in any specific town, we use the formula:

In the 17th town, i.e. n=17

The number of villagers that can be created will be found by substituting n=17 into the formula above.

Since number of villagers cannot be fractional, the number of villagers he can create to live in the 17th village is 28.