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# Vieta's Formulas

Vieta's Formulas, otherwise called Viète's Laws, are a set of equations relating the roots and the coefficients of polynomials.

## Introduction

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 get

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

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 formula's 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 $k$th 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 symmetric sum, then we can write those formulas more compactly as $\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}$, for $1\le k\le {n}$.

## Problems

• Let $r_1,r_2,$ and $r_3$ be the three roots of the cubic $x^3 + 3x^2 + 4x - 4$. Find the value of $r_1r_2+r_1r_3+r_2r_3$. Yeah
• Suppose the polynomial $5x^3 + 4x^2 - 8x + 6$ has three real roots $a,b$, and $c$. Find the value of $a(1+b+c)+b(1+a+c)+c(1+a+b)$.
• Let $m$ and $n$ be the roots of the quadratic equation $4x^2 + 5x + 3 = 0$. Find $(m + 7)(n + 7)$.

### Intermediate

• Let $a$, $b$, and $c$ be positive real numbers with $a such that $a+b+c=12$, $a^2+b^2+c^2=50$, and $a^3+b^3+c^3=216$. Find $a+2b+3c$.
• (USAMTS 2010) Find $c>0$ such that if $r$, $s$, and $t$ are the roots of the cubic $$f(x)=x^3-4x^2+6x-c,$$ then $$1=\dfrac1{r^2+s^2}+\dfrac1{s^2+t^2}+\dfrac1{t^2+r^2}.$$
• (HMMT 2007) The complex numbers $\alpha_1$, $\alpha_2$, $\alpha_3$, and $\alpha_4$ are the four distinct roots of the equation $x^4+2x^3+2=0$. Determine the unordered set $$\{\alpha_1\alpha_2+\alpha_3\alpha_4,\,\alpha_1\alpha_3+\alpha_2\alpha_4,\,\alpha_1\alpha_4+\alpha_2\alpha_3\}.$$

[2008 AIME II] Let $r$, $s$, and $t$ be the three roots of the equation $8x^3 + 1001x + 2008 = 0$. Find $(r + s)^3 + (s + t)^3 + (t + r)^3$.