# Vieta's Formulas Is a formula that relates a [[Polynomials|polynomial's]] coefficients to the sum and product of its roots. Consider any polynomial of degree $n$ with roots $r_1..r_n$ of form $ f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 $ Then the sum of its roots is: $ \sum_{i=1}^n r_i = -\frac{a_{n-1}}{a_n} $ And the sum of its roots is: $ \prod_{i=1}^n r_i = (-1)^n\frac{a_0}{a_n} $ ## For [[Quadratics]] $x_1 + x_2 = -\frac{b}{a} \qquad x_1x_2 = \frac{c}{a}$ ### Simple proof of Vieta's Formula for [[Quadratics]] Recall that the standard form of a quadratic is $y = ax^2 + bx + c$ and the factored form is $y= a(x-x_1)(x-x_2)$. $ \begin{align} y &= a(x-x_1)(x-x_2) \\ y &= a(x^2-x_1x-x_2x+x_1x_2) \\ y &= a(x^2 - (x_1+x_2)x + x_1x_2) \\ y &= ax^2 - a(x_1+x_2)x + ax_1x_2 \end{align} $ From there $ \begin{align} b &= -a(x_1+x_2) & c &= ax_1x_2 \\ x_1 + x_2 &= -\frac{b}{a} & x_1x_2 &= \frac{c}{a} \end{align} $ ### Use for solving monic quadratics Given [[Polynomials#Monic Polynomial|monic]] quadratic $x^2+bx+c=0$, find two numbers $j$ and $k$ such that $kj=c$ and $k+j=b$. The two numbers are the quadratic's roots multiplied by -1. $ x^2+bx+c=(x+j)(x+k) $