Vieta's Formulas
Introduction
Let ,
where the coefficient of
is
. As a consequence of the Fundamental Theorem of Algebra, we can also write
, where
are the roots of
. We thus have that
![$a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 = a_n(x-r_1)(x-r_2)\cdots(x-r_n).$](http://latex.artofproblemsolving.com/4/4/b/44b0375f3fff54d4440ac27ed32a29ea26fa6f10.png)
Expanding out the right hand side gives us
![$a_nx^n - a_n(r_1+r_2+\cdots+r_n)x^{n-1} + a_n(r_1r_2 + r_1r_3 + \cdots + r_{n-1}r_n)x^{n-2} + \cdots + (-1)^na_n r_1r_2\cdots r_n.$](http://latex.artofproblemsolving.com/4/6/6/46673b40dbd29e986828df0a67c5e366f4ca5a77.png)
We can see that the coefficient of will be the
th symmetric sum. The
th symmetric sum is just the sum of the roots taken
at a time. For example, the 4th symmetric sum is
Notice that every possible combination of four roots shows up in this sum.
We now have two different expressions for . These must be equal. However, the only way for two polynomials to be equal for all values of
is for each of their corresponding coefficients to be equal. So, starting with the coefficient of
, we see that
![$a_n = a_n$](http://latex.artofproblemsolving.com/b/0/b/b0bdcbc9e71069ce32a5b6c5a3137b1832620d2c.png)
![$a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)$](http://latex.artofproblemsolving.com/e/9/8/e98bba0737992471e152a486d4e12fc9d38f8803.png)
![$a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)$](http://latex.artofproblemsolving.com/e/5/7/e5734e66af5754c9f39706aafe68f3a0eb2bff8c.png)
![$\vdots$](http://latex.artofproblemsolving.com/6/9/c/69cdf7ec84ff918cbe7f21acf7543bf1bd243080.png)
![$a_0 = (-1)^n a_n r_1r_2\cdots r_n$](http://latex.artofproblemsolving.com/1/3/f/13f11a26327cb547c656475c515683b949d4823a.png)
More commonly, these are written with the roots on one side and the on the other (this can be arrived at by dividing both sides of all the equations by
).
If we denote as the
th symmetric sum, then we can write those formulas more compactly as
, for
.