## Vieta's formulas in elementary algebra

### Introduction

By the fundamental theory of algebra, a general polynomial of the form,

$$\tag{1} f(x) = a_{n}x^{n} + a_{n-1}x^{n-1} + \cdots + a_{1}x + a_{0} = 0$$

has at least $n$ complex roots (not necessarily unique). Therefore, given a root $\alpha$ of the polynomial, we can rewrite $(1)$ as,

$$f(x) = (x - \alpha) f^’(x) = 0$$

where $f^’(x)$ is a polynomial of degree $(n - 1)$. Now, by extending this idea to completely factor out $f^’(x)$, we would have,

$$\tag{2} f(x) = (x - \alpha_{1}) (x - \alpha_{2}) (x - \alpha_{3}) \cdots (x - \alpha_{n}) = 0$$

where $\alpha_{k}$ for $k = 1, 2, \ldots, n$ is the count of the not necessarily distinct roots of $f(x)$. Note that due to the possibility of having non-distinct roots, then $(2)$ may also be written as,

$$f(x) = (x - \alpha_{1})^{m_{1}} (x - \alpha_{2})^{m_{2}} (x - \alpha_{3})^{m_{3}} \cdots (x - \alpha_{p})^{m_{p}} = 0$$

where $m_{i}$ is the number of times a root $\alpha_{i}$ is repeated and is known as the multiplicity of $\alpha_{i}$ and $p \le n$.

### Vieta’s formulas

Vieta’s formulas establish an interesting relationship between the coefficients of a polynomial and its roots which we will derive here by comparing $(1)$ to the expanded version of $(2)$. For the sake of clarity, we will include as much of the calculations as is possible.

We start by rewriting $(1)$ so as to have the component of the polynomial with the highest power with a factor of $1$. The reason for this will be clear later in the process.

$$\tag{3} f(x) = a_{n} (x^{n} + \frac{a_{n-1}}{a_{n}}x^{n-1} + \frac{a_{n-2}}{a_{n}}x^{n-2} + \cdots + \frac{a_{1}}{a_{n}}x + \frac{a_{0}}{a_{n}}) = 0$$

Now, assuming we had factored out the $a_{n}$ in $(2)$ just as we have done above and expanded it by multiplying out the first term, we will obtain,

$$f(x) = a_{n}\big(x^2 - (\alpha_{1} + \alpha_{2})x + \alpha_{1}\alpha_{2}\big) (x - \alpha_{3}) \cdots (x - \alpha_{n}) = 0$$

By further expanding this by multiplying out the third term, we will obtain,

$$f(x) = a_{n}\big(x^3 - (\alpha_{1} + \alpha_{2} + \alpha_{3})x^2 + (\alpha_{1}\alpha_{2} + \alpha_{1}\alpha_{3} + \alpha_{2}\alpha_{3})x + \alpha_{1}\alpha_{2}\alpha_{3}\big) (x - \alpha_{4}) \cdots (x - \alpha_{n}) = 0$$

From the above equation, there’s already a pattern that can be obtained although not entirely obvious. If we were to expand this further all through, we will have,

\begin{aligned} \tag{4} \frac{f(x)}{a_{n}} =\; & \; x^n \\ \; & - (\alpha_{1} + \alpha_{2} + \cdots + \alpha_{n})x^{n-1} \\ \; & + (\alpha_{1}\alpha_{2} + \alpha_{1}\alpha_{3} + \alpha_{1}\alpha_{4} + \cdots + \alpha_{n-1}\alpha_{n})x^{n-2} \\ \; & - (\alpha_{1}\alpha_{2}\alpha_{3} + \alpha_{1}\alpha_{2}\alpha_{4} + \cdots + \alpha_{n-2}\alpha_{n-1}\alpha_{n})x^{n-3} \\ \; & \; \vdots \\ \; & + (-1)^n(\alpha_{1}\alpha_{2}\alpha_{3}\cdots\alpha_{n}) \\ \; = 0 \end{aligned}

We will now compare the coefficients of $(3)$ and $(4)$. Note that since these two are equal, the coefficients of $x^i$ should essentially be the equal. For the term with the highest degree, they both have a coefficient equal to $1$ and therefore not very interesting. For the other terms though, we have:

\begin{aligned} \frac{a_{n-1}}{a_{n}}x^{n-1} & = \; - (\alpha_{1} + \alpha_{2} + \cdots + \alpha_{n})x^{n-1} \\ \\ \; \frac{a_{n-2}}{a_{n-1}}x^{n-2} & = \; + (\alpha_{1}\alpha_{2} + \alpha_{1}\alpha_{3} + \alpha_{1}\alpha_{4} + \alpha_{2}\alpha_{3} + \cdots + \alpha_{n-1}\alpha_{n})x^{n-2} \\ \; & \; \vdots \\ \; \frac{a_{0}}{a_{1}} & = \; (-1)^n \alpha_{1}\alpha_{2}\alpha_{3}\cdots\alpha_{n} \end{aligned}

This beautiful pattern is essentially what Vieta’s formulas express and may be written as,

$$\frac{a_{n-k}}{a_{n}} = (-1)^n \sum_{1 \leq i_{1} \lt i_{2} \cdots \lt i_{k} \leq n}^{n}\Big( \prod_{j=1}^k \alpha_{i_{j}} \Big)$$

### Conclusion

I hope this helps give you the intuition behind the beautiful formulas!

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