Memorization and Generalization of Vieta's formulas

Consider the quadratic equation $$a_2x^2+a_1x+a_0=0$$

If $$x_1$$ and $$x_2$$ are the roots of the quadratic then its Vieta's formulas are : \left\{ \begin{align} x_1 + x_2 &= -\dfrac{a_1}{a_2}\\ x_1 \cdot x_2 &= \dfrac{a_0}{a_2} \end{align} \right.

In a similar manner if $$x_1,x_2,x_3$$ are the three roots of $$a_3x^3+a_2x^2+a_1x+a_0=0$$ then its Vieta's formulas are : \left\{ \begin{align} x_1 + x_2 + x_3 &= -\dfrac{a_2}{a_3}\\ x_1 \cdot x_2 \cdot x_3 &= -\dfrac{a_0}{a_3}\\ x_1x_2+x_1x_3+x_2x_3 &= \dfrac{a_1}{a_3} \end{align} \right.

Should I do one more? If you consider $$a_4x^4+a_3x^3+a_2x^2+a_1x+a_0=0$$ and $$x_1,x_2,x_3,x_4$$ are the 4 roots of the equation then it's possible to give such 4 formulas, you may call those Vieta's foemulas but my question is what's the use of it?

For any n degree polynomial there exists such n formulas. Should I remember those all the times to solve some handcrafted problems?

And how many of those you do remember?

2 Answers

An infinity.

The formulas are pretty regular. You form the sums of all distinct products of $$n-k$$ coefficients and equate them to the ratio of the $$k^{th}$$ coefficient and the leading one.

From scratch, for $$n=4$$,

$$\begin{cases}x_1+x_2+x_3+x_4=-\frac{a_3}{a_4}, \\x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4=\frac{a_2}{a_4}, \\x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4=-\frac{a_1}{a_4}, \\x_1x_2x_3x_4=\frac{a_0}{a_4}.\end{cases}$$

The number of terms follows the Binomial distribution, $$(1),4,6,4,1$$.

• Would you please write those formulas for $n=6$ and give me a hint about why the number of terms follow the Binomial distribution? – ARahman Oct 11 '18 at 16:32
• @ARahman: nope, your job. Think of combinations. – Yves Daoust Oct 11 '18 at 17:58

First, it's easier if you assume the leading coefficient (i.e. the coefficient of $$x^n)$$ is $$1,$$ which you can always arrange in practice by factoring out the leading coefficient. Then the formulas follow a simple pattern, and you only have to remember the pattern.

$$-\left(\text{sum of products-one-at-a-time of the roots}\right)=$$ coefficient of $$x^{n-1}$$

$$+\left(\text{sum of products-two-at-a-time of the roots}\right)=$$ coefficient of $$x^{n-2}$$

$$-\left(\text{sum of products-three-at-a-time of the roots}\right)=$$ coefficient of $$x^{n-3}$$

$$\cdots$$

$$(-1)^{n-1}\cdot\left(\text{sum of products-}(n-1)\text{-at-a-time of the roots}\right)=$$ coefficient of $$x^{n-(n-1)} = x^1$$

$$(-1)^{n}\cdot\left(\text{sum of products-}n\text{-at-a-time of the roots}\right)=$$ coefficient of $$x^{n-n} = x^0$$ (constant term)