how to memorize the sum and product of roots for an $n^{th}$ degree equation

Question

For my exams I need to know the following equations by heart:

for a polynomial equation: $a_nx+a_{n-1}x^{n-1}+...+a_1x+a_0,$ the sum and product of the roots are given by

$$\textrm{Sum}=-\frac{a_{n-1}}{a_n}$$ $$\textrm{Product}=(-1)^n\frac{a_0}{a_n}.$$

I have never been able to memorize these, and for some reason, they are not on the formula booklet. If anyone has any mnemonic or trick of some sort for memorizing them, it would be very useful to me.

Thank you very much in advance!

Answer 1

Write $$a_n x^n + a_{n-1}x^{n-1} + ... + a_1 x + a_0 = a_n(x-r_1)...(x-r_n)$$

and develop. You see immediately what is the constant term and the term of degree $n-1$.

Answer 2

Hint:

Write the polynomial as $$(x-r_1)(x-r_2)...(x-r_n),$$ where the roots are $r_1, r_2, ... r_n.$

Answer 3

Consider $a_n(x-r_1)(x-r_2)...(x-r_n)=a_n x^n-a_n(r_1+r_2+r_3...)x^{n-1}+...+(-1)^na_nr_1r_2...r_n$. And so product of roots are $\frac{(-1)^na_0}{a_n}$ and sum of roots is $\frac{-a_{n-1}}{a_n}$.

Answer 4

Since the OP is looking for an artificial device to recall the correct equations under duress rather than a way to derive the formulas quickly, consider the following suggestion.

Rewrite the equations as follows

\begin{eqnarray} a_nS&=&-a_{n-1}\\ a_nP&=&(-1)^na_0 \end{eqnarray}

Next, make some observations about these equations:

1. $a_n$ occurs only on the left side of each equation.
2. The variable $n$ occurs only once on the right side of each equation.
3. The Sum formula has a minus sign in front and the $n$ occurs in the Subscript as $n-1$.
4. The Product formula has a factor in front, $(-1)^n$, containing the $n$.

Answer 5

An interesting observation can be made, regarding the sum and product of polynomial roots.

Consider a quadratic equation, $ax^2+bx+c=0$, having the roots $\alpha$, and $ \beta$. Here, we observe

• $\alpha + \beta = -\frac{b}a$
• $\alpha \beta = \frac{c}a$

Now let us observe the same results for the cubic equation, $ax^3+bx^2+cx+d=0$, having roots $\alpha, \beta, \gamma$.

• $\alpha + \beta + \gamma = -\frac{b}a$
• $\alpha\beta + \beta\gamma + \alpha\gamma = \frac{c}a$
• $\alpha \beta \gamma = -\frac{d}a$

Also, let us observe a polynomial equation of the 4th power, $ax^4+bx^3+cx^2+dx+e=0$, and roots $\alpha, \beta, \gamma, \delta$:

• $\alpha + \beta + \gamma + \delta = -\frac{b}a$
• $\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta = \frac{c}a$
• $\alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta = -\frac{d}a$
• $\alpha \beta \gamma\delta = \frac{e}a$

Thus for an nth powered polynomial, say, $a_nx^n + a_{n-1}x^{n-1} + ... + a_{1}x + a_{0} = 0$, we will have $n$ roots. Say each root is denoted by $r_i$ we can derive the following result:-

1. $\sum_{k=1}^n (r_i) = -\frac{b}a$
2. $\prod_{k=1}^n (r_i) = (-1)^n\frac{a_0}{a_n}$

However the important thing to observe from the above series is that the alternate negative fraction (starting from the 1st -ve term) of each successive term with that of the first term of the polynomial is equal to the sum of the products of $(i-1)^{th}$ number of roots taken together.

So the ratio of the $2^{nd}$ to the first, will be the sum of products of $(2-1) = 1$ roots taken together.

Similarily the fraction of the ($(n+1)^{th}$ term) with the first term will be the sum of products of $(n+1)-1 = n$ roots taken together.

Note: You might be intersted in reading about these formulas called Vieta's Formulas here.