Relationship between coefficients and polynomial roots using induction proof?

Question

let $p(x)$ be a polynomial such that :

$ P(x) = a_{n}x^n + a_{n-1}x^{n-1} +\cdots+a_{1}x + a_{0} $ with roots $ r_{1}, r_{2}, \cdots , r_{n} $ , My question here is how to show that ($1$) is true using induction proof

$$\frac{1}{r_{1}} + \frac{1}{r_{2}} + \cdots + \frac{1}{r_{n}} = -\frac{a_{1}}{a_{0}}\tag{1}$$ ?.

Note: Really my attempt failed in the first step for initial condition

Answer 1

If you don't mind having a non-inductive proof, here it goes: let$$Q(x)=a_0x^n+a_1x^{n-1}+\cdots+a_n.$$Then the roots of $Q(x)$ are the inverses of the roots of $P(x)$, that is, they are $\frac1{r_1},\frac1{r_2},\ldots,\frac1{r_n}$. And the sum of the roots of $Q(x)$ is $-\frac{a_1}{a_0}$.

Answer 2

HINT

Base case

• $n=1\implies P(x)=a_1x+a_0=a_1(x-r_1)\implies a_0=-a_1r_1 \quad \frac1{r_1}=-\frac{a_0}{a_1}$

Inductive step

Assume true for $n$ that

$$\frac{1}{r_{1}} + \frac{1}{r_{2}} + \cdots + \frac{1}{r_{n}} = -\frac{a_{1}}{a_{0}}$$

and let

$$P_{n+1}(x)=a_{n+1}(x-r_{n+1})...(x-r_1)=(x-r_{n+1})P_n(x)$$

then indicating with $\bar a_1$ and $\bar a_0$ the coefficients for $P_{n+1}(x)$ we have

• $\bar a_1=a_0-r_{n+1}a_1$
• $\bar a_0=-r_{n+1}a_0$

and then

$$\frac{\bar a_1}{\bar a_0}=\frac{a_0-r_{n+1}a_1}{-r_{n+1}a_0}=-\frac1{r_{n+1}}+\frac{a_1}{a_0}$$