If $\alpha$ and $\beta$ are the two roots of equation of $ax^2 + bx + c$ .How is $$cx^2 + bx + a = c(x-1/\alpha)(x-1/\beta)$$
Later Edit:
How to deduce $$c(x-1/\alpha)(x-1/\beta)$$ from $$cx^2 + bx + a$$ , when we dont know if $1/\alpha$ and $1/\beta$ are the roots of equation $cx^2 + bx + a$
We have
$c \frac{1}{\alpha^2}+b\frac{1}{\alpha}+a=\frac{1}{\alpha^2}(c +b \alpha+a \alpha^2)=0.$
Hence $\frac{1}{\alpha^2}$ is a root of the equation $cx^2 + bx + a =0.$
By Vieta's formulas $$\alpha+\beta = \frac{-b}{a} \ \ \ \text{ and } \ \ \ \ \alpha\beta = \frac{c}{a}$$ Then $$c\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right) = c\left(x^2-\frac{\alpha+\beta}{\alpha\beta}x + \frac{1}{\alpha\beta}\right) = c\left(x^2 + \frac{b}{c}x + \frac{a}{c}\right) = cx^2+bx+a$$
Later edit: Here is the general situation:
Let $f(x) = a_n x^n + a_{n-1} x^{n-1}+\ldots a_1 x + a_0$ with $n \ge 1$ and $a_n \neq 0$. Assume also that $a_0 \neq 0$. This is the same as assuming that $0$ is not a root of $f$. Let $\alpha_1, \ldots, \alpha_n$ be the roots of $f$ (they are not necessarily distinct). Note that $$f(x) = a_n (x-\alpha_1) \ldots (x-\alpha_n)$$ Let $g(x)=f\left(\frac{1}{x}\right)$. Note that the roots of $g$ are $\frac{1}{\alpha_1}, \ldots, \frac{1}{\alpha_n}$. Since $$g(x)=f\left(\frac{1}{x}\right) = \frac{1}{x^n}\left(a_0 x^n + a_1 x^{n-1} + \ldots a_{n-1} x + a_n \right)$$ the roots of the polynomial $a_0 x^n + a_1 x^{n-1} + \ldots a_{n-1} x + a_n$ are precisely $\frac{1}{\alpha_1}, \ldots, \frac{1}{\alpha_n}$ and hence $$a_0 x^n + a_1 x^{n-1} + \ldots a_{n-1} x + a_n = a_0 \left(x- \frac{1}{\alpha_1}\right) \ldots \left(x - \frac{1}{\alpha_n} \right)$$
