Question

if alpha and beta are the zeros of the polynomial x2+3x+2 form the polynomial whose zeroes are (1+beta/alpha) and (1+alpha/beta)

Answer 1

We can factorize the quadratic as

`x^2 + 3x + 2 = (x - \alpha) (x - \beta)`

and expanding the right side leads to

`x^2 + 3x + 2 = x^2 - (\alpha + \beta) x + \alpha \beta \implies \begin{cases} \alpha + \beta = -3 \\ \alpha \beta = 2 \end{cases}`

The polynomial we want will have the factorization and expanded form

`\left(x - \left(1 + \frac\beta\alpha\right)\right) \left(x - \left(1 + \frac\alpha\beta\right)\right) = x^2 - \left(2 + \frac\beta\alpha + \frac\alpha\beta\right) x + \left(1 + \frac\beta\alpha\right) \left(1 + \frac\alpha\beta\right)`

and notice that the constant term is actually the same as (but has the opposite sign of) the coefficient of the
`x` term :

`\left(1 + \frac\beta\alpha\right) \left(1 + \frac\alpha\beta\right) = 1 + \frac\beta\alpha + \frac\alpha\beta + 1 = 2 + \frac\beta\alpha + \frac\alpha\beta`

Rewrite the coefficient as

`2 + \frac\beta\alpha + \frac\alpha\beta = 2 + (\alpha^2 + \beta^2)/(\alpha\beta) = 2 + \frac{\alpha^2 + \beta^2}2`

Now, observe that

`(\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2 \implies \alpha^2 + \beta^2 = (-3)^2 - 2^2 = 5`

Then the coefficient is simply
`2 + \frac52 = \frac92`, so the polynomial we want is

`\boxed{x^2 - \frac92 x + \frac92}`