Question

If zeroes of the polynomials `f(x)=x^3-3px^2+qx-r` are in AP then what is the relation between p, q and r?

Answer 1

`2p^3+pq+r = 0`

Explanation:

Denoting the three zeros by `alpha, beta, gamma`, we have:

`f(x) = x^3-3px^2+qx-r`

`color(white)(f(x)) = (x-alpha)(x-beta)(x-gamma)`

`color(white)(f(x)) = x^3-(alpha+beta+gamma)x^2+(alphabeta+betagamma+gammaalpha)x-alphabetagamma`

In particular, equating the coefficient of `x^2` we have:

`alpha+beta+gamma = 3p`

If `alpha, beta` and `gamma` are in arithmetic progression with common difference `delta`, then:

`alpha = beta-delta`

`gamma = beta+delta`

So:

`3 beta = (beta-delta)+beta+(beta+delta) = alpha+beta+gamma = 3p`

So:

`beta = p`

That is: `p` is one of the zeros of `f(x)`

So:

`0 = f(p) = color(blue)(p^3)-3pcolor(blue)(p)^2-qcolor(blue)(p)-r = -2p^3-pq-r`

Inverting the signs, that is:

`2p^3+pq+r = 0`