Question

If zeros of the polynomial f(x) = x³ − 3px² + qx − r are in A.P., then:

Options

• 2p³ = pq − r

• 2p³ = pq + r

• p³ = pq − r

• None of these

Answer 1

2p³ = pq − r

Explanation:

Let a - d, a, a + d be the zeros of the polynomial f(x) = x³ − 3px² + qx − r then

sum of zero = `(-  "coefficient of"  x^2)/("coefficient of"  x^3)`

`(a - d) + a(a +d)= (-(-3p))/1`

`a - cancel(d)+a+a+cancel(d)= 3p`

3a = 3p

a = `3/3p`

a = p

Since a is a zero of the polynomial f(x)

Therefore,

f(a) = 0

a³ - 3pa² + qa - r = 0

Substituting a = p we get

p³ - 3p(p)² + q × p - r = 0

p³ - 3p³ + qp - r = 0

-2p³ + qp - r = 0

qp - r = 2p³