Find the condition that the zeroes of the polynomial f(x) = x³ + 3px² + 3qx + r may be in A.P.

Video Explanation

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Solution

Given polynomial:

f(x) = x³ + 3px² + 3qx + r

Let the zeroes of the polynomial be in A.P.

∴ Let the zeroes be: a − d, a, a + d

Step 1: Use the Relationship Between Zeroes and Coefficients

For a cubic polynomial x³ + Ax² + Bx + C:

Sum of zeroes = −A
Sum of products of zeroes taken two at a time = B
Product of zeroes = −C

Comparing f(x) = x³ + 3px² + 3qx + r with x³ + Ax² + Bx + C:

A = 3p,   B = 3q,   C = r

Step 2: Form Equations Using A.P. Zeroes

Sum of zeroes:

(a − d) + a + (a + d) = 3a

∴ 3a = −3p

∴ a = −p

Sum of products of zeroes taken two at a time:

(a − d)a + a(a + d) + (a − d)(a + d)

= 3a² − d²

∴ 3a² − d² = 3q

Substituting a = −p:

3p² − d² = 3q

∴ d² = 3(p² − q)

Step 3: Use the Product of Zeroes

Product of zeroes:

(a − d)a(a + d) = a(a² − d²)

= −r

Substituting a = −p and d² = 3(p² − q):

(−p)[p² − 3(p² − q)] = −r

(−p)(−2p² + 3q) = −r

∴ 2p³ − 3pq = −r

∴ r = 3pq − 2p³

Final Answer

The required condition is:

r = 3pq − 2p³

Conclusion

Thus, the zeroes of the polynomial f(x) = x³ + 3px² + 3qx + r will be in arithmetic progression if and only if r = 3pq − 2p³.