Condition for Zeroes of a Cubic Polynomial to Be in Arithmetic Progression

Video Explanation

Question

Find the condition that the zeroes of the polynomial

\[ f(x) = x^3 + 3px^2 + 3qx + r \]

may be in arithmetic progression.

Solution

Step 1: Assume the Zeroes in A.P.

Let the three zeroes of the polynomial be

\[ a-d,\; a,\; a+d \]

where \(a\) is the middle term and \(d\) is the common difference.

Step 2: Use Relations Between Zeroes and Coefficients

For the cubic polynomial \[ x^3 + 3px^2 + 3qx + r, \]

comparing with \(x^3 + bx^2 + cx + d\), we get

\[ b = 3p,\quad c = 3q,\quad d = r. \]

(i) Sum of the zeroes

\[ (a-d) + a + (a+d) = 3a \]

But,

\[ \text{Sum of zeroes} = -\frac{b}{a} = -3p \]

So,

\[ 3a = -3p \Rightarrow a = -p \]

(ii) Sum of the products of zeroes taken two at a time

\[ (a-d)a + a(a+d) + (a-d)(a+d) \]

\[ = a^2 – ad + a^2 + ad + (a^2 – d^2) = 3a^2 – d^2 \]

But,

\[ \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = 3q \]

So,

\[ 3a^2 – d^2 = 3q \]

Substituting \(a = -p\):

\[ 3p^2 – d^2 = 3q \Rightarrow d^2 = 3(p^2 – q) \]

(iii) Product of the zeroes

\[ (a-d)a(a+d) = a(a^2 – d^2) \]

But,

\[ \alpha\beta\gamma = -r \]

So,

\[ a(a^2 – d^2) = -r \]

Substituting \(a = -p\) and \(d^2 = 3(p^2 – q)\):

\[ (-p)\left(p^2 – 3(p^2 – q)\right) = -r \]

\[ (-p)(-2p^2 + 3q) = -r \]

\[ 2p^3 – 3pq = -r \]

\[ \Rightarrow \boxed{2p^3 – 3pq + r = 0} \]

Conclusion

The condition that the zeroes of the polynomial \[ x^3 + 3px^2 + 3qx + r \] are in arithmetic progression is:

\[ \boxed{2p^3 – 3pq + r = 0} \]