Finding Zeroes of a Cubic Polynomial in Arithmetic Progression

Video Explanation

Question

If the zeroes of the polynomial

\[ f(x) = 2x^3 – 15x^2 + 37x – 30 \]

are in arithmetic progression, find them.

Solution

Step 1: Assume the Zeroes in A.P.

Let the three zeroes 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 \[ 2x^3 – 15x^2 + 37x – 30, \]

we have

\[ a = 2,\quad b = -15,\quad c = 37,\quad d = -30. \]

(i) Sum of the zeroes

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

But,

\[ \text{Sum of zeroes} = -\frac{b}{a} = -\frac{-15}{2} = \frac{15}{2} \]

So,

\[ 3a = \frac{15}{2} \Rightarrow a = \frac{5}{2} \]

(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,

\[ \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{37}{2} \]

So,

\[ 3a^2 – d^2 = \frac{37}{2} \]

Substitute \(a = \frac{5}{2}\):

\[ 3\left(\frac{5}{2}\right)^2 – d^2 = \frac{37}{2} \]

\[ \frac{75}{4} – d^2 = \frac{37}{2} \]

\[ \frac{75}{4} – \frac{74}{4} = d^2 \Rightarrow d^2 = \frac{1}{4} \]

\[ d = \frac{1}{2} \]

(iii) Product of the zeroes (verification)

\[ (a-d)a(a+d) = a(a^2 – d^2) \]

\[ = \frac{5}{2}\left(\frac{25}{4} – \frac{1}{4}\right) = \frac{5}{2}\cdot \frac{24}{4} = \frac{5}{2}\cdot 6 = 15 \]

\[ -\frac{d}{a} = -\frac{-30}{2} = 15 \]

Hence, the product relation is also verified.

Step 3: Write the Zeroes

\[ a-d = \frac{5}{2} – \frac{1}{2} = 2, \]

\[ a = \frac{5}{2}, \]

\[ a+d = \frac{5}{2} + \frac{1}{2} = 3 \]

Conclusion

The zeroes of the polynomial \[ 2x^3 – 15x^2 + 37x – 30 \] are

\[ \boxed{2,\; \frac{5}{2},\; 3} \]