If the zeroes of the polynomial f(x) = x³ − 12x² + 39x + k are in A.P., find the value of k

Video Explanation

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Solution

Given polynomial:

f(x) = x³ − 12x² + 39x + k

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 − 12x² + 39x + k with x³ + Ax² + Bx + C:

A = −12,   B = 39,   C = k

Step 2: Find the Value of a

Sum of zeroes:

(a − d) + a + (a + d) = 3a

3a = −A = 12

∴ a = 4

Step 3: Find the Value of d

Sum of products of zeroes taken two at a time:

(a − d)a + a(a + d) + (a − d)(a + d)

= 3a² − d²

3a² − d² = B = 39

Substituting a = 4:

3(16) − d² = 39

48 − d² = 39

∴ d² = 9

∴ d = 3

Step 4: Find the Value of k Using Product of Zeroes

Product of zeroes:

(a − d)a(a + d) = a(a² − d²)

= 4(16 − 9)

= 28

But,

Product of zeroes = −C = −k

∴ −k = 28

∴ k = −28

Final Answer

The value of k = −28.

Conclusion

Thus, if the zeroes of the polynomial f(x) = x³ − 12x² + 39x + k are in arithmetic progression, then the value of k is −28.