If the zeroes of the polynomial f(x) = 2x³ − 15x² + 37x − 30 are in A.P., find them

Video Explanation

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Solution

Given polynomial:

f(x) = 2x³ − 15x² + 37x − 30

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 Zeros and Coefficients

For a cubic polynomial ax³ + bx² + cx + d:

Sum of zeroes = −b/a
Sum of products of zeroes taken two at a time = c/a
Product of zeroes = −d/a

Comparing f(x) = 2x³ − 15x² + 37x − 30 with ax³ + bx² + cx + d:

a = 2,   b = −15,   c = 37,   d = −30

Step 2: Find the Value of a

Sum of zeroes:

(a − d) + a + (a + d) = 3a

3a = −b/a = 15/2

∴ a = 5/2

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 = c/a = 37/2

Substituting a = 5/2:

3(25/4) − d² = 37/2

75/4 − d² = 74/4

∴ d² = 1/4

∴ d = 1/2

Step 4: Find the Zeroes

a − d = 5/2 − 1/2 = 2

a = 5/2

a + d = 5/2 + 1/2 = 3

Step 5: Verify the Product of Zeroes

Product of zeroes:

2 × 5/2 × 3 = 15

−d/a = −(−30)/2 = 15

✔ Verified

Final Answer

The zeroes of the given polynomial are:

2, 5/2 and 3

Conclusion

Thus, the zeroes of the polynomial f(x) = 2x³ − 15x² + 37x − 30 which are in arithmetic progression are 2, 5/2 and 3.