Question

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

Answer 1

Answer:

The given polynomial is 2x3 - 15x2 + 37x - 30.

Since the roots of the polynomial are in AP, so let the roots be a - d, a, a + d.

Now, using the relation between zeroes and the coefficients of the given cubic polynomial, we have:

Sum of roots = a - d + a + a + d = - (-15)/ 2 = 15/ 2

So, 3a = 15/ 2

a = 5/ 2 = 2.5

Now, product of roots = (a - d) (a) (a + d) = -(-30)/ 2 = 15

2.5(a2 - d2) = 15

(2.5)2 - d2 = 6

d2 = 0.25

d = 0.5

Thus, the zeroes of the given polynomial are 2, 2.5, and 3.

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Answer 2

$\huge\fbox \red{Solution:}$

Let the zeros of the given polynomial be α, β and γ. (3 zeros as it’s a cubic polynomial)

And given, the zeros are in A.P.

So, let’s consider the roots as

α = a – d, β = a and γ = a +d

Where, a is the first term and d is the common difference.

From given f(x), a= 2, b= -15, c= 37 and d= 30

⇒ Sum of roots = α + β + γ = (a - d) + a + (a + d) = 3a = (-b/a) = -(-15/2) = 15/2

So, calculating for a, we get 3a = 15/2 ⇒ a = 5/2

⇒ Product of roots = (a - d) x (a) x (a + d) = a(a2 –d2) = -d/a = -(30)/2 = 15

⇒ a(a2 –d2) = 15

Substituting the value of a, we get

⇒ (5/2)[(5/2)2 –d2] = 15