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