If the zeroes of cubic polynomial p(x) = x³ - 6x² + 3x + 10 are of the form a, a +b, a + 2b for some real numbers a and b, find the zeroes of the polynomial and the values of a and b.
Hint: Obtain a + b = 2 from the sum of b a zeros =-b/a =6=3(a + b). This gives one of the zeroes as 2. Now, factorise p (x) and proceed.
Answers
Step-by-step explanation:
Let P(x) = x3 - 6x2 + 3x + 10
And (a), (a + b) and (a + 2b) are the zeroes of P(x).
We know, Sum of the zeroes = - (coefficient of x2) ÷ coefficient of x3
α + β + γ = - b/a
a + (a + b) + (a + 2b) = - (- 6)
3a + 3b = 6
a + b = 2
⇒ a = 2 - b ---------- ---- (1)
Product of all the zeroes = - (constant term) ÷ coefficient of x3
αβγ = - 10
a(a + b)(a + 2b) = - 10
(2 - b) (2) (2 + b) = - 10
(2 - b) (2 + b) = - 5
4 - b2 = - 5
⇒ b2 = 9 ⇒ b = +-3
When b = 3, a = 2 - 3 = - 1 (from (1))
⇒ a = - 1
When b = - 3,a = 2 - (- 3) = 5 (from(1))
⇒ a = 5
Case 1: when a = - 1 and b = 3
The zeroes of the polynomial are: a = - 1
a + b = - 1 + 3 = 2
a + 2b = - 1 + 2(3) = 5
⇒ - 1, 2 and 5 are the zeroes Case 2: when a = 5,b = - 3
The zeroes of the polynomial are:
a = 5
a + b = 5 - 3 = 2
a + 2b = 5 - 2(3) = - 1
⇒ - 1, 2 and 5 are the zeroes
By both the cases the zeroes of the polynomial are - 1, 2, 5
HOPE YOU WILL UNDERSTAND
Answer:
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Step-by-step explanation:
HOPE YOU ARE SATISFIED BY THIS SOLUTION.
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