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If one of the zeroes of cubic polynomial x3+ px2+ q+
r is - 1, then the product of the other two zeroes is a) p q 1 b) p-q-1 c) q-p 1 d) q-p-1?
r is - 1, then the product of the other two zeroes is a) p q 1 b) p-q-1 c) q-p 1 d) q-p-1?
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If one of the zeroes of cubic polynomial x3+ px2+ q+r is - 1, then the...
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If one of the zeroes of cubic polynomial x3+ px2+ q+r is - 1, then the...
Explanation:
A cubic polynomial is of the form ax^3 + bx^2 + cx + d where a, b, c, and d are constants.
Let α, β, and γ be the zeroes of the cubic polynomial x^3 + px^2 + qx + r. Then, by the factor theorem, we have:
- x^3 + px^2 + qx + r = (x - α)(x - β)(x - γ)
Given that one of the zeroes is -1, we can substitute this value in the equation above to get:
- 1 - p + q + r = 0
Therefore, we have:
- p - q - r = 1
Let the other two zeroes be α and β. Then, by Vieta's formulas, we have:
- αβγ = -r
- α + β + γ = -p
- αβ + βγ + γα = q
Since one of the zeroes is -1, we can rewrite the above equations as:
- αβ = -r/γ
- α + β + γ = -p
- αβ + βγ + γα = q
- αβγ = -r
- αβγ = γ(-r/γ) = -rγ
- αβ + βγ + γα = αβ + γ(-1) + α(-1) = αβ - γ - α
Substituting the values of αβγ and αβ + βγ + γα in the equation αβγ = -r, we get:
- (αβ - γ - α)γ = -r
- αβγ - γ^2 - αγ = -r
- -r - γ^2 - αγ = -r
- γ^2 + αγ = 0
- γ(γ + α) = 0
Since γ = -1, we have:
- αβ - γ - α = αβ + 1 + α = α(α + β + 1) = -r/γ = r
Substituting α + β = -γ - p = 1 - p and simplifying, we get:
- αβ = r - α - β = r - (1 - p) = p - r + 1
Therefore, the product of the other two zeroes is p - r + 1.
Hence, the answer is option (d) q - p - 1.
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If one of the zeroes of cubic polynomial x3+ px2+ q+r is - 1, then the product of the other two zeroes is a) p q 1 b) p-q-1 c) q-p 1 d) q-p-1?
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