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If -1and -2are the two zeroes of the polynomial p(x)=x^4+ 11x^3+ 41x^2 +61x+ 30 then fund the other two zeroes of the polynomial ? Plz ans. Fast.?
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If -1and -2are the two zeroes of the polynomial p(x)=x^4+ 11x^3+ 41x^2...
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If -1and -2are the two zeroes of the polynomial p(x)=x^4+ 11x^3+ 41x^2...
Answer:
Given, p(x) = x^4 - 11x^3 + 41x^2 - 61x + 30
We know that if α is a zero of p(x), then (x-α) is a factor of p(x).
So, (x+1) and (x+2) are factors of p(x).
We can find the other two zeroes by factorizing the polynomial using the factor theorem.
p(x) = x^4 - 11x^3 + 41x^2 - 61x + 30
p(x) = (x+1)(x+2)(ax^2 + bx + c) (Let the other two factors be ax^2 + bx + c)
Expanding the above expression, we get:
p(x) = ax^4 + (b+2a)x^3 + (c+2b+a)x^2 + (2c+b-61)x + 2c
Comparing the coefficients of x^4 and x^3, we get:
a = 1
b+2a = -11
b = -13
Comparing the coefficients of x^2 and x, we get:
c+2b+a = 41
c = 8
2c+b-61 = -61
Substituting the values of a, b, and c in the factorized expression of p(x), we get:
p(x) = (x+1)(x+2)(x^2 - 13x + 8)
The other two zeroes can be found by solving the quadratic equation:
x^2 - 13x + 8 = 0
Using the quadratic formula, we get:
x = [13 ± √(169 - 32)]/2
x = [13 ± √137]/2
Therefore, the four zeroes of the polynomial p(x) are -1, -2, (13 + √137)/2, and (13 - √137)/2.
Given, p(x) = x^4 - 11x^3 + 41x^2 - 61x + 30
We know that if α is a zero of p(x), then (x-α) is a factor of p(x).
So, (x+1) and (x+2) are factors of p(x).
We can find the other two zeroes by factorizing the polynomial using the factor theorem.
Factorizing p(x):
p(x) = x^4 - 11x^3 + 41x^2 - 61x + 30
p(x) = (x+1)(x+2)(ax^2 + bx + c) (Let the other two factors be ax^2 + bx + c)
Expanding the above expression, we get:
p(x) = ax^4 + (b+2a)x^3 + (c+2b+a)x^2 + (2c+b-61)x + 2c
Comparing the coefficients of x^4 and x^3, we get:
a = 1
b+2a = -11
b = -13
Comparing the coefficients of x^2 and x, we get:
c+2b+a = 41
c = 8
2c+b-61 = -61
Substituting the values of a, b, and c in the factorized expression of p(x), we get:
p(x) = (x+1)(x+2)(x^2 - 13x + 8)
Calculating the other two zeroes:
The other two zeroes can be found by solving the quadratic equation:
x^2 - 13x + 8 = 0
Using the quadratic formula, we get:
x = [13 ± √(169 - 32)]/2
x = [13 ± √137]/2
Therefore, the four zeroes of the polynomial p(x) are -1, -2, (13 + √137)/2, and (13 - √137)/2.
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If -1and -2are the two zeroes of the polynomial p(x)=x^4+ 11x^3+ 41x^2 +61x+ 30 then fund the other two zeroes of the polynomial ? Plz ans. Fast.?
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If -1and -2are the two zeroes of the polynomial p(x)=x^4+ 11x^3+ 41x^2 +61x+ 30 then fund the other two zeroes of the polynomial ? Plz ans. Fast.? for Class 10 2025 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about If -1and -2are the two zeroes of the polynomial p(x)=x^4+ 11x^3+ 41x^2 +61x+ 30 then fund the other two zeroes of the polynomial ? Plz ans. Fast.? covers all topics & solutions for Class 10 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If -1and -2are the two zeroes of the polynomial p(x)=x^4+ 11x^3+ 41x^2 +61x+ 30 then fund the other two zeroes of the polynomial ? Plz ans. Fast.?.
If -1and -2are the two zeroes of the polynomial p(x)=x^4+ 11x^3+ 41x^2 +61x+ 30 then fund the other two zeroes of the polynomial ? Plz ans. Fast.? for Class 10 2025 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about If -1and -2are the two zeroes of the polynomial p(x)=x^4+ 11x^3+ 41x^2 +61x+ 30 then fund the other two zeroes of the polynomial ? Plz ans. Fast.? covers all topics & solutions for Class 10 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If -1and -2are the two zeroes of the polynomial p(x)=x^4+ 11x^3+ 41x^2 +61x+ 30 then fund the other two zeroes of the polynomial ? Plz ans. Fast.?.
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