Class 9 Exam > Class 9 Questions > Assertion (A): 3x2 + x – 1 = (x + 1)(3x – 2x)...
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Assertion (A): 3x2 + x – 1 = (x + 1)(3x – 2x) + 1.
Reason(R) If p(x) and g(x) are two polynomials such that degree of p(x) ≥ degree of g(x)
and g(x) ≥ 0 then we can find polynomials q(x) and r(x) such that p(x) = g(x) q(x) + r(x) ,
where r(x) = 0 of degree of r(x)
?
Reason(R) If p(x) and g(x) are two polynomials such that degree of p(x) ≥ degree of g(x)
and g(x) ≥ 0 then we can find polynomials q(x) and r(x) such that p(x) = g(x) q(x) + r(x) ,
where r(x) = 0 of degree of r(x)
?
Verified Answer
Assertion (A): 3x2 + x – 1 = (x + 1)(3x – 2x) + 1.Reason(R) If p(x) an...
The assertion (A) is false, and the reason (R) is true.
The equation 3x^2 + x – 1 = (x + 1)(3x – 2x) + 1 is not true, as the left-hand side is a second-degree polynomial and the right-hand side is not equivalent to it.
However, the reason (R) is true. The statement is a special case of the Euclidean division algorithm, which states that if p(x) and g(x) are two polynomials such that the degree of p(x) is greater than or equal to the degree of g(x) and g(x) is not equal to 0, then there exist unique polynomials q(x) and r(x) such that p(x) = g(x)q(x) + r(x), where the degree of r(x) is less than the degree of g(x), and r(x) is either equal to 0 or has a leading coefficient different from 0.
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Most Upvoted Answer
Assertion (A): 3x2 + x – 1 = (x + 1)(3x – 2x) + 1.Reason(R) If p(x) an...
Assertion (A):
3x^2 - x - 1 = (x + 1)(3x - 2x - 1)
Reason (R):
If p(x) and g(x) are two polynomials such that the degree of p(x) ≥ the degree of g(x) and g(x) ≥ 0, then we can find polynomials q(x) and r(x) such that p(x) = g(x) q(x) + r(x), where r(x) = 0 or the degree of r(x) is less than the degree of g(x).
Explanation:
To prove the assertion, we need to show that the given polynomial can be expressed as the product of two polynomials such that the degree of the divisor is greater than or equal to the degree of the dividend.
We have the polynomial 3x^2 - x - 1 on the left-hand side. To factorize it, let's try to express it as the product of two binomials, (x + a) and (3x + b):
3x^2 - x - 1 = (x + a)(3x + b)
Now, we expand the product on the right-hand side:
3x^2 + bx + 3ax + ab = 3x^2 + (b + 3a)x + ab
Comparing the coefficients of the like terms on both sides, we get:
bx + 3ax = -x
b + 3a = 0 ----(1)
ab = -1 ----(2)
From equation (1), we have b = -3a. Substituting this value in equation (2), we get:
ab = -1
(-3a)a = -1
-3a^2 = -1
a^2 = 1/3
a = ±√(1/3)
So, we have two possible values for a: a = √(1/3) or a = -√(1/3).
Now, substituting the value of a in equation (1), we can find the corresponding values of b:
For a = √(1/3):
b + 3(√(1/3)) = 0
b = -3(√(1/3))
b = -√3
For a = -√(1/3):
b + 3(-√(1/3)) = 0
b = 3(√(1/3))
b = √3
Therefore, we have two possible factorizations of the given polynomial:
1. 3x^2 - x - 1 = (x + √(1/3))(3x - √3)
2. 3x^2 - x - 1 = (x - √(1/3))(3x + √3)
In both cases, the degree of the divisor (x + √(1/3)) or (x - √(1/3)) is equal to 1, which is greater than the degree of the dividend (3x^2 - x - 1), which is equal to 2.
Hence, the assertion (A) is true and the reason (
3x^2 - x - 1 = (x + 1)(3x - 2x - 1)
Reason (R):
If p(x) and g(x) are two polynomials such that the degree of p(x) ≥ the degree of g(x) and g(x) ≥ 0, then we can find polynomials q(x) and r(x) such that p(x) = g(x) q(x) + r(x), where r(x) = 0 or the degree of r(x) is less than the degree of g(x).
Explanation:
To prove the assertion, we need to show that the given polynomial can be expressed as the product of two polynomials such that the degree of the divisor is greater than or equal to the degree of the dividend.
We have the polynomial 3x^2 - x - 1 on the left-hand side. To factorize it, let's try to express it as the product of two binomials, (x + a) and (3x + b):
3x^2 - x - 1 = (x + a)(3x + b)
Now, we expand the product on the right-hand side:
3x^2 + bx + 3ax + ab = 3x^2 + (b + 3a)x + ab
Comparing the coefficients of the like terms on both sides, we get:
bx + 3ax = -x
b + 3a = 0 ----(1)
ab = -1 ----(2)
From equation (1), we have b = -3a. Substituting this value in equation (2), we get:
ab = -1
(-3a)a = -1
-3a^2 = -1
a^2 = 1/3
a = ±√(1/3)
So, we have two possible values for a: a = √(1/3) or a = -√(1/3).
Now, substituting the value of a in equation (1), we can find the corresponding values of b:
For a = √(1/3):
b + 3(√(1/3)) = 0
b = -3(√(1/3))
b = -√3
For a = -√(1/3):
b + 3(-√(1/3)) = 0
b = 3(√(1/3))
b = √3
Therefore, we have two possible factorizations of the given polynomial:
1. 3x^2 - x - 1 = (x + √(1/3))(3x - √3)
2. 3x^2 - x - 1 = (x - √(1/3))(3x + √3)
In both cases, the degree of the divisor (x + √(1/3)) or (x - √(1/3)) is equal to 1, which is greater than the degree of the dividend (3x^2 - x - 1), which is equal to 2.
Hence, the assertion (A) is true and the reason (
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Assertion (A): 3x2 + x – 1 = (x + 1)(3x – 2x) + 1.Reason(R) If p(x) and g(x) are two polynomials such that degree of p(x) ≥ degree of g(x)and g(x) ≥ 0 then we can find polynomials q(x) and r(x) such that p(x) = g(x) q(x) + r(x) ,where r(x) = 0 of degree of r(x) ?
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Assertion (A): 3x2 + x – 1 = (x + 1)(3x – 2x) + 1.Reason(R) If p(x) and g(x) are two polynomials such that degree of p(x) ≥ degree of g(x)and g(x) ≥ 0 then we can find polynomials q(x) and r(x) such that p(x) = g(x) q(x) + r(x) ,where r(x) = 0 of degree of r(x) ? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about Assertion (A): 3x2 + x – 1 = (x + 1)(3x – 2x) + 1.Reason(R) If p(x) and g(x) are two polynomials such that degree of p(x) ≥ degree of g(x)and g(x) ≥ 0 then we can find polynomials q(x) and r(x) such that p(x) = g(x) q(x) + r(x) ,where r(x) = 0 of degree of r(x) ? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Assertion (A): 3x2 + x – 1 = (x + 1)(3x – 2x) + 1.Reason(R) If p(x) and g(x) are two polynomials such that degree of p(x) ≥ degree of g(x)and g(x) ≥ 0 then we can find polynomials q(x) and r(x) such that p(x) = g(x) q(x) + r(x) ,where r(x) = 0 of degree of r(x) ?.
Assertion (A): 3x2 + x – 1 = (x + 1)(3x – 2x) + 1.Reason(R) If p(x) and g(x) are two polynomials such that degree of p(x) ≥ degree of g(x)and g(x) ≥ 0 then we can find polynomials q(x) and r(x) such that p(x) = g(x) q(x) + r(x) ,where r(x) = 0 of degree of r(x) ? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about Assertion (A): 3x2 + x – 1 = (x + 1)(3x – 2x) + 1.Reason(R) If p(x) and g(x) are two polynomials such that degree of p(x) ≥ degree of g(x)and g(x) ≥ 0 then we can find polynomials q(x) and r(x) such that p(x) = g(x) q(x) + r(x) ,where r(x) = 0 of degree of r(x) ? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Assertion (A): 3x2 + x – 1 = (x + 1)(3x – 2x) + 1.Reason(R) If p(x) and g(x) are two polynomials such that degree of p(x) ≥ degree of g(x)and g(x) ≥ 0 then we can find polynomials q(x) and r(x) such that p(x) = g(x) q(x) + r(x) ,where r(x) = 0 of degree of r(x) ?.
Solutions for Assertion (A): 3x2 + x – 1 = (x + 1)(3x – 2x) + 1.Reason(R) If p(x) and g(x) are two polynomials such that degree of p(x) ≥ degree of g(x)and g(x) ≥ 0 then we can find polynomials q(x) and r(x) such that p(x) = g(x) q(x) + r(x) ,where r(x) = 0 of degree of r(x) ? in English & in Hindi are available as part of our courses for Class 9.
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Assertion (A): 3x2 + x – 1 = (x + 1)(3x – 2x) + 1.Reason(R) If p(x) and g(x) are two polynomials such that degree of p(x) ≥ degree of g(x)and g(x) ≥ 0 then we can find polynomials q(x) and r(x) such that p(x) = g(x) q(x) + r(x) ,where r(x) = 0 of degree of r(x) ?, a detailed solution for Assertion (A): 3x2 + x – 1 = (x + 1)(3x – 2x) + 1.Reason(R) If p(x) and g(x) are two polynomials such that degree of p(x) ≥ degree of g(x)and g(x) ≥ 0 then we can find polynomials q(x) and r(x) such that p(x) = g(x) q(x) + r(x) ,where r(x) = 0 of degree of r(x) ? has been provided alongside types of Assertion (A): 3x2 + x – 1 = (x + 1)(3x – 2x) + 1.Reason(R) If p(x) and g(x) are two polynomials such that degree of p(x) ≥ degree of g(x)and g(x) ≥ 0 then we can find polynomials q(x) and r(x) such that p(x) = g(x) q(x) + r(x) ,where r(x) = 0 of degree of r(x) ? theory, EduRev gives you an
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