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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) .

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. Download more important topics, notes, lectures and mock test series for Class 9 Exam by signing up for free.

Here you can find the meaning 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) defined & explained in the simplest way possible. Besides giving the explanation 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) , 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 ample number of questions to practice 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) tests, examples and also practice Class 9 tests.