Solution: Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (-z + x-2y). = [(a + b + c)3 – a3] – (b3+ c3) Solution: (iv) Degree of polynomial y3(1-y4) or y3 – y7 is seven, because the maximum exponent of y is seven. (i) Firstly, determine the factor by using splitting method. Exercise 2.1: Multiple Choice Questions (MCQs). (d) not defined e.g., Let f(x) = x5 + 2 and g(x) = -x5 + 2x2 ⇒ 8m = 8 Important questions in Polynomials with video lesson. All questions with solutions of polynomials will help all the students to revise complete syllabus and score more marks in examinations. Hence, the value of a is 3/2. This chapter consists of problems based on polynomial in one variable, Zeroes of a Polynomial, Remainder Theorem, Factorisation of Polynomials and Algebraic Identities. If a + b + c = 9 and ab + bc + ca = 26, find a2 + b2 +c2. (vi) Polynomial 2 + x is a linear polynomial, because its degree is 1. (x-2 + x+2)(x-2-x-2) = 0 [using identity, a2-b2 =(a-b)(a + b)] =» (2x)(-4) = 0. (ii) Polynomial 27a – a = 15 – 41 . Solution: Question 24: (i) We have, 2X3 – 3x2 – 17x + 30 = 2a(2a + 3) -1 (2a + 3) = (2a – 1)(2a + 3) Hence, possible length = 2a -1 and breadth = 2a + 3, Question 1: Question 5. (i) x3 +y3 -12xy + 64,when x+y = -4. Hence, p-1 is a factor of g(p). If x51 + 51 is divided by x + 1, then the remainder is NCERT solutions for class 9 Maths will help you to understand and solve complex problems easily. For what value of m is x3 -2mx2 +16 divisible by x + 2? Now, p(1) = 2(1)4 – 5(1)3 + 2(1)2 -1 + 2 =2-5+2-1+2=6-6=0 All the chapterwise questions with solutions to help you to revise complete CBSE syllabus and score more marks in Your board examinations. Solution: Question 3: (a) -3 (b) 4 (c) 2 (d)-2 (a) x2 + y2 + 2 xy (v) -3 is a zero of y2 +y-6 (x) Polynomial √2x-1 is a linear polynomial, because maximum exponent of xis 1. (ii) Further, use any of the identities i.e., a3 + b3 =(a + b)(a2+b2-ab) and a3 -b3 =(a-b)(a2 + b2 + ab), then simplify it, to get the factor. p(x) = x – 4 (i) False Each exponent of the variable x is a whole number. (i) 2x3 – 3x2 – 17x + 30 (d) 10x (b) 6 Factorise the following Find the following products: Write whether the following statements are true or false. Question 20: Here, zero of g(x) is 1/2. Solution: Question 17: x + 1 is a factor of the polynomial The factorization of 4x2 + 8x+ 3 is (b) x² + 5 [polynomial and also a binomial]. In this method firstly check the values of a + b+ c, then . Question 8: Hence, the remainder is 50. Solution: [using identity, a2 – b2 = (a – b)(a + b)] Solution: (v) 3 NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12. Here are all questions are solved with a full explanation and available for free to download. (c) any real number Question 10: (i) We have, = 27a3 – 8b3 – 18ab(3a – 2b) Solution: (i) 9x2 +4y2 + 16z2 +12xy-16yz -24xz (iii) Degree of polynomial x3 – 9x + 3x5 is 5, because the maximum exponent of x is 5. (d) -2 Solution: Factorise We have, 6x2 + 7x – 3 = 6x2 + 9x – 2x – 3 (a) (x + 1) (x + 3) ⇒ y = 2 and y = -3 ⇒ -2a + 3 = 0 27a+41 = 15+a Question 8. The factorization of 4x2 + 8x+ 3 is We know that, a3 + b3 + c3 – 3 abc = (a + b + c)(a2 + b2 + c2 -ab-bc-ca) (c) 3abc Question 1. (b) Given, p(x) = x2 – 2√2x + 1 …(i) (ii) 25x2 + 16y2 + 4Z2 – 40xy +16yz – 20xz Question 8. ⇒ y(y + 3) – 2(y + 3) = 0 37 (c) Let p(x) = 2x2 + kx (a) 0 (b) 1 (c) any real number (d) not defined 2x= 7 => x =7/2 Also, find the remainder when p(x) is divided by x + 2. = 32 – 40 + 8 = 40 – 40 = 0 =(-5x)2 + (4y)2 + (2z)2 + 2(-5x)(4y) + 2(4y)(2z) + 2(2z)(-5x) Question 4. e.g., Let us consider zero polynomial be 0(x – k), where k is a real number For determining the zero, put x – k = 0 ⇒ x = k Find the value of the polynomial 3x3 – 4x2 + 7x – 5, when x = 3 and also when x = -3. NCERT Class 9 New Books for Maths Chapter 2 Polynomials includes all the questions given in CBSE syllabus. And to make that base strong students are advised to solve NCERT Exemplar class 9 Maths. If x + 2a is a factor of a5 – 4a2x3 + 2x + 2a + 3, then find the value of a. If p (x) = x + 3, then p(x) + p(- x) is equal to (i) p(x) = 10x – 4x2 – 3 (ii) p(y) = (y + 2)(y – 2) (ii) Polynomial y3 – 5y is a one variable polynomial, because it contains only one variable i.e., y. (v) Polynomial 3 is a constant polynomial, because the exponent of variable is 0. = -81 – 36 – 21 – 5 = -143 p1(3) = p2(3) (a) x2 + y2 + 2 xy (b) x2 + y2 – xy (c) xy2 (d) 3xy Question 1. NCERT Exemplar Class 9 Maths Unit 2 Polynomials. p(x) = x- 4 (i) monomial of degree 1. It is not a polynomial because it is a rational function. Since, x + 1 is a factor of p(x), then p(-1) = 0 Question 1. Question 7: (iv) The constant term in given polynomial is 1/5. (iv) Polynomial x2 – Zxy + y2 +1 is a two variables pplynomial, because it contains two variables x and y. (d) 2abc (iii) x3 + x2 – 4x – 4 (iv) 3x3 – x2 – 3x +1 and p(2) = 2(2)4 – 5(2)3 + 2(2)2 – 2 + 2 = 2x16-5x8+2x4+ 0 = 32 – 40 + 8 = 40 – 40 =0 (i) Firstly, adjust the given number either in the farm 0f a3 + b3 or a3 -b3 Solution: Download the NCERT Exemplar Problem Solutions for Class 9 Maths Chapter 2 - Polynomials solved by Mathematics Expert Teachers at Mathongo.com as per CBSE (NCERT) Book guidelines. (ii) lf p(a) = Q then p(x) is a multiple of g(x) and f(p(a) # Q then p(x) is not a multiple of g(x) where ‘a’ is a zero of g (x)). If the polynomials az3 +4z2 + 3z-4 and z3-4z + o leave the same remainder when divided by z – 3, find the value of a. [ ∴ If a + b + c = 0, then a3 + b3 + c3 = 3abc] = -0.018, Question 38. (vii) y³ – y (i) a3 -8b3 -64c3 -2Aabc = 3 x (-1) = -3 (-1)3 + (-1)2 + (-1) + 1 = 0 (b) 2x NCERT Exemplar Class 9 for Maths Chapter 2 – Polynomials. Factorise the following Solution: If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is (b) 1 (i) 9x2 + 4y2+16z2+12xy-16yz-24xz Expand the following Solution: Hence, the zero of polynomial is 0, Question 12: (d) Now, a3 + b3 + c3 = (a + b + c) (a2 + b2 + c2 – ab – be – ca) + 3abc ⇒ x = 0 On putting x = -1 in Eq. (i) We have, (3a – 2b)3 = 2x2(x – 2) + x(x – 2) – 15(x – 2) Because every polynomial is not a binomial. (a) 4 (b) 5 (c) 3 (d) 7 Because a binomial has exactly two terms. So, the degree of the polynomial is 3. = 50x2 + 10x = 10x (5x+ 1) Question 16. Fi nd (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz). ∴ (0.2)³ + (-0.3)³ + (0.1)3 = 3(0.2) (-0.3) (0.1) Solution: Question 36: ⇒ -8 – 8m + 16 = 0 (d) Now, 2492 – 2482 = (249 + 248) (249 – 248) [using identity, a2 – b2 = (a – b)(a + b)] Question 1: (ii) x3 -8y3 -36xy-216,when x = 2y + 6. = -20- 4 × 4 – 3 = -20 – 16 – 3 = -39 The type of questions that will be asked from NCERT Class 9 Maths Unit 2 are displayed in the below provided NCERT Exemplar Class 9 Maths Unit 2. Without actually calculating the cubes, find the value of 36xy-36xy = 0 (ii) -1/3 is a zero of 3x + 1 (iv) True (i) Since, x + y + 4 = 0, then (a) 4 (ii) -10 (i) We have, 9x2 – 12x + 3 = 3(3x2 – 4x + 1) Since, x + 2a is a factor of p(x), then put p(-2a) = 0 (b) (2x + 1) (2x + 3) Hence, the zeroes of t² – 2t are 0 and 2. (ii) The two different values of zeroes put in biquadratic polynomial. As we know that the degree of a polynomial is equal to the highest power of variable x. = x3 – 8y3 – z3 – 6xyz. Hence, the value of k is 2. SOLUTION: (i) The given polynomial is 5x³ + 4x² + 7x The highest power of the variable x is 3. (a) 3 (b) 2x (c) 0 (d) 6 (c) 3abc (d) 2abc ∴ p(3) = (3)³ – 3(3)² + 4(3) + 50 = 4a2 + 6a – 2a – 3 x + 1 is a factor of the polynomial (i) Let p(x) = x3 – x2 + 11x + 69 2a =3 Solution: (d) x4 + 3x3 + 3x2 + x + 1 (ii) Given, polynomial is Again, putting p = 1 in Eq. (i), we get (iii) x3 + x2-4x-4 = 3 x (-1) = -3 5 Questions. = 10-4-3= 10-7= 3 Solution: (iv) 0 and 2 are the zeroes of t2 – 2t 8x4 +4x3 -16x2 +10x+07. = 8 – 20 + 8 – 3 = – 7 CBSE Maths notes, CBSE physics notes, CBSE chemistry notes. Solution: ⇒ -a + 1 + 2 + 4a – 9 = 0 ⇒ k = 2 Factorise the following 2 – ( x3 + y3 ) exactly two terms 2a +3, prove... New academic year 2020-21 ncert exemplar class 9 maths polynomials cublic polynomial, because the sum of any two each! 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