(ii) Polynomial ⇒ a2 + b2 + c2 = 25 – 20 polynomial is divided by the second polynomial x4 + 1 and x-1. lf a+b+c=9 and ab + bc + ca = 26, find a2 + b2 + c2. BeTrained.in has solved each questions of NCERT Exemplar very thoroughly to help the students in solving any question from the book with a team of well experianced subject matter experts. (i) We have, p(x) = x³ – 2x² – 4x – 1 and g(x) = x + 1 Determine which of the following polynomial has x – 2 a factor (i) 3x2 + 6x – 24 (ii) 4x2+ x – 2 (c) -1 Solution: Question 25: (b) ½ NCERT Exemplar for Class 9 Maths Chapter 2 with Solutions by Swiflearn are by far the best and most reliable NCERT Exemplar Solutions that you can find on the internet. One of the zeroes of the polynomial 2x2 + 7x – 4 is Hence, zero of 3x + 1 is -1/5. We have, p(x) = x4 – 2x3 + 3x2 – ox + 3a – 7 Firstlyadjust the given number into two number such that one is a multiple of 10 and use the proper identity. = 4x3 + 2x + 2 which is not a polynomial of degree 4. (ii) x3 – 8y3 – 36xy-216,when x = 2y + 6. Now, x2-3x+2 = x2-2x-x+2 [by splitting middle term] By remainder theorem, find the remainder when p(x) is divided by g(x) (iv) ‘p(x) = x3-6x2+2x-4, g(x) = 1 -(3/2) x (iii) We have, p(x) = 4x³ – 12x² + 14x – 3 and g(x) = 2x -1 Question 2. Because the sum of any two polynomials of same degree has not always same degree. Solution: Polynomials | Maths | NCERT Exemplar Solutions | Class 9. Substituting x = 2 in (1), we get (c) Zero of the zero polynomial is any real number. Show that p-1 is a factor of p10 -1 and also of p11 -1. iii) 16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz (iii) 2x2– 7x – 15 The highest power of … Solution: [using identity, a3 + b3 = (a + b)(a2 -ab+ b2)] = (x+ y)[(x+ y)2 -(x2 -xy+ y2)] Now, p(x) = x3 – 5x2 + 4x – 3 Since, p(x) is divisible by (x+2), then remainder = 0 Which one of the following is a polynomial? (ii) Given, polynomial isp(y) = (y+2)(y-2) (iii) xy+ yz +zx (iv) x2 – Zxy + y2 +1 NCERT Exemplar Polynomials Class 9 (Part - 2) Nov 10, 2020 • 1h . (i) 1033 (ii) 101 x 102 (iii) 9992 (A) 0 = 27a3 – 54a2b + 36ab2 – 8b3. (d) 7 Vivek Patriya. Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. Check whether p(x) is a multiple of g(x) or not (iv) x2 – Zxy + y2 + 1 (d) Let p(x) = x51 + 51 . NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials Exercise 2.1 Question 1. Hence, one of the factor of given polynomial is 10x. Find the zeroes of the polynomial in each of the following, Hence, one of the zeroes of the polynomial p(x) is ½. Solution: We have, a + b + c = 9 (d) 8 √2 +1 So, x = -1 is zero of x3 + x2 + x+1 Find the following products We hope that our NCERT Class 9 New Books for Maths helped with your studies! Using long division method Solution: Hence, the degree of a polynomial is 4. g(x) = 3-6x Which of the following is a factor of (x+ y)3 – (x3 + y3)? You can also Download NCERT Solutions for class 9 Maths in Hindi to help you to revise complete Syllabus and score more marks in your examinations. Here, zero of g(x) is -1. For zero of polynomial, put p(x) = x-4 = 0 = 5(5 -10) = 5(-5) = – 25 = R.H.S. (iv) Polynomial Factorise As we know that the degree of a polynomial is equal to the highest power of variable x. Question 5: Solution: Question 30: (v) Polynomial 3 = 3x° is a constant polynomial, because its degree is 0. Solution: Question 5: ∴ p(-1) = (-1)³ – 2(-1)² – 4(-1) -1 (i) Let p(x) = x3 – x2 + 11x + 69 (d) -2 Find the following products: NCERT Exemplar Class 9 Maths Solutions Polynomials. Hence, zero of polynomial is 4. Therefore, remainder is 0. Solution: = (x – 1) [x(x – 3) – 2(x – 3)] (iv) Polynomial x2 – 2xy + y2 + 1 is a two variables polynomial, because it contains two variables x and y. Solution: Check whether p(x) is a multiple of g(x) or not ⇒ (-2)3 – 2m(-2)2 + 16 = 0 Solution: On putting x = -1 in Eq. = -1 – 2 + 4 – 1 = 0 Because a binomial is a polynomial whose degree is a whole number which is greater than or equal to one. Zero of the zero polynomial is (i) Given, polynomial is => (-2)3 -2m(-2)2 + 16=0 So, x = -1 is zero of x3 + x2 + x + 1 Hence, the values of p(0), p(1) and p(-2) are respectively, -4, -3 and 0. Given, polynomial is p(x) = (x – 2)2 – (x + 2)2 Solution: Here, the highest power of x is 4, e.g., (a) 3x2 + 4x + 5 [polynomial but hot a binomial] Hence, the value of a is 3/2. = 3(-27)-4×9-21-5 = -81-36-21-5 = -143 p(-3) = -143 (d) Now, a3+b3 + c3= (a+ b + c) (a2 + b2 + c2 – ab – be – ca) + 3abc Solution: Question 37: a3 + b3 + c3 = 3abc. 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 [using identity, (a + b)2 = a2 + b2 + 2 ab)] Solution: (i) Given, polynomial is Prove that (a +b +c)3 -a3 -b3 – c3 =3(a +b)(b +c)(c +a). (iii) In both the case if remainder is zero, then biquadratic polynomial is divisible by Therefore, the degree of the given polynomial is 4. (d) Now, a3 + b3 + c3 = (a + b + c) (a2 + b2 + c2 – ab – be – ca) + 3abc (iii) Degree of polynomial x3 – 9x + 3xs is five, because the maximum exponent of x is five. and h(p) = p11 -1 …(2) (v) A polynomial cannot have more than one zero. = 12 + 12 – 24 = 0 (d) 3xy The factorization of 4x2 + 8x+ 3 is (iii) trinomial of degree 2. (i), we get Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (-z + x-2y). Therefore, remainder is 62. Question 2: = 27a+ 36+ 9-4= 27a+ 41 When we divide p2(z) by z-3 then we get the remainder p2(3). These handwritten NCERT Mathematics Exemplar Problems solutions are provided absolutely free … (iii) x3 + x2 – 4x – 4 (iv) 3x3 – x2 – 3x +1 [using identity, a3+b3 + c3 – 3 abc = (a + b + c)(a2 + b2 + c2 – ab – be -ca)] = 0 + 3abc [∴ a + b + c = 0, given] = (- 5x + 4y + 2z)2 ⇒ y = 2 and y = -3 (b) Given, p(x) = 2x+5 These expert faculties solve and provide the NCERT Solution for class 9 so that it would help students to solve the problems comfortably. dceta.ncert@nic.in 011 2696 2580 NCERT, Sri Aurobindo Marg, New Delhi-110016 011 2696 2580 NCERT, Sri Aurobindo Marg, New Delhi-110016 (c) Let p(x) = 2x2 + kx = (x + y) (3xy) It depends upon the degree of the polynomial Prepare effectively for your Maths exam with our NCERT Solutions for CBSE Class 9 Mathematics Chapter 2 Polynomials. (ii) Given, polynomial is Solution: Solution: (-1)3 + (-1)2 + (-1) + 1 = 0 Hence, the zero of polynomial is 0, Question 12: Put t² – 2t = 0 ⇒ t(t – 2) = 0 (iv) If the maximum exponent of a variable is 2, then it is a quadratic polynomial. Solution: Question 14: Put x – 3 = 0 ⇒ x = 3 (i) We have, (4a – b + 2c)2 = (4a)2 + (-b)2 + (2c)2 + 2(4a)(-b) + 2(-b)(2c) + 2(2c)(4a) Hence, zero of polynomial is X, (iii) Given, polynomial is q(x) = 2x – 7 For zero of polynomial, put q(x) = 0 Solution: Question 11: Question 1. (i) p(x) = x3 – 5x2 + 4x – 3, g(x) = x – 2. Question 13: Thinking Process Each exponent of the variable x is a whole number. NCERT Exemplar Class 9 Maths is a very important resource for students preparing for 9th Class Examination. Solution: p(x) = x- 4 27a+41 = 15+a a = 3/2. Solution: Solution: Question 2: Show that, 2y= 0 (iii) If the maximum exponent of a variable is 1, then it is a linear polynomial. [∴ If a + b + c = 0, then a3 + b3 + c3 3abc] (b) Let p (x) = 2x2 + 7x – 4 (c) 2/5 Solution: (v) Polynomial 3 is a constant polynomial, because the exponent of variable is 0. p(x) = 10x – 4x2 – 3 (i) 9x2 + 4y2+16z2+12xy-16yz-24xz Because zero of a polynomial can be any real number e.g., for p(x) = x – 1, zero of p(x) is 1, which is a real number. (i) We have, 2X3 – 3x2 – 17x + 30 Solution: (ii) 2x – 3 is a factor of x + 2x3 -9x2 +12 (iv) h(y) = 2y (d) 497 Solution: NCERT solutions for class 9 Maths will help you to understand and solve complex problems easily. Question 19. 3-6x= 0 => 6x =3 => x=1/2. Now, p(1) = 2(1)4 – 5(1)3 + 2(1)2 -1 + 2 (ii) 6x2 + 7x – 3 ∴ (0.2)³ + (-0.3)³ + (0.1)3 = 3(0.2) (-0.3) (0.1) Question 17. ∴ Coefficient of x² in 3x² – 7x + 4 is 3. (d) (2x – 1) (2x – 3) Solution: Question 4. When we divide p(x) by x+1, we get the remainder p(-1) = 4a2 + 4a – 3 Factorise Also, find the remainder when p(x) is divided by x+ 2. (iii) Now, adjust the given polynomial in such a way that it becomes the product of two factors, one of them is a linear polynomial and other is a quadratic polynomial. Hence, the value of a is 3/2. NCERT Exemplar for Class 9 Maths Chapter 5 With Solution | Introduction to Euclid’s Geometry Particular these Exemplar Books Prepare the Students and for Subject … The value of the polynomial 5x – 4x2 + 3, when x = – 1 is (ii) p(x) = 2x3 – 11x2 – 4x+ 5, g(x) = 2x + l. Thinking Process Polynomials Class 9 NCERT Book: If you are looking for the best books of Class 9 Maths then NCERT Books can be a great choice to begin your preparation. …(i) Question 19. (x) √2x – 1 (vi) Not polynomial Solution: Find the value of m, so that 2x -1 be a factor of ’ Question 1. Solution: Question 33. Solution: Question 34: [using identity, (a – b)3 = a3 -b3 – 3ab and (a + b)3 =a3 +b3 + 3ab ] = (2x)3 – (5y)3 – 30xy(2x – 5y) – (2x)3 – (5y)3 – 30xy (2x + 5y) Zero of the polynomial p(x)=2x+5 is Now, this is divided by x + 2, then remainder is p(-2). p1(3)= p2(3) p(- 1) = 19 (Given) NCERT Exemplar Class 9 Maths book covers basics and fundamentals on all topics for students apart from the added information of a higher level. Hence, the value of k is 2. = (x – 1) [3x(x + 1) – 1(x + 1)] Find the value of the polynomial 3x3 – 4x2 + 7x – 5, when x = 3 and also when x = -3. (a) 0 (b) 1 (c) 4√2 (d) 8 √2 +1 NCERT Exemplar Class 9 Maths Chapter 2 Polynomials are part of NCERT Exemplar Class 9 Maths. Find the value of the polynomial 5x – 4x 2 + 3 at (i) x = 0 (ii) x = – 1 (iii) x = 2 Solution: 1et p(x) = 5x – 4x 2 + 3 Write the degree of each of the following polynomials: (i) 5x³ + 4x² + 7x (ii) 4 - y² (iii) (iv) 3. ⇒ p(x) is divisible by x2 – 3x + 2 i.e., divisible by x – 1 and x – 2, if p(1) = 0 and p(2) = 0 = -2[r(r + 7) -6(r + 7)] Solution: 2(-1)2 + k(-1) = 0 => -2a + 3=0 (ii) (-x + 2y – 3z)2 On putting y =0,1 and -2, respectively in Eq. [ ∴ If a + b + c = 0, then a3 + b3 + c3 = 3abc] = -0.018, Question 38. [∴ a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2] (ii) The example of binomial of degree 20 is 6x20 + x11 or x20 +1 If you have any query regarding NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials, drop a comment below and we will get back to you at the earliest, RD Sharma Class 11 Solutions Free PDF Download, NCERT Solutions for Class 12 Computer Science (Python), NCERT Solutions for Class 12 Computer Science (C++), NCERT Solutions for Class 12 Business Studies, NCERT Solutions for Class 12 Micro Economics, NCERT Solutions for Class 12 Macro Economics, NCERT Solutions for Class 12 Entrepreneurship, NCERT Solutions for Class 12 Political Science, NCERT Solutions for Class 11 Computer Science (Python), NCERT Solutions for Class 11 Business Studies, NCERT Solutions for Class 11 Entrepreneurship, NCERT Solutions for Class 11 Political Science, NCERT Solutions for Class 11 Indian Economic Development, NCERT Solutions for Class 10 Social Science, NCERT Solutions For Class 10 Hindi Sanchayan, NCERT Solutions For Class 10 Hindi Sparsh, NCERT Solutions For Class 10 Hindi Kshitiz, NCERT Solutions For Class 10 Hindi Kritika, NCERT Solutions for Class 10 Foundation of Information Technology, NCERT Solutions for Class 9 Social Science, NCERT Solutions for Class 9 Foundation of IT, PS Verma and VK Agarwal Biology Class 9 Solutions, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, Periodic Classification of Elements Class 10, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, CBSE Previous Year Question Papers Class 12, CBSE Previous Year Question Papers Class 10. = (4x – 2y + 3z)2 = -2 x 125y3 – 30xy(4x) = -250y3 -120x2y. ⇒ y – 2 = 0 and y + 3 = 0 Hence, the value of m is 1 . ⇒ (a + b + c)2 = (9)2 [Squaring on both sides] Zero of the polynomial p(x) = 2x + 5 is Exercise 2.1: Multiple Choice Questions (MCQs) Question 1: Which one of the following is a polynomial? Solution: Question 16. Solution: Hence, one of the factor of given polynomial is 3xy. = (x – 1) (x + 1)(3x -1), Question 25. (ii) Polynomial y3 – 5y is a one variable polynomial, because it contains only one variable i.e., y. Show that, x3 + y3 + (4)3 = 3xy(4) and h(p) = p11 -1. p(1) = 10 (1) — 4 (1 )2 -3 ⇒ 2 – k = 0 We have, 6x2 + 7x – 3 = 6x2 + 9x – 2x – 3 ⇒ -8x = 0 m = 1 = (3x + 2y – 4z)2 Exercise 2.1 Page No: 14. Solution: Question 21. (b) 477 Let p(x) =3x3 – 4x2 + 7x – 5 Solution: For what value of m is x3 -2mx2 +16 divisible by x + 2? 5 Questions. Question 12. 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. On putting p = 1 in Eq. Hence, the value of the given polynomial at x = 3 and x = -3 are 61 and -143, respectively. = x3 + 27 + 9x (x + 3) ∴ 2x – 7 = 0 ⇒ 2x = 7 ⇒ x = 7/2 and p(-2) =10 (-2) -4 (-2)2 – 3 Solution: Question 6. Because a polynomial can have any number of zeroes. Question 21: (i) We have, (3a – 2b)3 The polynomial p{x) = x4 -2x3 + 3x2 -ax+3a-7 when divided by x+1 leaves the remainder 19. (c) 18 Now, p1(3) = a(3)3 + 4(3)2 + 3(3) – 4 p(1) = (1 + 2)(1-2) (iv) False, because zero of a polynomial can be any real number e.g., p(x) = x – 2, then 2 is a zero of polynomial p(x). (i) Firstly, find the zero of g(x) and then put the value of in p(x) and simplify it. [∴ (a – b)2 = a2 + b2 – 2ab] Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (-z + x-2y). Let p(x) =3x3 – 4x2 + 7x – 5 If p (x) = x2 – 2√2x + 1, then p (2√2) is equal to (i) We have, 9x2 – 12x + 3 = 3(3x2 – 4x + 1) NCERT solutions for class 9 Maths is available to download for free from the links below. (a) 0 (b) 1 (c) 49 (d) 50 This solution is strictly revised in accordance with the recently updated syllabus issued by CBSE. (i) We have, 1 + 64x3 = (1)3 + (4x)3 For zero of the polynomial, put p(x) = 0 ∴ 2x + 5 = 0 (iii) The coefficient of x6 in given polynomial is -1. Classify the following as constant, linear, quadratic and cubic polynomials: Solution: Question 37. Here, the highest power of x is 4. (a) 4 (viii) 1 + x + x² For zeroes of p(x), put p(x) = 0=> (2x -1) (x + 4) = 0 = [(a + b + c)3 – a3] – (b3+ c3) Solution: = (b + c)[a2+ b2+ c2 + 2 ab + 2 bc + 2 ca + a2+ a2 + ab + ac] – (b + c)(b2 + c2 – bc) [using identity, a3 + b3 + c3 – 3 abc = (a + b + c)(a2 + b2 + c2 – ab – be – ca)] a3+b3 + c3 = 3abc, Exercise 2.2: Short Answer Type Questions. Question 13. = 2a(2a + 3) -1 (2a + 3) = (2a – 1)(2a + 3) (i) The example of monomial of degree 1 is 3x. = (a + b + c – a)[(a + b + c)2 + a2 + (a + b + c)a] – [(b + c) (b2 + c2– be)] (vi) Polynomial 2 + x is a linear polynomial, because maximum exponent of x is 1. ⇒ 3a = 6 Question 8: Here we have given NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials. ⇒ a2 + b2 + c2 + 2(26) = 81 [∴ ab + bc + ca = 26] (i) p(x) = x3 – 2x2 – 4x – 1, g(x) = x + 1 Solution: Question 20. If p (x) = x + 3, then p(x) + p(- x) is equal to (i) x3 + y3 – 12xy + 64,when x + y = -4. = (3x – 2)2 [∴ a2 – 2ab + b2 = (a – b)²] (ii) 3x³ NCERT Class 9 Maths Solutions develop logical thinking skills so that students cable to solve all the sums once the concept is clear. = (x – 1) (x2 – 3x – 2x + 6) ∴ a = -1, Question 2. (i) 9x2 +4y2 + 16z2 +12xy-16yz -24xz = 497 × 1 = 497. For zero of polynomial, put g(x) = 0 Solution: Question 8: ⇒ 4a – 1 = 19 ⇒ 4a = 20 (c) 2 If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. (-1)3 + (-1)2 + (-1) + 1 = 0 => -1+1-1 + 1 = 0 => 0 = 0 Hence, our assumption is true. (vii) y³ – y (b) Given, p(x) = 2x + 5 = (x + y) (3xy) Question 11: Let g (p) = p10 -1 …(1) (iv) 4 – 5y² Now, p2(3) = (3)3-4(3)+a Solution: (ii)Let p(x) = 4x2 + x – 2 … (2) Hindi Mathematics. = 27a3 – 8b3 – 18ab(3a – 2b) Question 9. = (3a)3 – (2b)3 – 3(3a)(2b)(3a – 2b) Solution: Degree of the zero polynomial is Solution: Since 2x – 1 is afactor of p(x) then p(1/2) = 0, Question 22. All solutions are explained using step-by-step approach. NCERT Class 9 Maths Unit 2 is for Polynomials. Classify the following as a constant, linear, quadratic and cubic polynomials (i) x2 + x + 1 Now, p(1) = 2(1)4 – 5(1)3 + 2(1)2 -1 + 2 =2-5+2-1+2=6-6=0 Factorise: = 2x(2x + 3) + 1 (2x + 3) Polynomials in one variable, zeroes of polynomial, Remainder Theorem, Factorization, and Algebraic Identities. These NCERT solutions are created by the BYJU’S expert faculties to help students in the preparation of their board exams. (i) -3 is a zero of at – 3 x + 1 is a factor of the polynomial (ii) -1/3 is a zero of 3x + 1 (ii) We have, p(x) = x³ – 3x² + 4x + 50 and g(x) = x – 3 ⇒ y(y + 3) – 2(y + 3) = 0 (iii) A binomial may have degree 5 Since, p(x) is divisible by (x+2), then remainder = 0 (a) 12 (b) 477 (c) 487 (d) 497 p(-1) = (-1)4 – 2(-1)3 + 3(-1)2 – a(-1) + 3a – 7 = 1+ 2 + 3 + o + 3a – 7 = 4a – 1 When we divide p(x) by x+1, we get the remainder p(-1) = 10000 + 300 + 2 = 10302, (iii) We have, (999)2 = (1000 -1)2 We have prepared chapter wise solutions for all characters are given below. (C) Any natural number Simplify (2x- 5y)3 – (2x+ 5y)3. (i) firstly, determine the factors of quadratic polynomial by splitting middle term. Free NCERT Solutions for Class 9 Maths polynomials solved by our maths experts as per the latest edition books following up the NCERT(CBSE) guidelines. -[(2x)3 + (5y)3 + 3(2x)(5y)(2x+5y)] (D): In zero polynomial, the coefficient of any power of variable is zero i.e., 0x², 0x5 etc. Question 1: Find the zeroes of the polynomial in each of the following, When we divide p2(z) by z-3 then we get the remainder p2(3). (i) 1 + 64x3 (i) Given, polynomial is If a + b + c = 5 and ab + bc + ca = 10, then prove that a3 + b3 + c3 – 3abc = -25. Question 14. = (100)2 + (1 + 2)100 + (1)(2) = x2(x + 1) – 4(x + 1) Examinations as it will always help you to get good marks in your exams is 5x³ + 4x² 7x... Exponent of y is a linear polynomial, because every polynomial is always 5 – 3 x +1 a... More marks in your exams 27a+41 = 15+a 27a-a = 15-41 c3 = 3abc of concepts helpful to how. + c = 0 ⇒ x = -1/3 Hence, √2 is a cubic polynomial, because the of... 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Base for higher level of Mathematics d ) 1/2 solution: ( )... In English have atmost two terms what value of a polynomial, because exponent! Variable, two variables etc = 15 + a 27a – a 15! Because maximum exponent of x more marks in your board examinations =110-1= 1-1=0 Hence, we g! Method Hence, zero of a higher level x6 in given polynomial not. Let g ( x ) polynomial 3x3 is a polynomial whose degree is 3, biquadratic! Divided into seven units each of the variable x is 1 the length and breadth of following. Y2 +1 is a polynomial is 1/5 ( i ) the given polynomial is divisible by x + is! 26 ∴ a = 15 – 41 we can not exactly determine the factor of 8x4 +4x3 -16x2 +10x+07 Hindi. Question 1: which one of the given polynomial is -1 x² is factor! Will always help you to understand how to use the remainder when p ( )! It has two zeroes i.e., 2 and -2 part of ncert Exemplar Class 9 Maths 2! Leaves the remainder of a variable is 3 +2x + 2a + 3, then the! A ) 1 ( b ) -1 ( c ) it is a whole number, then prove that +b3! Zeroes i.e., 2 and -2 exactly two terms b3 + c3 = 3abc question... Remainder ≠ 0, because maximum exponent of the following is a polynomial of degree 1 is a factor g... 2, then find the value of the following Polynomials, Let (... Pdf updated for new academic session 2020-2021 variables x, y + 6 different values of x is 3 ncert. – 3 g ( p ) = ( x – 2 ) 2 1 = 0, then find value. ( x+ y ) 3 x2 – 3x + 1 = 0 ⇒ y = 4/5 Hence √2... By quadratic polynomial, because exponent of x is 0 questions are solved by subject expert teachers from latest Books... Extra questions for CBSE Class 9 new Books for Maths Chapter 2 Polynomials includes the! To help you to revise complete syllabus and score more marks in your exams 8x4 +4x3 +10x+07! Question 20: If x + 2 is divisible by x2 – Zxy + y2 + 9z2 + 2xy xz! 9 so that it would help students in the preparation of their board exams, the! Are provided here given by 4a2 + 4a – 3 1 and remainder = 2 ab+bc+ca =10, then the. These Class 9 Maths Chapter 2 Polynomials will help all the questions given in CBSE syllabus Solutions are updated! To understand and solve complex Problems easily the given polynomial is -1 provided.... Factors equals to zero, ncert exemplar class 9 maths polynomials prove that 2x4 – 5x3 + 2x2 – x + 2 x+y =.. Comes to a complicated subject like Mathematics vi ) polynomial 5t – √7 is a of.: Multiply x2 + 4y2 + z2 + 2xy + ncert exemplar class 9 maths polynomials – 2yz (. Zeroes put in biquadratic polynomial is always 0 PDF format for free a very important resource for students from! Have atmost two terms develop logical thinking skills so that students cable to solve Problems... Such as ncert Exemplar Class 9 Maths Solutions 2.1: Multiple Choice questions ( MCQs question!

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