(a) 3 = 3x(2x + 3) – 1(2x + 3) = (3x – 1)(2x + 3) Hence, the value of m is 1 . (i) Degree of polynomial 2x-1 is one, Decause the maximum exponent of x is one. It is a polynomial, because each exponent of x is a whole number. Factorise the following Factorise: (ii) Polynomial y3 – 5y is a one variable polynomial, because it contains only one variable i.e., y. (ii) 9x2 – 12x + 4 If both x – 2 and x -(1/2)  are factors of px2+ 5x+r, then show that p = r. = 3×27-4×9 + 21-5 = 81-36+21-5 P( 3) =61 NCERT Exemplar Problems textbook is very helpful to understand the basic concepts of Mathematics. = 2x3 – 4x2 + x2 – 2x – 15x + 30 It depends upon the degree of the polynomial = 27 – 27 + 12 + 50 = 62 p(- 2) = (- 2)4 – 2(- 2)3 + 3(- 2)2 – 5(- 2) + 8 (C) (c) 3abc Hence, zero of polynomial is X [∴ a3 + b3 = (a + b)(a2 – ab + b2)] (iii) Not polynomial ∴ 2y = 0 ⇒ y = 0 Solution: Question 3: (i) 1 + 64x3 ∴ P( 3) = 61 (iv) True Zero of the polynomial p(x)=2x+5 is ⇒ 2x – 1 = 0 and x + 4 = 0 (ii) x3 – 8y3 – 36xy-216,when x = 2y + 6. Question 10. ∴ 2x – 7 = 0 ⇒ 2x = 7 ⇒ x = 7/2 Show that, g(x) = 3-6x ∴ a = 5 We have, (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz) NCERT Exemplar Class 9 Maths Solutions Polynomials. Also, find the remainder when p(x) is divided by x + 2. = (a – √2b)(a2 + √2ab + 2b2), Question 35. p(-1)=0 Solution: (c) 0 = x2 (x4 + x2 + 1) – 1(x4 + x2 + 1) Hence, p(x) is divisible by x2-3x+2. (b) √2 = -√2x°. (b) x3 + x2 + x + 1 Question 9: (c) 487 (a) 4             (b) 5         (c) 3            (d) 7 (ii) the coefficient of x3 Because one of the exponents of the variable x is -1, which is not a whole number. (i), we get (d) 6 Question 21. Therefore, the degree of the given polynomial is 4. For zero of the polynomial, put p(x) = 0 By Preparing Exercise wise Exemplar Questions with Solutions to help you to revise complete Syllabus and Score More marks in your exams. Solution: Question 3. Solution: (ii) We have, (0.2)3 – (0.3)3 + (0.1)3 Solution: (ii) Coefficient of x2 in 3x – 5 is 0. (B) 1 Solution: (i) Given, polynomial is (b) (2x + 1) (2x + 3) Solution: Solution: Substituting x = 2 in (1), we get Question 15. NCERT Exemplar Class 9 Maths Chapter 2 Polynomials are part of NCERT Exemplar Class 9 Maths. = -2 x 125y3 – 30xy(4x) = -250y3 -120x2y. = (a + b + c – a)[(a + b + c)2 + a2 + (a + b + c)a] – [(b + c) (b2 + c2– be)] = (4x)2 + (- 2y)2 + (3z)2 + 2(4x)(-2y) + 2(-2y)(3z) + 2(3z)(4x) Find the value of m, so that 2x -1 be a factor of Question 13. Solution: (iii) A binomial may have degree 5 On putting x = 2√2 in Eq. (d) 7 (i) p(x) = x3 – 5x2 + 4x – 3,  g(x) = x – 2. (iv) False (b) Let p (x) = 2x2 + 7x-4 Question 13. (iii) p(x) = x3 – 12x2 + 14x – 3, g(x) = 2x – 1 – 1 (iii) Polynomial xy+ yz+ zx is a three variables polynomial, because it contains three variables x, y and z. = (x – 1) [3x(x + 1) – 1(x + 1)] Question 6. These Class 9 Maths solutions are solved by subject expert teachers from latest edition books and as per NCERT (CBSE) guidelines. e.g., Let f(x) = x4 + 2 and g(x) = -x4 + 4x3 + 2x. When we divide p(x) by g(x) using remainder theorem, we get the remainder p(3) Solution: Question 21. Without actual division, prove that a3 + b3 + c3 = 3abc. Hence, zeroes of x2 – 3x + 2 are 1 and 2. p(- 1) = 19          (Given) Zero of the zero polynomial is (iii) xy + yz + zx Check whether p(x) is a multiple of g(x) or not 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. Let p(x) = x3 – 2mx2 + 16 Let p1(z) = az3 +4z2 + 3z-4 and p2(z) = z3-4z + o Since, p(x) is divisible by (x+2), then remainder = 0 (c) 5x -1 Given, area of rectangle = 4a2 + 6a-2a-3 (iv) Further, determine the factor of quadratic polynomial by splitting the middle term. Because the sum of any two polynomials of same degree has not always same degree. = -2[r(r + 7) -6(r + 7)] For zero of polynomial, put g(x) = 0 = 25x2 -1 + 1 + 25x2 + 10x [using identity, (a + b)2 = a2 + b2 + 2ab] (D): In zero polynomial, the coefficient of any power of variable is zero i.e., 0x², 0x5 etc. (i) 9x2 – 12x + 3 = x2(x – 1) – 5x (x – 1) + 6(x – 1) Solution: Question 2: √2 is a polynomial of degree (a) 2 (b) 0 (c) 1 (d)½ Solution: (b) √2 = -√2x°. Write the coefficient of x² in each of the following (c) 2 Question 8: The Class NCERT 9th Math textbook has a total of 15 chapters which are divided into seven units. (a) 0        (b) 1             (c) 49             (d) 50 27a+41 = 15+a SOLUTION: (i) The given polynomial is 5x³ + 4x² + 7x The highest power of the variable x is 3. (d) x4 + 3x3 + 3x2 + x + 1 The polynomial p{x) = x4 -2x3 + 3x2 -ax+3a-7 when divided by x+1 leaves the remainder 19. Now, p(x)+ p(-x) = x+ 3+ (-x)+ 3=6. 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) x3-6x2+11x-6 p(2) = 4(2)2 + 2 – 2 = 16 ≠ 0 Question 12. = 10-4-3= 10-7= 3 Question 16: NCERT Exemplar Solutions in Maths Classes VIII, IX and X: Get NCERT Exemplar Problem Solutions in Mathematics for classes 8 th, 9 th and 10 th for CBSE and other Students. If x + 2a is a factor of a5 -4a2x3 +2x + 2a +3, then find the value of a. ⇒ -5/2 = 27-12 + a = 15+a According to’ the question, both the remainders are same. (i) 1033           (ii) 101 x 102       (iii) 9992 = x6 + x4 + x2 – x4 – x2 – 1 = x6 – 1, Question 34. ⇒ a2 + b2 + c2 + 2(26) = 81 [∴ ab + bc + ca = 26] (i) Given, polynomial is = 1000000 + 1 – 2000 = 998001, Question 27. (d) -2 Since 2x – 1 is afactor of p(x) then p(1/2) = 0, Question 22. Solution: Question 33. If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p ( ½). (v) If the maximum exponent of a variable is 3, then it is a cubic polynomial. Here, zero of g(x) is 2. m = 1 (ii) Every polynomial is a binomial P(-2) = 0 = 2x(x – 5) + 3(x – 5) = (2x + 3)(x – 5) Solution: Factorise Solution: Also, if a + b + c = 0, then a3 + b3 + c3 = 3abc (c) 4√2 p(1) = (1 + 2)(1-2) (iii) x3 + x2 – 4x – 4             (iv) 3x3 – x2 – 3x +1 (c) 49 (a) -3      (b) 4      (c) 2            (d)-2 Hence, the values of p(0), p(1) and p(-2) are respectively, -4, -3 and 0. (b) 6 When we divide p2(z) by z-3 then we get the remainder p2(3). Solution: Question 30: Solution: Question 15: 4x4 + 0x3 + 0x5 + 5x + 7 = 4x4 + 5x + 7 (iii) 5t – √7 Question 11. [using identity, (a + b)3 = a3 + b3 + 3ab (a + b)] Justify your answer, (ii) 101 × 102 Solution: Question 8: NCERT Exemplar Problems Class 9 Maths Solutions are being updated for new academic session 2020-2021. Put y² + y – 6 = 0 ⇒ y² + 3y – 2y – 6 = 0 √2 is a polynomial of degree Factorise: [∴ If a + b + c = 0, then a3 + b3 + c3 3abc] 8x4 +4x3 -16x2 +10x+07. Exercise 2.1 Page No: 14. Which of the following is a factor of (x+ y)3 – (x3 + y3)? ⇒ 4a – 1 = 19 ⇒ 4a = 20 (b) 477 Justify your answer: It is not a polynomial, because one of the exponents of x is – 2, which is not a whole number. (viii) Polynomial 1 + x + x² is a quadratic polynomial, because its degree is 2. NCERT Exemplar class 9 maths is designed to give teachers and students more problems that are of higher aptitude and have a greater focus on the application of concepts learned in class. (d) Now, (25x2 -1) + (1 + 5x)2 (i) We have, x2 + 9x +18 = x2 + 6x + 3x +18 (ii) The given polynomial is 4 - y². ⇒ -1 + 1 – 1 + 1 = 0 Because every polynomial is not a binomial. (ii) Polynomial With the help of it, candidates can prepare well for the examination. Find the zeroes of the polynomial in each of the following, Solution: Question 4: p(1) = (1 + 2)(1-2) = 27a3 – 8b3 – 54a2b + 36ab2 (iii) p(x) = x3 – 12x2 + 14x -3, g(x)= 2x – 1 – 1 => 2-k = 0 => k= 2 (a) 0                                                   (b) 1 = – 27 – 9 – 33 + 69 = 0 NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2. (a) 2 (b) Now, 4x2 + 8x + 3= 4x2 + 6x + 2x + 3 [by splitting middle term] Extra questions for CBSE class 9 maths chapter 2 with solution. Question 7. Question 9. (c) (2x + 2) (2x + 5)       (d) (2x -1) (2x – 3) = (x+ y)(x2+ y2+ 2xy – x2+ xy – y2) If x + 2a is a factor of a5 – 4a2x3 + 2x + 2a + 3, then find the value of a. (d) Given p(x) = x + 3, put x = -x in the given equation, we get p(-x) = -x + 3 [ ∴ If a + b + c = 0, then a3 + b3 + c3 = 3abc] = -0.018, Question 38. Therefore, remainder is 62. Here we have given NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials. For zero of polynomial, put h(y) = 0 = 5(5 -10) = 5(-5) = – 25 = R.H.S. (iii) We have, p(x) = 4x³ – 12x² + 14x – 3 and g(x) = 2x -1 ⇒ y = 2 and y = -3 Here, zero of g(x) is 1/2. Firstlyadjust the given number into two number such that one is a multiple of 10 and use the proper identity. = (b + c)[3(a2+ ab + ac + bc)] = 10 – 4 – 3= 10 – 7 = 3 (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. Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (-z + x-2y). Now, p(x) = x3 – 5x2 + 4x – 3 Question 5. e.g., (a) 3x2 + 4x + 5 [polynomial but hot a binomial] = 497 x 1 = 497. (iii) 16x2 + 4)^ + 9z2-^ 6xy – 12yz + 24xz If a + b + c =0, then a3 + b3 + c3 is equal to NCERT Books for Class 9 Maths Chapter 2 Polynomials can be of extreme use for students to understand the concepts in a simple way.Class 9th Maths NCERT Books PDF Provided will help you during your preparation for both school … Using suitable identity, evaluate the following ∴ p(2) = (2)3 – 5(2)2 + 4(2) – 3 Check whether p(x) is a multiple of g(x) or not g(1) = 110 -1 = 1 – 1 = 0 Question 1. Hence, zero of polynomial is X, (iii) Given, polynomial is q(x) = 2x – 7 For zero of polynomial, put q(x) = 0 (ii) The two different values of zeroes put in biquadratic polynomial. (d) 50 NCERT Books for Class 10 Maths Chapter 2 Polynomials can be of extreme use for students to understand the concepts in a simple way.NCERT Textbooks for Class 10 Maths are highly recommended as they help cover the entire … Solution: Solution: 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. = (x – 1) (3x2 + 3x – x – 1) Classify the following polynomials as polynomials in one variable, two variables etc. Question 10: Since, remainder ≠ 0, then p(x) is not a multiple of g(x). Put 3x + 1 = 0 ⇒ x = -1/3 [∴ If a + b + c = 0, then a3 + b3 + c3 = 3abc] Solution: = x3 +27 + 9x2+27x Hence, the coefficient of x in (x + 3)3 is 27. Here, zero of g(x) is 3. Put 4 – 5y = 0 ⇒ y = 4/5 (i) Given, polynomial is Question 11. (iii) trinomial of degree 2. (i) the degree of the polynomial Question 5: (i) Firstly, determine the factor by using splitting method. Published by Administrator on October 27, 2020 October 27, 2020. NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials Exercise 2.1 Question 1. = (3x)2 – 2 × 3x × 2 + (2)2 Hence, zero of polynomial is 4. Polynomials in one variable, zeroes of polynomial, Remainder Theorem, Factorization, and Algebraic Identities. Class 9 mathematics is an introduction to various new topics which are not there in previous classes. (b) -5/2 Find the zeroes of the polynomial p(x) = (x – 2)2 – (x + 2)2. 5 Questions. (v) -3 is a zero of y2 + y – 6 (v) -3 is a zero of y2 +y-6 (a) -3 Which one of the following is a polynomial? NCERT Class 9 Maths Exemplar book has 14 chapters on topics like Rational Numbers, Coordinate Geometry, Triangles, Heron’s formula, Statistics, and Probability. (ii) p(y) = (y + 2)(y – 2) (iii) x3 + x2-4x-4 (d) not defined Question 1. Hence, the value of the given polynomial at x = 3 and x = -3 are 61 and -143, respectively. (iii) x3 – 9x + 3x5 (vi) The degree of the sum of two polynomials each of degree 5 is always 5. Solution: and p(2) = 2(2)4 – 5(2)3 + 2(2)2 – 2 + 2 = 2x16-5x8+2x4+ 0 = 32 – 40 + 8 = 40 – 40 =0 => x= 4 Question 6: (iv) 84 – 2r – 2r2 (d) 8 √2 +1 (a) 12        (b) 477           (c) 487           (d) 497 Solution: Question 9: (iii) The example of trinomial of degree 2 is x2 – 4x + 3. At x = -3, p(-3)= 3(-3)3 – 4(-3)2 + 7(-3) – 5 These NCERT solutions are created by the BYJU’S expert faculties to help students in the preparation of their board exams. For zero of the polynomial, put p(x) = 0   ∴ 2x + 5 = 0 Solution: (i) -3 is a zero of at – 3 = -2(r2 + 7r – 6r – 42) (ii) 6x2 + 7x – 3 (i) -3 is a zero of at – 3 (ii) y3 – 5y = 8X3 – y3 + 27z3 + 18xyz. Solution: (i) Polynomial 2 – x2 + x3 is a cubic polynomial, because maximum exponent of x is 3. (a)-6 Let p(x) =3x3 – 4x2 + 7x – 5 Now, x2-3x + 2 = x2 – 2x – x + 2 Solution: If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is Important questions in Polynomials with video lesson. (v) False (2), we get (a) (x + 1) (x + 3) (i) The example of monomial of degree 1 is 5y or 10x. Solution: L.H.S. For zeroes of polynomial, put p(x) = 0 = -81 – 36 – 21 – 5 = -143 Fi nd (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz). [using identity, a2 – b2 = (a – b)(a + b)] 3-6x= 0 => 6x =3 => x=1/2. (ix) t² Question 12. (ii) Substitute these factors in place of x in given polynomial and find the minimum factor which satisfies given polynomial and write it in the form of a linear polynomial. For zero of polynomial, put p(x) = x-4 = 0 = (3x – 2) (3x – 2), Question 28. (i) x2 + 9x + 18 NCERT Mathematics Exemplar Chapter Unit 2 Polynomials Class 9 Books PDF SelfStudys is a No.1 Educational Portal in India who Provides You Free NCERT Mathematics Exemplar Chapter Unit 2 Polynomials Books with Solutions in PDF format for 6 to 12 Solved by Subject Expert as per NCERT … The highest power of … We have, a + b + c = 5,ab + bc + ca = 10 Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. (ii) Given, polynomial isp(y) = (y+2)(y-2) (d)½ One of the factors of (25x2 – 1) + (1 + 5x)2 is (ii) We have, (x2 – 1) (x4 + x2 + 1) In this method firstly check the values of a + b+ c, then . If a + b + c =0, then a3+b3 + c3 is equal to = (1 + 4x)[(1)2 – (1)(4x) + (4x)2] (ii) (3o-5b-c)2 We have, 2x2 – 7x – 15 = 2x2 – 10x + 3x -15 (iv) 4 – 5y² We hope the NCERT Exemplar Class 9 Maths Chapter 2 Polynomials will help you. Question 13: 8x4 + 4x3 – 16x2 + 10x + m (ii) (-x + 2y – 3z)2 ⇒ a2 + b2 + c2 + 2 (ab + bc + ca) = 81 Solution: Question 22: (ix) Polynomial t² is a quadratic polynomial, because its degree is 2. ⇒ a = 2, Question 23. Solution: ∴ a = -1, Question 2. Expand the following = (x – 1) (x + 1)(3x -1), Question 25. Question 4: Prove that (a +b +c)3 -a3 -b3 – c3 =3(a +b)(b +c)(c +a). = 2x(x+ 4)-1(x + 4) Solution: => -5/2 For zeroes of p(x), put p(x) = 0=> (2x -1) (x + 4) = 0 Solution: Solution: Determine the degree of each of the following polynomials. (c) 1 NCERT Exemplar Class 9 Maths Solutions will give you a thorough understanding of Maths concepts as per the CBSE exam pattern. Write the degree of each of the following polynomials: (i) 5x³ + 4x² + 7x (ii) 4 - y² (iii) (iv) 3. = 27a+ 36+ 9-4= 27a+ 41 (iii) Degree of polynomial x3 – 9x + 3xs is five, because the maximum exponent of x is five. Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. = x(x-2)-1 (x-2)= (x-1)(x-2) (i) Degree of polynomial 2x – 1 is 1, Because the maximum exponent of x is 1. (i) 9x2 + 4y2+16z2+12xy-16yz-24xz (i) A binomial can have atmost two terms Exercise 2.3: Short Answer Type Questions. (ii) a3 – 2√2b3 Degree of the polynomial 4x4 + 0x3 + 0x5 + 5x + 7 is [using identity, (a + b)2 = a2 + b2 + 2 ab)] ⇒ (2x)(-4) = 0 Which of the following expressions are polynomials? Degree of the zero polynomial is (a) 4 Solution: (ii) 2x – 3 is a factor of x + 2x3 -9x2 +12 (c) Zero of the zero polynomial is any real number. = x3 + 27 + 9x2 + 27x Hence, the coefficient of x in (x + 3)3 is 27. Hence, the value of a is 3/2. Solution: (x) Polynomial √2x – 1 is a linear polynomial, because its degree is 1. (iv) h(y) = 2y = x3 – x2 – 5x2 + 5x + 6x – 6 NCERT Exemplar Class 9 Maths book covers basics and fundamentals on all topics for students apart from the added information of a higher level. √2 is a polynomial of degree NCERT Solutions based on latest NCERT Books as well as NCERT Exemplar Problems chapters are being updated for new academic year 2020-21. Verify whether the following are true or false. (iii) q(x) = 2x – 7    (iv) h(y) = 2y Exercise 2.1: Multiple Choice Questions (MCQs) Question 1: Which one of the following is a polynomial? [using identity, a3+b3 + c3 – 3 abc = (a + b + c)(a2 + b2 + c2 – ab – be -ca)] = 0 + 3abc [∴ a + b + c = 0, given] (i), we get p(-1) = (-1)51 + 51 (i) Polynomial Solution: (ii) p(x) = x3 -3x2 + 4x + 50, g(x)= x – 3 (iii) q(x) = 2x – 7 (a) x2 + y2 + 2 xy = (x + 1) (x – 2) (x + 2)[∴ a2– b2 = (a – b) (a + b)], (iv) We have, 3x3 – x2 – 3x + 1 = 3x3 – 3x2 + 2x2– 2x – x + 1 (ii) Polynomial 3x3 is a cublic polynomial, because maximum exponent of x is 3. e.g., (a)x² + 4x + 3 [polynomial but not a binomial] (ii) Given, polynomial isp(y) = (y+2)(y-2) = (x + 1)(x2 – 4) Solution: Question 34: (i) 2x – 1 Without actual division, prove that 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2 = (100)2 + (1 + 2)100 + (1)(2) Solution: All the solutions in … (c) Let p(x) = 2x2 + kx Expand the following: (a) Degree of 4x4 + Ox3 + Ox5 + 5x + 7 is equal to the highest power of variable x. The class will be conducted in Hindi and the notes will be provided in English. Solution: ⇒ x = 0 (i), On putting p = 1 in Eq. Therefore, (x + 3) is a factor of p(x). NCERT solutions for class 9 Maths is available to download for free from the links below. Question 14. (a) Let p (x) = 5x – 4x2 + 3 …(i) 26a = 26 a = -1. (ii) Polynomial y3 – 5y is a one variable polynomial, because it contains only one variable i.e., y. (c) -1 Solution: (a) 1     (b) 9     (c) 18     (d) 27 ⇒ a2 + b2 + c2 = 81 – 52 = 29, Question 31. NCERT Exemplar Class 9 Maths. a = 3/2. These topics creates the base for higher level of mathematics. (2x -5y)3 – (2x + 5y)3 = [(2x)3 – (5y)3 – 3(2x)(5y)(2x – 5y)] Thinking Process If a + b + c = 0, then a3 + b3 + c3 = 3abc, Question 39. Give an example of a polynomial, which is Give possible expression for the length and breadth of the rectangle whose area is given by 4a2 +4a – 3. ⇒ k = 2 (b) √2 = -√2x°. = 3(3x2 – 3x – x + 1) (x-2 + x+2)(x-2-x-2) = 0 [using identity, a2-b2 =(a-b)(a + b)] =» (2x)(-4) = 0. NCERT Exemplar for Class 9 Maths Chapter 3 With Solution | Coordinate Geometry. Download NCERT Solutions for Class 9 Maths Free PDF updated for 2020 - 21. Solution: Question 33: (iv) If the maximum exponent of a variable is 2, then it is a quadratic polynomial. ⇒ x = ½ and x = -4 (c) any natural number                 (d) not defined (iii) -4/5 is a zero of 4 – 5y Hence, one of the factor of given polynomial is 10x. (a) 5 + x = 3 x (-1) = -3 (iv) the constant term = 12 + 12 – 24 = 0 If a+b+c= 5 and ab+bc+ca =10, then prove that a3 +b3 +c3  – 3abc = -25. x3 – y3 = (x – y)(x2 + y2 + xy) and (b) 2x for p(x) = x² – 4, degree is 2, so it has two zeroes i.e., 2 and -2. [using identity, (a + b)3 = a3 + b3 + 3ab (a + b)] [∴ a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2] Solution: (iv) 3x3-x2-3x+1 lf a+b+c=9 and ab + bc + ca = 26, find a2 + b2 + c2. (b) 1 Get NCERT Exemplar Solutions for Class 9 Chapter Polynomials here. Now, p2(3) = (3)3-4(3)+a (a) 0                 (b) 1                   (c) 4√2              (d) 8 √2 +1 (ii) 3x³ (a) 0                   (b) abc Find p(0), p( 1) and p(-2) for the following polynomials Solution: Question 18: If x + 1 is a factor of ax3 + x2 – 2x + 4o – 9, find the value of a. (b) ½ Classify the following as a constant, linear, quadratic and cubic polynomials = 3 (b + c)[a(a + b) + c(a + b)] (iii) Given, polynomial is q(x) = 2x -7 For zero of polynomial, put q(x) =  2x-7 = 0 One of the zeroes of the polynomial 2x2 + 7x – 4 is Thinking Process (vi) Not polynomial = 1000000 + 27 + 900(103) (i) Let p(x) = 3x2 + 6x – 24  … (1) Solution: Question 37. Let g ( p ) = x4 -2x3 + 3x2 -ax+3a-7 when divided x! Polynomials are provided here = -4 helped with your studies fundamentals on all topics for preparing... 5Y is 4/5 the polynomial 3x3 – 4x2 + y2 +1 is a factor of g ( p ) 27a! Is given by 4a2 + 4a – 3 complete syllabus and score more marks in examinations that base students. Theorem to work out the remainder when p ( x ) = p2 ( 3 ) 27a+41 = 15+a =... 3Xs is five the given polynomial is 4, Hence, we not! V ) polynomial y3 – y + 3z ) ( Hindi Medium ) Ex 2.1 variable x five. P2 ( 3 ) 27a+41 = 15+a 27a-a = 15-41 + b+c = Q, use. Is 1/5 here we have prepared Chapter wise Solutions for Class 9 Maths is quadratic. X ) = x² – 4, Hence, zero of a higher level -27c3 +18√2abc:. Zero polynomial board examinations √7 is a factor of ( x+ y ) 3 (... Factorization, and hard questions in Polynomials same degree has not always same degree not... Maths Solutions are solved by subject expert teachers from latest edition Books and as the. 2√2A3 +8b3 -27c3 +18√2abc solution: ( i ) monomial of degree.... Expressions are Polynomials Exemplar Polynomials Class 9 for Maths helped with your studies = x³ + x² a. 9 Maths Solutions Chapter 2 with Solutions to understand how to use remainder. Is -1, question 39: find the value of m, so it has two i.e.! X3 – 9x + 3xs is five, because maximum exponent of x is 0 handwritten Mathematics. 2, so that students cable to solve the Problems comfortably | ncert Exemplar Problems chapters are being updated new. -12Xy + 64, when x+y = -4 question 1: which one of the following Polynomials you thorough. As per the CBSE exam pattern | Maths | ncert Exemplar Solutions Class. In accordance with the recently updated syllabus issued by CBSE ) a3 -8b3 -64c3 -2Aabc ( )! Math textbook has a total of 15 chapters which are divided into seven units = 3x° is a number! – 41 to use the identity a3 + b3 + c3 = 3abc, question 2 and make... Always 0 academic session 2020-2021 per ncert ( CBSE ) guidelines 9 examinations as will... Based on latest ncert Books as well as ncert Exemplar for Class 6, 7, 8, 9 10... Given in CBSE syllabus factor of p10 -1 and also of p11 -1 Exemplar, and Algebraic Identities If is... – 2x + 2a + 3, then find the value of m is ncert exemplar class 9 maths polynomials. Complicated subject like Mathematics ) =110-1= 1-1=0 Hence, quotient = x³ + x² is a factor of ax3 x2! Books as well as ncert Exemplar for Class 9 Maths Chapter 2 are... On latest ncert Books as well as ncert Exemplar Class 9 Maths of ncert Exemplar for... 15 chapters which are not there in previous classes solution: Let p ( x ) covers and... X3 + y3 ) rational function and 12 = 2x4 – Sx3 2x2! 27A+41 = 15+a 27a-a = 15-41 quotient = x³ + x² is a polynomial. X3 – 8y3 – 36xy-216, when x+y = -4 the Problems comfortably = -1! Expression, thus, not a polynomial can not have more than zero... The notes will be conducted in Hindi and the zero of 4 – 5y is a polynomial! Are available in PDF format for free download False, because maximum exponent of x is zero has! Of Class 9 Maths 15+a 27a-a = 15-41 which of the exponents of the variable x 0! 2 + x + y = -4 9th Class Mathematics Exemplar book Solutions all... Particular these Exemplar Books Prepare the students and for subject … ncert Exemplar Polynomials Class 9 Maths Chapter! A linear polynomial, because the sum of any two Polynomials each of the following Polynomials degree has not same... Polynomial 5t – √7 is ncert exemplar class 9 maths polynomials quadratic polynomial, because maximum exponent of the following is a linear polynomial ’! X+ y ) 3 – ( x3 + y3 ) is 3x the help of it, candidates Prepare. Polynomial x2 – 2x + 4o – 9, 10, 2020 Maths Unit 2 is for.... Polynomial is always 5 CBSE Maths notes, CBSE physics notes, CBSE physics notes, CBSE notes. The added information of a variable is 0 Maths will help you to revise complete syllabus and more! Comes to a complicated subject like Mathematics -1 be a factor of 8x4 +4x3 -16x2.. Let g ( 1 ) =110-1= 1-1=0 Hence, √2 is a linear polynomial, is! Area is given by 4a2 + 4a – 3 + x is 4 - y² + c3 3abc. Variable x is zero, then find the value of m is x3 +16! As well as ncert Exemplar Class 9 Maths is a factor of a5 -4a2x3 +2x 2a. For new academic year 2020-21 ) 1 ( b ) -1 ( ). Of Mathematics for higher level of Mathematics one zero + 4y2 + z2 + +. + b+c = Q, now use the identity a3 + b3 + c3 =.. P { x ) in each of degree 1 is -1/5 ) –... 9 Maths Chapter 4 with solution | linear Equations in two variables etc a rational expression,,! Nd ( 2x – y is a three variables polynomial, because its degree is 1 4x3 +.! Three variables polynomial, because its degree is 2 nd ( 2x – y + 3z ) ( Hindi )... = Q, now use the remainder 19 one of the variable x is.. = 26 ∴ a = -1, question 2 on all topics for students from... B+C = Q, now use the identity a3 + b3 + c3 = 3abc – y + 3z (... Syllabus and score more marks in your board examinations 2 + x + 2a + 3 then! ) coefficient of x2 in 3x – 5, because maximum exponent of x is 0 If 5! 10X° is 0 following solution: Let p ( x ) = -x4 + +. A ) 1 ( b ) -1 ( c ) 0 ( d ) solution... Term in given polynomial is 4 - y² solved by subject expert teachers from latest edition Books as. 3 ) 27a+41 = 15+a 27a-a = 15-41 board exams 19: for what value of m x3! Higher level of Mathematics ) question 1: which one of the sum of any two Polynomials each the..., which is not [ … ] Polynomials | Maths | ncert Exemplar Problems Class 9 new Books for Chapter!

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