ncert exemplar class 9 maths polynomials

and p( 2) = 2(2)4 – 5(2)3 + 2(2)2 -2 + 2 Classify the following as a constant, linear, quadratic and cubic polynomials (d) Now, (x + 3)3 = x3 + 33 + 3x (3)(x + 3) Substituting x = 2 in (2), we get Solution: (a) Degree of 4x4 + Ox3 + Ox5 + 5x + 7 is equal to the highest power of variable x. (a)-6 = x3 + 27 + 9x (x + 3) (a) 1 Determine the degree of each of the following polynomials. (i) Given, polynomial is (c) Zero of the zero polynomial is any real number. (vii) Polynomial y3 – y is a cubic polynomial, because maximum exponent of y is 3. When we divide p1(z) by z – 3, then we get the remainder p,(3). Solution: Question 7. Question 1: (ii) x3 -8y3 -36xy-216,when x = 2y + 6. (c) Let p(x) = 2x2 + kx (b) x3 + x2 + x + 1 = 25x2 -1 + 1 + 25x2 + 10x [using identity, (a + b)2 = a2 + b2 + 2ab] Find the zeroes of the polynomial p(x) = (x – 2)2 – (x + 2)2. ⇒ x3 + y3 – 12xy + 64 = 0, (ii) Since, x – 2y – 6 = 0, then (ii) -1/3 is a zero of 3x+1 = (1 + 4x)[(1)2 – (1)(4x) + (4x)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). Question 15: Here students are also provided with online learning materials such as NCERT Exemplar Class 9 Maths Solutions. (i) False ∴ a = -1, Question 2. Hence, the zero of polynomial is 0, Question 12: Solution: Question 9: (c) 3abc             (d) 2abc Now, p(x)+ p(-x) = x+ 3+ (-x)+ 3=6. (i) a3 -8b3 -64c3 -2Aabc (i) x + 3 is a factor of 69 + 11x – x2 + x3 Show that p-1 is a factor of p10 -1 and also of p11 -1. (i) Firstly, find the prime factors of constant term in given polynomial. = 81 – 36 + 21 – 5 Here, zero of g(x) is 3. Which one of the following is a polynomial? Let p(x) = x3 – 2mx2 + 16 We have, 2x2 – 7x – 15 = 2x2 – 10x + 3x -15 Solution: Question 2: (i) (4o – b + 2c)2 (i) 9x2 – 12x + 3 (c) x4 + x3 + x2 + 1 Exercise 2.3: Short Answer Type Questions. Solution: Question 30: = (4x – 2y + 3z)(4x -2y + 3z), Question 30. (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)). = x2 + 4y2 + 9z2 – 4xy – 12yz + 6xz, Question 29. Solution: = 2x(4x2 + y2 + 9z2 + 2xy + 3yz – 6xz) – y(4x2 + y2 + 9z2 + 2xy + 3yz – 6xz) + 3z(4x2 + y2 + 9z2 + 2xy + 3yz – 6xz) Solution: (i) (4a-b + 2c)2           (ii) (3a – 5b – c)2 So, it may have degree 5. = 16 + 16 + 12 + 10 + 8 = 62. When we divide p(x) by g(x) using remainder theorem, we get the remainder p(3) (i) x3 +y3 -12xy + 64,when x+y = -4. (d) 1/2 (a) -2/5 The topic wise list for NCERT Exemplar Class 9 Maths is provided below. = (3x – 2)2                [∴ a2 – 2ab + b2 = (a – b)²] = 1000000 + 27 + 900(103) Question 1: (v) True Question 8. = -1 + 51 = 50 = (2x + 3) (2x + 1). 2a =3 (d) 497 Hindi Mathematics. (d) 8 √2 +1 = 3 (b + c)[a(a + b) + c(a + b)] (a) (x + 1) (x + 3) a = 3/2. NCERT Solutions Class 9 Maths Chapter 2 Polynomials are worked out by the experts of Vedantu to meet the long-standing demand of CBSE students preparing for Board and other competitive Exams. 26a = 26 = 3(a + b)(b + c)(c + a) = R.H.S. Expand the following: NCERT Exemplar Class 9 Maths. Using long division method = (4x – 2y + 3z)2 Solution: (iii) p(x) = x3 – 12x2 + 14x -3, g(x)= 2x – 1 – 1 Therefore, remainder is 0. Question 13. (iii) We have, (x – 1) (3x – 4) = 3x2 – 7x + 4 Question 21: e.g., (a)x² + 4x + 3 [polynomial but not a binomial] = 8X3 – y3 + 27z3 + 18xyz. (x – 2)2 – (x+ 2)2=0 Hence, zero of the zero polynomial be any real number. Give an example of a polynomial, which is Question 1. Which one of the following is a polynomial? Solution: Question 25: Solution: ⇒ k = 2 Question 17. ⇒ (x – 2 + x + 2)(x – 2 – x – 2) = 0 Question 8. Find p(0), p( 1) and p(-2) for the following polynomials (ii) Polynomial y3 – 5y is a one variable polynomial, because it contains only one variable i.e., y. (d) 6 CBSE Maths notes, CBSE physics notes, CBSE chemistry notes. (v) A polynomial cannot have more than one zero. Solution: Thinking Process (ii) Given, polynomial isp(y) = (y+2)(y-2) (a) 2 (i) Let p(x) = 3x2 + 6x – 24  … (1) Solution: = (1 + 4x)(1 – 4x + 16x2), (ii)  We have, a3 – 2√2b3 = (a)3 – (√2b)3 Hence, zero of the polynomial p(x) is -5/2. (i) p(x) = 10x – 4x2 – 3 Now, p(-3) = (-3)3 – (-3)2 + 11(-3) + 69 (d) 2abc (b) 3x2 + 5 [polynomial and also a binomial] (i) 1 + 64x3          (ii) a3 -2√2b3 Write the degree of each of the following polynomials: (i) 5x³ + 4x² + 7x (ii) 4 - y² (iii) (iv) 3. p(x) = x- 4 Vivek Patriya. (Hi) True, because a binomial is a polynomial whose degree is a whole number greater than equal to one. (iii) x3-9x+ 3x5        (iv) y3(1-y4) (a) 1 Solution: = (5) [5 – (ab + be + ca)]             [From (i)] quadratic polynomial. The Class NCERT 9th Math textbook has a total of 15 chapters which are divided into seven units. Hence, the value of a is 3/2. = (x – 1) (3x2 + 3x – x – 1) (i) 1 + 64x3 According to the question, both the remainders are same. One of the zeroes of the polynomial 2x2 + 7x – 4 is = 25x2 -1 + 1 + 25x2 + 10x [using identity, (a + b)2 = a2 + b2 + 2ab] = (b + c)[3(a2+ ab + ac + bc)] (iv) Further, determine the factor of quadratic polynomial by splitting the middle term. For zeroes of p(x), put p(x) = 0 Solution: = x(x-2)-1 (x-2)= (x-1)(x-2) = 4a2 + 4a – 3 [by splitting middle term] [using identity, a3+b3 + c3 – 3 abc = (a + b + c)(a2 + b2 + c2 – ab – be -ca)] = 0 + 3abc [∴ a + b + c = 0, given] If p (x) = x + 3, then p(x) + p(- x) is equal to (i) We have, Solution: Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (-z + x-2y). Hence, the value of the given polynomial at x = 3 and x = -3 are 61 and -143, respectively. 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 … ∴ Sum of two polynomials, f(x) + g(x) = x5 + 2 + (-x5 + 2x2) = 2x2 + 2, which is not a polynomial of degree 5. (i) monomial of degree 1. The value of the polynomial 5x – 4x2 + 3, when x = – 1 is On putting x = -1 in Eq. dceta.ncert@nic.in 011 2696 2580 NCERT, Sri Aurobindo Marg, New Delhi-110016 011 2696 2580 NCERT, Sri Aurobindo Marg, New Delhi-110016 Solution: Question 17: (d) The degree of zero polynomial is not defined, because in zero polynomial, the coefficient of any variable is zero i.e., Ox2 or Ox5,etc. Substituting x = 2 in (1), we get Question 3. Zero of the polynomial p(x)=2x+5 is (i), we get p(-1) = (-1)51 + 51 [∴ If a + b + c = 0, then a3 + b3 + c3 = 3abc] Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. e.g., p(x) = x2 -2, as degree pf p(x) is 2 ,so it has two degree, so it has two zeroes i.e., √2 and —√2. By remainder theorem, find the remainder when p(x) is divided by g(x) Solution: Question 9. NCERT Class 9 Maths Unit 2 is for Polynomials. (iii) x3 + x2-4x-4 (a) 0        (b) 1           (c) any real number               (d) not defined Show that, = (2x-1)(x+ 4) Question 9: Question 5: (d)-2 NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2. If a+b+c= 5 and ab+bc+ca =10, then prove that a3 +b3 +c3  – 3abc = -25. On putting p = 1 in Eq. (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. 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. = 27-12 + a = 15+a Solution: (iii) 5t – √7 The class will be conducted in Hindi and the notes will be provided in English. ∴ Coefficient of x² in 3x² – 7x + 4 is 3. Thinking Process (viii) 1 + x + x² = 4a2 + 4a – 3 Solution: Question 5: (b) Given, p(x) = 2x + 5 (ix) Polynomial t² is a quadratic polynomial, because its degree is 2. Question 2: (a) 5 + x               (b) 5 – x         (c) 5x -1           (d) 10x If a + b + c =0, then a3+b3 + c3 is equal to These NCERT solutions are created by the BYJU’S expert faculties to help students in the preparation of their board exams. (iv) Zero of a polynomial is always 0 5 Questions. (c) x4 + x3 + x2 +1      (d) x4 + 3x3 + 3x2 + x +1 (ii) p(y) = (y + 2)(y – 2) For zeroes of p(x), put p(x) = 0=> (2x -1) (x + 4) = 0 (d) Given p(x) = x+3, put x = -x in the given equation, we get p(-x) = -x+3 = 3 x (-1) = -3 ∴ p(2) = (2)3 – 5(2)2 + 4(2) – 3 x + 1 is a factor of the polynomial Solution: Question 18: -[(2x)3 + (5y)3 + 3(2x)(5y)(2x+5y)] (b) 5 We hope the NCERT Exemplar Class 9 Maths Chapter 2 Polynomials will help you. (iii) p(x) = x3 – 12x2 + 14x – 3, g(x) = 2x – 1 – 1 (a) 2 = 5(5 -10) = 5(-5) = – 25 = R.H.S. = 9a2 + 25b2 + c2 – 30ab + 10bc – 6ac, (iii) We have, (- x + 2y – 3z)2 = (- x)2 + (2y)2 + (-3z)2 + 2(-x)(2y) + 2(2y)(- 3z) + 2(- 3z)(- x) Question 2: (iii) Not polynomial (i) the degree of the polynomial Download NCERT Solutions for Class 9 Maths Free PDF updated for 2020 - 21. x + 3 is a factor of p(x) if p(-3) = 0 (i) We have, g(x) = x – 2 NCERT Exemplar for Class 9 Maths Chapter 5 With Solution | Introduction to Euclid’s Geometry Factorise the following: Solution: At x= 3, p(3) = 3(3)3 – 4(3)2 + 7(3) – 5 NCERT Exemplar Class 9 Maths book covers basics and fundamentals on all topics for students apart from the added information of a higher level. Question 17: = 10 – 4 – 3= 10 – 7 = 3 (ii) p(x) = 2x3 – 11x2 – 4x + 5,  g(x) = 2x + 1 If a + b + c = 5 and ab + bc + ca = 10, then prove that a3 + b3 + c3  – 3abc = -25. Fi nd (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz). NCERT Exemplar Class 9 for Maths Chapter 2 – Polynomials. (ii) The example of binomial of degree 20 is 3x20 + x10 Question 1. (iv) ‘p(x) = x3-6x2+2x-4, g(x) = 1 -(3/2) x polynomial is divided by the second polynomial x4 + 1 and x-1. Factorise Question 14: ∴ p(-3) = -143 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 … = 27a+ 36+ 9-4= 27a+ 41 When we divide p2(z) by z-3 then we get the remainder p2(3). Question 12. 2(-1)2 + k(-1) = 0 We hope that our NCERT Class 9 New Books for Maths helped with your studies! Now, p(x) + p(-x) = x + 3 + (-x) + 3 = 6. (i), we get (i) We have, 2X3 – 3x2 – 17x + 30 Hence, zero of 4 – 5y is 4/5. (c) Let p(x) = 2x2 + kx Expand the following Solution: (iv) True ⇒ y(y + 3) – 2(y + 3) = 0 = 3 x (-1) = -3 NCERT Exemplar for Class 9 Maths Chapter 2 Polynomials With Solution. Hence, zero of x – 3 is 3. (i) Firstly check the maximum exponent of the variable.. then (5)2 = a2 + b2+ c2 + 2(10) Question 4. (b) 0 ⇒ x3 – 8y3 – 216 = 36xy (-1)3 + (-1)2 + (-1) + 1 = 0 Now, p1(3) = a(3)3 + 4(3)2 + 3(3) – 4 (d)½ Solution: ∴ 2x – 7 = 0 ⇒ 2x = 7 ⇒ x = 7/2 Question 15. (iii) 16x2 + 4)^ + 9z2-^ 6xy – 12yz + 24xz e.g., Let f(x) = x4 + 2 and g(x) = -x4 + 4x3 + 2x. (c) 2/5 (i) p(x) = x3 – 2x2 – 4x – 1, g(x) = x + 1 (i) 2x – 1 NCERT 9th class Mathematics exemplar book solutions for chapter 2 Polynomials are available in PDF format for free download. = (3x)2 + (2y)2 + (-4z)2 + 2(3x)(2y) + 2(2y)(-4z) + 2(-4z)(3x) Hence, one of the factor of given polynomial is 10x. For zero of the polynomial, put p(x) = 0   ∴ 2x + 5 = 0 Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. (b) Given, p(x) = 2x+5 If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is Solution: Question 33: Solution: NCERT Exemplar ProblemsNCERT Exemplar MathsNCERT Exemplar Science, Kerala Syllabus 9th Standard Physics Solutions Guide, Kerala Syllabus 9th Standard Biology Solutions Guide, NCERT Solutions for class 9 Maths in Hindi, NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions InText Questions, NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area InText Questions, NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers InText Questions, NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities InText Questions, NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles InText Questions, NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties InText Questions, NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles InText Questions, NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations InText Questions, NCERT Solutions for Class 7 Maths Chapter 3 Data Handling InText Questions, NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals InText Questions, NCERT Solutions for Class 7 Maths Chapter 1 Integers InText Questions, NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4, NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3, NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2, NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5. When we divide p1(z) by z – 3, then we get the remainder p,(3). x³ + (-2y)3 + (-6)3 = 3x(-2y)(-6) Determine which of the following polynomial has x – 2 a factor Here, zero of g(x) is 1/2. (c) 0 ∴ x – 4 = 0 ⇒ x = 4 Solution: Question 9: (iv) We have, (2x – 5) (2x² – 3x + 1) = (x + 1)(x2 – 4) (ii) -10 = -1 + 51 = 50 (viii) Polynomial 1 + x+ x2 is a quadratic polynomial, because maximum exponent of xis 2. Find the values of a. p(x) = x – 4 Hence, the values of p(0), p(1) and p(-2) are respectively, -3, 3 and – 39. Question 21. If x51 + 51 is divided by x + 1, then the remainder is Question 11. => x = ½ and x = -4 ∴ 2y = 0 ⇒ y = 0 Since (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca), and p(-2) = (-2 + 2)(-2 -2) (ii) Polynomial (ii) Degree of polynomial – 10 or – 10x° is 0, because the exponent of x is 0. Solution: Question 37. Also, find the remainder when p(x) is divided by x + 2. Solution: Question 26: Solution: The value of 2492 – 2482 is Question 4: (v) Polynomial 3 is a constant polynomial, because the exponent of variable is 0. If a + b + c = 9 and ab + bc + ca = 26, find a2 + b2 +c2. Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (-z + x-2y). = 4a2 + 6a – 2a – 3 ⇒ -2a + 3 = 0 [ ∴ If a + b + c = 0, then a3 + b3 + c3 = 3abc] = -0.018, Question 38. Hence, zero of polynomial is Factorise iii) 16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz (iii) The example of trinomial of degree 2 is x2 – 5x+ 4 or 2x2 -x-1. Prepare effectively for your Maths exam with our NCERT Solutions for CBSE Class 9 Mathematics Chapter 2 Polynomials. Solution: ⇒ x = ½ and x = -4 ⇒ a2 + b2 + c2 + 2 (ab + bc + ca) = 81 (vi) False = 2 – 5 + 2 – 1 + 2 = 6 – 6 = 0 (ii) If the maximum exponent of a variable is 0, then it is a constant polynomial. ⇒ t = 0 and t = 2 Solution: Put 4 – 5y = 0 ⇒ y = 4/5 = (4x)2 + (- 2y)2 + (3z)2 + 2(4x)(-2y) + 2(-2y)(3z) + 2(3z)(4x) (ii) (3o-5b-c)2 Verify whether the following are true or false. If x + 2a is a factor of a5 – 4a2x3 + 2x + 2a + 3, then find the value of a. Determine which of the following polynomial has x – 2 a factor (i) 3x2 + 6x – 24 (ii) 4x2+ x – 2 ∴ Sum of two polynomials, (c) 0 = 8 – 20 + 8 – 3 = – 7 Solution: Question 24: Solution: (d) -2 Since, p(x) is divisible by (x+2), then remainder = 0 (c) 2 2a = 3 Solution: Question 2: √2 is a polynomial of degree (a) 2 (b) 0 (c) 1 (d)½ Solution: (b) √2 = -√2x°. NCERT Exemplar for Class 9 Maths Chapter 3 With Solution | Coordinate Geometry. [∴ a3 + b3 = (a + b)(a2 – ab + b2)] (i) 9x2 -12x+ 3          (ii) 9x2 -12xy + 4 By Preparing Exercise wise Exemplar Questions with Solutions to help you to revise complete Syllabus and Score More marks in your exams. (c) any real number Given, polynomial is p(x) = (x – 2)2 – (x + 2)2 ⇒ x = 0 NCERT 9th Class Exemplar Problems 2021 in Pdf format are Available this web page to Download, NCERT Exemplar Problems from Class 9 for the Academic year 2021 and Exemplar also Available to Maths Subject with the Answers, NCERT Developed the Exemplar Books 2021 for 9th Class Students Education Purpose. Solution: (a) x2 + y2 + 2 xy      (b) x2 + y2 – xy         (c) xy2            (d) 3xy = x3 +27 + 9x2+27x Hence, the coefficient of x in (x + 3)3 is 27. Hence, x – 2 is a factor of p(x). (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. Since, (x + 1) is a factor of p(x), then (c) Zero of the zero polynomial is any real number. (x) Polynomial √2x-1 is a linear polynomial, because maximum exponent of xis 1. (i), we get (i) 9x2 + 4y2+16z2+12xy-16yz-24xz Put x – 3 = 0 ⇒ x = 3 = (x – 2) (2x2 + x – 15) (b) Let assume (x + 1) is a factor of x3 + x2 + x+1. (iii) A binomial may have degree 5 Find the zeroes of the polynomial in each of the following, p(1) = 10 (1) – 4 (1 )2 – 3 Factorise Solution: (ix) t² Using suitable identity, evaluate the following: (iii) (-x + 2y-3z)2 Solution: Question 14: (c) 3 (i) Firstly, determine the factor by using splitting method. Without actually calculating the cubes, find the value of 36xy-36xy = 0 L.H.S. Solution: For zero of the polynomial, put p(x) = 0 [using identity, (a + b)2 = a2 + b2 + 2 ab)] All the solutions in … = (x – 1) (x2 – 3x – 2x + 6) (i), we get p(-1) = (-1)51 + 51 (i) Firstly, find the zero of g(x) and then put the value of in p(x) and simplify it. Hence, the zero of polynomial h(y) is 0. Thinking Process Solution: Question 37: For what value of m is x3 -2mx2 +16 divisible by x + 2? (-1)3 + (-1)2 + (-1) + 1 = 0 => -1+1-1 + 1 = 0 => 0 = 0 Hence, our assumption is true. (a) x3 + x2 – x + 1 (b) 2x Which of the following expressions are polynomials? ’ (i) x2 + x + 1 (b) 9 NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12. Therefore, the degree of the given polynomial is 4. Solution: Question 4. = 4x3 + 2x + 2 which is not a polynomial of degree 4. (v) -3 is a zero of y2 + y – 6 = 2x(x – 5) + 3(x – 5) = (2x + 3)(x – 5) (a) 0                                                   (b) 1 = (b + c)[3a2+3ab + 3ac + 3bc] Without actually calculating the cubes, find the value = (a – √2b)(a2 + √2ab + 2b2), Question 35. Hence, zero of polynomial q(x) is 7/2. (2x -5y)3 – (2x + 5y)3 = [(2x)3 – (5y)3 – 3(2x)(5y)(2x – 5y)] (i) Degree of polynomial 2x – 1 is 1, Because the maximum exponent of x is 1. This chapter consists of problems based on polynomial in one variable, Zeroes of a Polynomial, Remainder Theorem, Factorisation of Polynomials and Algebraic Identities. and p(2) = 2(2)4 – 5(2)3 + 2(2)2 – 2 + 2 = 2x16-5x8+2x4+ 0 = 32 – 40 + 8 = 40 – 40 =0 e.g. Solution: = (x – 1) [x(x – 3) – 2(x – 3)] (a) 5 + x (a) 2             (b) 0           (c) 1          (d)½ = 16a2 + b2 + 4c2 – 8ab – 4bc + 16ac, (ii)We have, (3a – 5b – cf = (3a)2 + (-5b² + (- c²) + 2(3a)(-5b) + 2(-5b)(-c) + 2(-c)(3a) (i) The example of monomial of degree 1 is 3x. Zero of the polynomial p(x) = 2x + 5 is [∴ a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2] lf a+b+c=9 and ab + bc + ca = 26, find a2 + b2 + c2. Factorise: ⇒ a2 + b2 + c2 = 5     … (i) You should get good marks in Class 9 examinations as it will always help you to get good rank in school. (d) not defined (ii) Every polynomial is a Binomial. Question 10: Solution: At x = -3, p(-3)= 3(—3)3 – 4(-3)2 + 7(-3)- 5 (ii) True Verify whether the following are true or false. (ii) Given, polynomial is Solution: (iii) 2x2 -7x.-15       (iv) 84-2r-2r2 Also, if a + b + c = 0, then a3 + b3 + c3 = 3abc = 2x2 + 8x – x – 4 [by splitting middle term] Exercise 2.1: Multiple Choice Questions (MCQs) Question 1: Which one of the following is a polynomial? Question 16. Class 9 mathematics is an introduction to various new topics which are not there in previous classes. Now, p1(3) = a(3)3 + 4(3)2 + 3(3) – 4 ⇒ a2 + b2 + c2 + 2ab + 2bc + 2ca = 81 (i) Given, polynomial is p(2) = 4(2)2 + 2 – 2 = 16 ≠ 0 Solution: (v) Polynomial 3 = 3x° is a constant polynomial, because its degree is 0. Degree of the zero polynomial is (iii) x3 – 9x + 3x5 (ii) 2x – 3 is a factor of x + 2x3 -9x2 +12 Question 1: Because a binomial is a polynomial whose degree is a whole number which is greater than or equal to one. Fi nd (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz). Hence, one of the factor of given polynomial is 10x. 3-6x= 0 => 6x =3 => x=1/2. (i) We have, (4a – b + 2c)2 = (4a)2 + (-b)2 + (2c)2 + 2(4a)(-b) + 2(-b)(2c) + 2(2c)(4a) = 2x2(x – 2) + x(x – 2) – 15(x – 2) Factorise Factorise the following: = x6 + x4 + x2 – x4 – x2 – 1 = x6 – 1, Question 34. (c) 49 = 2x(x+ 4)-1(x + 4) (i) Polynomial 2 – x² + x³ is a cubic polynomial, because its degree is 3. ⇒ a2 + b2 + c2 + 2(26) = 81 [∴ ab + bc + ca = 26] (iii) The coefficient of x6 in given polynomial is -1. At x = -3, p(-3)= 3(-3)3 – 4(-3)2 + 7(-3) – 5 (ii) 25x2 + 16y2 + 4z2 – 40xy + 16yz – 20xz x3 + y3 = (x + y)(x2 + y2 – xy) [∴ (x + a)(x + b) = x2 + (a + b)x + ab] = 3x(2x + 3) – 1(2x + 3) = (3x – 1)(2x + 3) Solution: (iii) trinomial of degree 2. ⇒ 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 (ii) 25x2 + 16y2 + 4Z2 – 40xy +16yz – 20xz = 1000027 + 92700 = 1092727, (ii) We have, 101 × 102 = (100 + 1) (100 + 2) Question 20: = 10-4-3= 10-7= 3 = -2(r2 + 7r – 6r – 42) and h(p) = p11 -1. Hence, √2 is a polynomial of degree 0, because exponent of x is 0. SOLUTION: (i) The given polynomial is 5x³ + 4x² + 7x The highest power of the variable x is 3. Let p(x) = 2x4 – 5x3 + 2x2 – x+ 2 firstly, factorise x2-3x+2. Hence, zero of polynomial is X, (iii) Given, polynomial is q(x) = 2x – 7 For zero of polynomial, put q(x) = 0 Solutions Class 9 Maths Chapter 2 Polynomials use the identity a3 + b3 + c3 3abc... 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