a3 + b3 + c3 – 3 abc = (a + b + c)(a2 + b2 + c2 -ab-bc-ca) ⇒ a2 + b2 + c2 = 81 – 52 = 29, Question 31. (i) Degree of polynomial 2x-1 is one, Decause the maximum exponent of x is one. Question 16. Solution: These NCERT solutions are created by the BYJU’S expert faculties to help students in the preparation of their board exams. Write whether the following statements are true or false. Here, zero of g(x) is -1. (ii) p(x) = 2x3 – 11x2 – 4x + 5,  g(x) = 2x + 1 Give possible expression for the length and breadth of the rectangle whose area is given by 4a2 + 4a – 3. NCERT Exemplar for Class 9 Maths Chapter 2 Polynomials With Solution. = (x-1)(x-2)(x-3), (iii) We have, x3 + x2 – 4x – 4 (viii) Polynomial 1 + x+ x2 is a quadratic polynomial, because maximum exponent of xis 2. ∴ a = 5 Solution: Question 29: NCERT solutions for class 9 Maths will help you to understand and solve complex problems easily. (a) 0 By remainder theorem, find the remainder when p(x) is divided by g(x) = (x – 1)(3x2 + 2x – 1) Question 32. Substituting x = 2 in (1), we get Hence, 0 of x2-3x+2 are land 2. (d) Now, (x+ y)3 – (x3 + y3) = (x + y) – (x + y)(x2– xy + y2) (a) 3               (b) 2x            (c) 0                   (d) 6 ⇒ (a + b + c)2 = (9)2 [Squaring on both sides] (iv) the constant term ⇒ -8 – 8m + 16 = 0 (a) -2/5 Fi nd (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz). Solution: = 2x3 – 4x2 + x2 – 2x – 15x + 30 = 2x(x+ 4)-1(x+ 4) = (x + 1)(x2 – 4) Solution: Hence, zeroes of x2 – 3x + 2 are 1 and 2. (c) 2/5 It is not a polynomial, because one of the exponents of x is – 2, which is not a whole number. (-1)3 + (-1)2 + (-1) + 1 = 0 => -1+1-1 + 1 = 0 => 0 = 0 Hence, our assumption is true. If x +1 is a factor of ax3 +x2 -2x+4a-9, then find the value of a. Question 13: [∴ a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2] Find the zeroes of the polynomial in each of the following, (i) 2x – 1 (a) 0                   (b) abc (c) 3 = a3 + b3 + c3 – 3abc (d) Now, a3+b3 + c3= (a+ b + c) (a2 + b2 + c2 – ab – be – ca) + 3abc (iv) The constant term in given polynomial is 1/5. = (y-2)(y + 3) = 0 (ii) We have, (x2 – 1) (x4 + x2 + 1) Hence, one of the factor of given polynomial is 10x. Question 2. 2y= 0 ⇒ 4a – 1 = 19 ⇒ 4a = 20 Solution: (iii) x3 + x2 – 4x – 4             (iv) 3x3 – x2 – 3x +1 Question 21: It depends upon the degree of the polynomial. (iv) Given polynomial h(y) = 2 y For zero of polynomial, put h(y) = 0 2x= 7 => x =7/2 Solution: Question 18: (a)-6          (b) 6           (c) 2                    (d) -2 (a) -3 (viii) 1 + x + x² = 3[3x(x – 1) – 1(x – 1)] = 3(3x – 1)(x – 1), (ii) We have, 9x2 – 12x + 4 = x3 + 27 + 9x2 + 27x Hence, the coefficient of x in (x + 3)3 is 27. ⇒ x = ½ and x = -4 (a)-6 ⇒ t = 0 and t – 2 = 0 (c) 5x -1 Because exponent of the variable x is 1/2, which is not a whole number. At x = -3, p(-3)= 3(—3)3 – 4(-3)2 + 7(-3)- 5 Firstlyadjust the given number into two number such that one is a multiple of 10 and use the proper identity. ⇒ -2a + 3 = 0 Question 13. Since (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca), Solution: (iv) Zero of a polynomial is always 0 x³ + (-2y)3 + (-6)3 = 3x(-2y)(-6) Classify the following as a constant, linear, quadratic and cubic polynomials Solution: Question 32: then (5)2 = a2 + b2+ c2 + 2(10) Solution: Question 24: Given, area of rectangle = (Length) × (Breadth) Give an example of a polynomial, which is 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 (i) x + 3 is a factor of 69 + 11c – x2 + x3 (ii) Given, polynomial is ∴ Coefficient of x² in 3x² – 7x + 4 is 3. (ii) The example of binomial of degree 20 is 6x20 + x11 or x20 +1 Degree of the zero polynomial is Therefore, (x – 2y)³ + (2y – 3z)³ + (3z – x)³ = 3(x – 2y)(2y – 3z)(3z – x). Solution: Because 8 = 8x°, then exponent of the variable x is 0, which is a whole number. = x2 + 4y2 + 9z2 – 4xy – 12yz + 6xz, Question 29. NCERT Exemplar for Class 9 Maths Chapter 2 With Solution | Polynomials. Also, if a + b + c = 0, then a3 + b3 + c3 = 3abc = 1000027 + 92700 = 1092727, (ii) We have, 101 × 102 = (100 + 1) (100 + 2) Question 1: Question 9: Show that p-1 is a factor of p10 -1 and also of p11 -1. Because each exponent of the variable a is a whole number. Solution: (C) It is not […] 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. (iii) x3-9x+ 3x5        (iv) y3(1-y4) (d) 3xy Solution: (a) x2 + y2 + 2 xy      (b) x2 + y2 – xy         (c) xy2            (d) 3xy Factorise: Question 12: ⇒ (-2)3 – 2m(-2)2 + 16 = 0 Hence, the zeroes of t² – 2t are 0 and 2. Show that, Here, zero of g(x) is 2. (i) The degree of the polynomial is the highest power of the variable x i.e., 6. (b) √2 = -√2x°. = 27 – 27 + 12 + 50 = 62 (ii) 101 × 102 Solution: ⇒ x3 + y3 + 64 = 12xy Factorise the following: 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. = -1 – 2 + 4 – 1 = 0 a = 3/2. Solution: Question 2: p(1) = (1 + 2)(1-2) (i), we get If both x – 2 and x -(1/2)  are factors of px2+ 5x+r, then show that p = r. (d) Now, (25x2 -1) + (1 + 5x)2 Let p(x) = x3 -2mx2 +16 (ii) 6x2 + 7x – 3 Thinking Process On putting x = 2√2 in Eq. (c) (2x + 2) (2x + 5) = 1000000 + 1 – 2000 = 998001, Question 27. (Hi) True, because a binomial is a polynomial whose degree is a whole number greater than equal to one. 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. (a) x3 + x2 – x +1         (b) x3 + x2 + x+1 Solution: Question 23: Factorise ∴ p(-3) = -143 (d) 1/2 = (x – 1) (3x2 + 3x – x – 1) So, it may have degree 5. Hence, zero of x – 3 is 3. (v) -3 is a zero of y2 +y-6 On putting x = -1 in Eq. Solution: Question 17: Degree of the polynomial 4x4 + Ox3 + Ox5 + 5x+ 7 is (ii) the coefficient of x3 (ii) Polynomial y3 – 5y is a one variable polynomial, because it contains only one variable i.e., y. NCERT solutions for class 9 Maths is available to download for free from the links below. Question 5. (a) -3      (b) 4      (c) 2            (d)-2 = -2[r(r + 7) -6(r + 7)] = (a – √2b)(a2 + √2ab + 2b2), Question 35. a(- 1)3+ (- 1)2 – 2(-1) + 4a – 9 = 0 = 10 – 4 – 3= 10 – 7 = 3 3-6x= 0 => 6x =3 => x=1/2. (d) not defined m = 1 (ii) p(y) = (y + 2)(y – 2) All the solutions in … (i) monomial of degree 1. [∴ (a + b)3 = a3 + b3 + 3ab(a + b)] Therefore zero of the polynomial is p(x) is 0. (a) Let p (x) = 5x – 4x2 + 3 …(i) ⇒ -a + 1 + 2 + 4a – 9 = 0 = 8 – 20 + 8 – 3 = – 7 (vi) 2 + x (a) 0 (iii) A binomial ipay have degree 5. Question 19: p(x) = x4 – 2x3 + 3x2 – 5x + 3(5) – 7 Solution: We have, 2x2 – 7x – 15 = 2x2 – 10x + 3x -15 = 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) Question 12. 2(-1)2 + k(-1) = 0 = -2 (5y)3 – 30xy(2x – 5y + 2x + 5y) Solution: p(-1)=0 (iii) 5t – √7 (i) x2 + x +1           (ii) y3 – 5y Expand the following (i) x + 3 is a factor of 69 + 11x – x2 + x3 = x2(x – 1) – 5x (x – 1) + 6(x – 1) Question 20. Solution: Question 2: (b) 1 Solution: (iii) Not polynomial Hence, the value of k is 2. ⇒ a2 + b2 + c2 + 2ab + 2bc + 2ca = 81 Solution: The coefficient of x in the expansion of (x + 3)3 is 27a + 41 = 15 + a (i), we get p(0) = (0+2)(0-2) = -4 ∴ Sum of two polynomials, f(x) + g(x) = x5 + 2 + (-x5 + 2x2) = 2x2 + 2, which is not a polynomial of degree 5. Without actually calculating the cubes, find the value of 36xy-36xy = 0 (i) We have, g(x) = x – 2 (iv) Polynomial  x2 – Zxy + y2 +1 is a two variables pplynomial, because it contains two variables x and y. Find the value of m, so that 2x -1 be a factor of Question 11: 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. 2x4 – Sx3 + 2x2 – x + 2 is divisible by x2 – 3x + 2. (d) Now, 2492 – 2482 = (249 + 248) (249 – 248) [using identity, a2 – b2 = (a – b)(a + b)] Given, area of rectangle = 4a2 + 6a-2a-3 (i) x3 + y3 – 12xy + 64,when x + y = -4. 27a+41 = 15+a (ii) The two different values of zeroes put in biquadratic polynomial. (ii) False = 2x2(x – 2) + x(x – 2) – 15(x – 2) Solution: (b) Given, p(x) = x2 – 2√2x + 1 …(i) (i) Firstly, determine the factor by using splitting method. Solution: Question 40: [∴ a3 + b3 = (a + b)(a2 – ab + b2)] = (x – 1) [3x(x + 1) – 1(x + 1)] Determine the degree of each of the following polynomials. Question 2. (iii) 2x2 -7x.-15       (iv) 84-2r-2r2 = 81 – 36 + 21 – 5 Question 8. (iii) trinomial of degree 2. If x + 2a is a factor of a5 – 4a2x3 + 2x + 2a + 3, then find the value of a. -[(2x)3 + (5y)3 + 3(2x)(5y)(2x+5y)] Which one of the following is a polynomial? One of the zeroes of the polynomial 2x2 + 7x – 4 is Let p1(z) = az3 + 4z2 + 3z – 4 and p2(z) = z3 – 4z + o (a) 12        (b) 477           (c) 487           (d) 497 (iii) We have, (x – 1) (3x – 4) = 3x2 – 7x + 4 (b) 4 (ii) Further, put the factors equals to zero, then determine the values of x. Hence, p(x) is divisible by x2-3x+2. Exercise 2.3: Short Answer Type Questions. (a) 3 Question 10. 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 … = 2x (2x² – 3x + 1) – 5(2x² – 3x + 1) ⇒ -32a5 + 32a5 – 4a + 2a + 3 = 0 (b) 5 Solution: Since 2x – 1 is afactor of p(x) then p(1/2) = 0, Question 22. 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: Question 16: Check whether p(x) is a multiple of g(x) or not Classify the following as  constant, linear, quadratic and cubic polynomials: (ii) True (C) Any natural number = (3a)3 – (2b)3 – 3(3a)(2b)(3a – 2b) Hence, the value of k is 2. For zero of polynomial, put p(x) = x-4 = 0 Hence, the values of p(0), p(1) and p(-2) are respectively, -3,3 and – 39. NCERT 9 Mathematics Exemplar Problem Text book Solutions. 8x4 +4x3 -16x2 +10x+07. When we divide p1(z) by z – 3, then we get the remainder p,(3). Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. (a) 0                                                   (b) 1 (ii) a3 – 2√2b3 polynomial is divided by the second polynomial x4 + 1 and x – 1. 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. So, it is a cubic polynomial. [∴ (x + a)(x + b) = x2 + (a + b)x + ab] (iii) p(x) = x3 – 12x2 + 14x – 3, g(x) = 2x – 1 – 1 ⇒ 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 NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials Exercise 2.1 Question 1. All questions with solutions of polynomials will help all the students to revise complete syllabus and score more marks in examinations. Now, x2-3x+2 = x2-2x-x+2 [by splitting middle term] = 3×27-4×9 + 21-5 = 81-36+21-5 P( 3) =61 (d) -2 (ii) Given, polynomial isp(y) = (y+2)(y-2) (b) x3 + x2 + x + 1 8x4 + 4x3 – 16x2 + 10x + m At x= 3, p(3) = 3(3)3 – 4(3)2 + 7(3) – 5 ⇒ a2 + b2 + c2 = 5     … (i) = (100)3 + (3)3 + 3(100)(3)(100 + 3) Question 21. (vi) The degree of the sum of two polynomials each of degree 5 is always 5. Fi nd (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz). Put 4 – 5y = 0 ⇒ y = 4/5 Question 4: NCERT Exemplar for Class 9 Maths Chapter 5 With Solution | Introduction to Euclid’s Geometry Since, x + 2a is a factor of p(x), then put p(-2a) = 0 Zero of the zero polynomial is Let p1(z) = az3 +4z2 + 3z-4 and p2(z) = z3-4z + o (i) We have, 2X3 – 3x2 – 17x + 30 (iv) 3x3-x2-3x+1 ⇒ 0 = 0 Hence, our assumption is true. NCERT Exemplar for Class 9 Maths Chapter 3 With Solution | Coordinate Geometry. Question 1. Solution: Question 2: Solution: ⇒ t = 0 and t = 2 If a + b + c =0, then a3 + b3 + c3 is equal to Hence, the values of p(0), p(1) and p(-2) are respectively, -4, -3 and 0. (i) Given, polynomial is Download NCERT Solutions for Class 9 Maths Free PDF updated for 2020 - 21. On putting y =0,1 and -2, respectively in Eq. Factorise: (i) False, because a binomial has exactly two terms. Solution: Question 18. Again, putting p = 1 in Eq. It is a polynomial, because each exponent of x is a whole number. [using identity, (a + b)3 = a3 + b3 + 3ab (a + b)] Question 1. = 27a+ 36+ 9-4= 27a+ 41 (c) 4√2 and p( 2) = 2(2)4 – 5(2)3 + 2(2)2 -2 + 2 (i) Firstly, find the zero of g(x) and then put the value of in p(x) and simplify it. e.g., (a) 3x2 + 4x + 5 [polynomial but hot a binomial] (iv) p(x) = x3 – 6x2 + 2x – 4, g(x) = 1 – (3/2) x NCERT Class 9 New Books for Maths Chapter 2 Polynomials includes all the questions given in CBSE syllabus. If x + 2a is a factor of a5 -4a2x3 +2x + 2a +3, then find the value of a. (vii) y³ – y = (2x + 3) (2x + 1). Hence, one of the zeroes of the polynomial p(x) is ½. Hence, the zero of polynomial h(y) is 0. = x(x-2)-1 (x-2)= (x-1)(x-2) Solution: (iv) Given polynomial h(y) = 2 y = (b + c)[3(a2+ ab + ac + bc)] Solution: Here, zero of g(x) is 3. Factorise the following: ∴ (x – 2)2 – (x + 2)2 = 0 = x3 – x2 – 5x2 + 5x + 6x – 6 Factorise the following Question 1. (i) 2x3 -3x2 -17x + 30         (ii) x3 -6x2 +11 x-6 (c) (2x + 2) (2x + 5)       (d) (2x -1) (2x – 3) (i) Firstly, adjust the given number either in the farm 0f a3 + b3 or a3 -b3 (ii) Coefficient of x2 in 3x – 5 is 0. P(-2) = 0 (viii) Polynomial 1 + x + x² is a quadratic polynomial, because its degree is 2. Solution: Question 15: (a) 4             (b) 5         (c) 3            (d) 7 NCERT Class 9 Maths Exemplar book has 14 chapters on topics like Rational Numbers, Coordinate Geometry, Triangles, Heron’s formula, Statistics, and Probability. Now, p(x)+ p(-x) = x+ 3+ (-x)+ 3=6. (i) 9x2 -12x+ 3          (ii) 9x2 -12xy + 4 Put t² – 2t = 0 ⇒ t(t – 2) = 0 Question 1: (i), we get we get p(0) = 10(0) – 4(0)2 – 3 = 0 – 0 – 3 = -3 (d) Now, (25x2 -1) + (1 + 5x)2 (iv) Degree of polynomial y3(1 – y4) or y3 – y7 is 7, because the maximum exponent of y is 7. = 497 × 1 = 497. p(-1) = (-1)4 – 2(-1)3 + 3(-1)2 – a(-1) + 3a – 7 = 1+ 2 + 3 + o + 3a – 7 = 4a – 1 On putting x = 0, 1 and – 2, respectively in Eq. Factorise the following Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (-z + x-2y). Solution: (i) We have, 9x2 – 12x + 3 = 3(3x2 – 4x + 1) (ii) The example of binomial of degree 20 is 3x20 + x10 (iv) 0 and 2 are the zeroes of t2 – 2t The factorization of 4x2 + 8x+ 3 is (iv) h(y) = 2y (iii) x3 – 9x + 3x5 (ii) Polynomial 3x³ is a cubic polynomial, because its degree is 3. (c) any real number These topics creates the base for higher level of mathematics. Solution: =(-5x)2 + (4y)2 + (2z)2 + 2(-5x)(4y) + 2(4y)(2z) + 2(2z)(-5x) = 2x2 + 8x-x-4 [by splitting middle term] (i) The example of monomial of degree 1 is 3x. Question 3. 2a = 3 p(x) = 10x – 4x2 – 3 Solution: Solution: Question 33. Question 19. Solution: (i) Polynomial 2 – x2 + x3 is a cubic polynomial, because maximum exponent of x is 3. Solution: ⇒ -5/2 = -20-4×4-3 =-20-16-3=-39 ⇒ x = 0 Thinking Process x3 + y3 = (x + y)(x2 + y2 – xy) = 3x(2x + 3) – 1(2x + 3) = (3x – 1)(2x + 3) Question 4. = x6 + x4 + x2 – x4 – x2 – 1 = x6 – 1, Question 34. (i) p(x)=x – 4           (ii) g(x)= 3 – 6x These handwritten NCERT Mathematics Exemplar Problems solutions are provided absolutely free … By actual division, find the quotient and the remainder when the first (C) = (0.2)3 + (- 0.3)3 + (0.1)3 (i) -3 is a zero of at – 3 (i) 1 + 64x3 One of the factors of (25x2 – 1) + (1 + 5x)2 is Solution: (b) 6 NCERT Exemplar Class 9 Maths Solutions will give you a thorough understanding of Maths concepts as per the CBSE exam pattern. For zero of polynomial, put h(y) = 0 (a) 2              (b)½        (c)-1              (d)-2 Solution: Question 38: Justify your answer: for p(x) = x² – 4, degree is 2, so it has two zeroes i.e., 2 and -2. Using long division method Question 8. (i) We have, (4a – b + 2c)2 = (4a)2 + (-b)2 + (2c)2 + 2(4a)(-b) + 2(-b)(2c) + 2(2c)(4a) (v) Polynomial 3 is a constant polynomial, because the exponent of variable is 0. On putting x = 2√2 in Eq. Solution: (d) 2abc (b) 5 – x It is not a polynomial, because exponent of x is 1/2 which is not a whole number. (B) 1 On putting p=1 in Eq. Hence, the values of p(0),p(1) and p(-2) are respectively,-4,-3 and 0. Thinking Process 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. Simplify (2x- 5y)3 – (2x+ 5y)3. Solution: (c) 487 Question 16: (ii) 9x2 – 12x + 4 Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (-z + x-2y). If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is = -2 x 125y3 – 30xy(4x) = -250y3 -120x2y. (iii) The coefficient of x6 in given polynomial is -1. Solution: 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. Give an example of a polynomial, which is (i) (4a-b + 2c)2           (ii) (3a – 5b – c)2 Thinking Process (a) 0        (b) 1           (c) any real number               (d) not defined Solution: Question 3: (c) any natural number                 (d) not defined Question 5. (i) Let p(x) = 3x2 + 6x – 24  … (1) (v) Polynomial 3 = 3x° is a constant polynomial, because its degree is 0. ∴ a = -1, Question 2. These Class 9 Maths solutions are solved by subject expert teachers from latest edition books and as per NCERT (CBSE) guidelines. Question 13: The class will be conducted in Hindi and the notes will be provided in English. = (x – 2) (2x2 + x – 15) (i) 2x3 – 3x2 – 17x + 30 Hence, √2 is a polynomial of degree 0, because exponent of x is 0. Factorise the following (d) 50 (i) False (iii) In both the case if remainder is zero, then biquadratic polynomial is divisible by If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is