5 Questions. Therefore zero of the polynomial is p(x) is 0. ⇒ a2 + b2 + c2 = 81 – 52 = 29, Question 31. (iv) False (a) (x + 1) (x + 3) Exercise 2.1 Page No: 14. Since, remainder ≠ 0, then p(x) is not a multiple of g(x). Solution: Question 4: 27a-a = 15-41 . (iv) the constant term (iv) h(y) = 2y Factorise [∴ a3 – b3 = (a – b)(a2 + ab + b2)] (i) p(x)= x – 4 ∴ (x – 2)2 – (x + 2)2 = 0 (ii) -1/3 is a zero of 3x+1 p(2) = 3(2)2 + 6(2) – 24 Also, if a + b + c = 0, then a3 + b3 + c3 = 3abc (C) Any natural number Since, x + 2a is a factor of p(x), then put p(-2a) = 0 Factorise: So, it is a cubic polynomial. (a) x3 + x2 – x + 1 Question 5: (i) We have, x2 + 9x +18 = x2 + 6x + 3x +18 935k watch mins. It is not a polynomial because it is a rational function. Hence, one of the zeroes of the polynomial p(x) is ½. = (x – 1) (x2 – 3x – 2x + 6) = -1 + 51 = 50 Solution: ∴ x – 4 = 0 ⇒ x = 4 Here we have given NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials. Question 17: = (1000)2 + (1)2 – 2(1000)(1) (a) 0 e.g., Let f(x) = x4 + 2 and g(x) = -x4 + 4x3 + 2x. If a + b + c = 5 and ab + bc + ca = 10, then prove that a3 + b3 + c3 – 3abc = -25. Because every polynomial is not a binomial. (c) xy2 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 (d) -2 Let p(x) =3x3 – 4x2 + 7x – 5 Solution: These Class 9 Maths solutions are solved by subject expert teachers from latest edition books and as per NCERT (CBSE) guidelines. (iii) q(x) = 2x – 7 (iv) h(y) = 2y Solution: Question 27: = 2x(2x+ 3) + 1 (2x+ 3) Let p(x) = ax3 + x2 – 2x + 4a – 9 The Class NCERT 9th Math textbook has a total of 15 chapters which are divided into seven units. Factorise the following: Here, zero of g(x) is -1. (iv) Zero of a polynomial is always 0. Number Systems | NCERT Exemplar Solutions | Class 9 Exercise 1.1 Page No: 2 Write the correct answer in each of the following: 1. ⇒ -a + 1 + 2 + 4a – 9 = 0 Solution: When we divide p(x) by g(x) using remainder theorem, we get the remainder p(-1) Therefore, we can not exactly determine the highest power of variable, hence cannot define the degree of zero polynomial. Solution: (i) p(x) = 10x – 4x2 – 3 (ii) p(y) = (y + 2)(y – 2) Solution: and p( 2) = 2(2)4 – 5(2)3 + 2(2)2 -2 + 2 (i), we get p(-1) = (-1)51 + 51 ⇒ a2 + b2 + c2 = 5 … (i) (ii), we get Solution: Question 26: (ii) a3 – 2√2b3 (ii) Polynomial y3 – 5y is a one variable polynomial, because it contains only one variable i.e., y. Solution: = -1 – 2 + 4 – 1 = 0 (ii) Polynomial 3x³ is a cubic polynomial, because its degree is 3. (a) 3 (b) 2x (c) 0 (d) 6 (ii) The example of binomial of degree 20 is 3x20 + x10 NCERT Exemplar Solutions for class 9 Mathematics Polynomials. (d) Let p(x) = x51 + 51 . (b) 4 a3 + b3 + c3 = 3abc. Solution: (ii) Every polynomial is a binomial For what value of m is x3 -2mx2 +16 divisible by x + 2? 4x4 + 0x3 + 0x5 + 5x + 7 = 4x4 + 5x + 7 (iii) We have, p(x) = 4x³ – 12x² + 14x – 3 and g(x) = 2x -1 Solution: Since, remainder ≠ 0, then p(x) is not a multiple of g(x). Question 17. (i) Degree of polynomial 2x-1 is one, Decause the maximum exponent of x is one. If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. 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. Here we have given NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials. Solution: Question 13. = 25x2 -1 + 1 + 25x2 + 10x [using identity, (a + b)2 = a2 + b2 + 2ab] (iii) Degree of polynomial x3 – 9x + 3xs is five, because the maximum exponent of x is five. h (1) = (1)11 —1 = 1 —1 = 0 Hence, p -1 is a factor of h(p). (ii) The coefficient of x3 in given polynomial is 1/5. p(1) = 10 (1) – 4 (1 )2 – 3 Question 4. 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. Question 14: Because given expression is a rational expression, thus, not a polynomial. (a) 1 Class 9 mathematics is an introduction to various new topics which are not there in previous classes. (v) False, because a polynomial can have any number of zeroes. P(-2) = 0 [using identity, (a + b)3 = a3 + b3 + 3ab (a + b)] Solution: Question 38: ⇒ 2 – k = 0 Question 36. = (x – 1) [x(x – 3) – 2(x – 3)] 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. = (x+ y)(x2+ y2+ 2xy – x2+ xy – y2) (i), we get p(-1) = (-1)51 + 51 (iii) 2x2 -7x.-15 (iv) 84-2r-2r2 = (x -1) (x2 – 5x + 6) (i) 1033 (ii) 101 x 102 (iii) 9992 (c) Let p(x) = 2x2 + kx If x +1 is a factor of ax3 +x2 -2x+4a-9, then find the value of a. Question 18: (i) 2x-1 (ii) -10 p(- 1) = 19 (Given) ⇒ -5/2 x + 1 is a factor of the polynomial g(1) = 110 -1 = 1 – 1 = 0 (x2 + 4y2 + z2 + 2xy + xz – 2yz)(- z + x – 2y) Given, polynomial is p(x) = (x – 2)2 – (x + 2)2 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. (ii) We have, p(x) = x³ – 3x² + 4x + 50 and g(x) = x – 3 With the help of the NCERT Exemplar Class 9 Maths, candidates can understand the level and type of questions that are asked in the exam. (2x -5y)3 – (2x + 5y)3 = [(2x)3 – (5y)3 – 3(2x)(5y)(2x – 5y)] p(- 2) = (- 2)4 – 2(- 2)3 + 3(- 2)2 – 5(- 2) + 8 When we divide p(x) by g(x) using remainder theorem, we get the remainder p(3) ⇒ a2 + b2 + c2 + 2ab + 2bc + 2ca = 81 Question 5. Hence, the zeroes of y² + y – 6 are 2 and – 3. Solution: = 1000000 + 1 – 2000 = 998001, Question 27. Because a binomial is a polynomial whose degree is a whole number which is greater than or equal to one. ∴ Sum of two polynomials, The polynomial p{x) = x4 – 2x3 + 3x2 – ax + 3a – 7 when divided by x + 1 leaves the remainder 19. Solution: (i) 9x2 +4y2 + 16z2 +12xy-16yz -24xz (A) 0 (v) A polynomial cannot have more than one zero Vivek Patriya. (ii) The example of binomial of degree 20 is 6x20 + x11 or x20 +1 (a) 0 (b) 1 (c) 4√2 (d) 8 √2 +1 NCERT Class 9 Maths Exemplar book has 14 chapters on topics like Rational Numbers, Coordinate Geometry, Triangles, Heron’s formula, Statistics, and Probability. We have, p(x) = x4 – 2x3 + 3x2 – ox + 3a – 7 = x2(x – 1) – 5x (x – 1) + 6(x – 1) (i) -3 is a zero of at – 3 (ii) Every polynomial is a Binomial. (i) x2 + x +1 (ii) y3 – 5y Therefore, remainder is 62. Now, this is divided by x + 2, then remainder is p(-2). Hence, zero of the zero polynomial be any real number. (i) 2 – x² + x³ ∴ p(-3) = -143 Substituting x = 2 in (1), we get Because one of the exponents of the variable x is -1, which is not a whole number. Solution: Now, x2-3x+2 = x2-2x-x+2 [by splitting middle term] Classify the following polynomials as polynomials in one variable, two variables etc. ∴ p(2) = (2)3 – 5(2)2 + 4(2) – 3 = x2 + 4y2 + 9z2 – 4xy – 12yz + 6xz, Question 29. (c) 2/5 Solution: => 2x-1 = 0 and x+4 = 0 Write whether the following statements are true or false. (i) Firstly, adjust the given number either in the farm 0f a3 + b3 or a3 -b3 (ii) 2√2a3 +8b3 -27c3 +18√2abc Question 21. The coefficient of x in the expansion of (x + 3)3 is Write the degree of each of the following polynomials: (i) 5x³ + 4x² + 7x (ii) 4 - y² (iii) (iv) 3. g(x) = 3-6x (d) 7 Polynomials Class 10 NCERT Book If you are looking for the best books of Class 10 Maths then NCERT Books can be a great choice to begin your preparation. Here are all questions are solved with a full explanation and available for free to download. On putting x = -1 in Eq. (ii) p(x) = 2x3 – 11x2 – 4x+ 5, g(x) = 2x + l. Thinking Process (b) Let assume (x + 1) is a factor of x3 + x2 + x+1. h (1) = (1)11 -1 = 1 – 1 = 0 Find the value of (i) 1 + 64x3 (ii) a3 -2√2b3 (i), we get p(0) = (0+2)(0-2) = -4 Which of the following expressions are polynomials? e.g., (a)x² + 4x + 3 [polynomial but not a binomial] (i) We have, 2X3 – 3x2 – 17x + 30 Solution: ∴ p(-1) = (-1)³ – 2(-1)² – 4(-1) -1 g(x) = 3 – 6x 37 For zero of polynomial, put g(x) = 0 (i) 3x2 + 6x – 24 Zero of the polynomial p(x) = 2x + 5 is (iv) The constant term in given polynomial is 1/5. …(i) = x3 +27 + 9x2+27x Hence, the coefficient of x in (x + 3)3 is 27. (iii) Given, polynomial is q(x) = 2x -7 For zero of polynomial, put q(x) = 2x-7 = 0 (ii) 25x2 + 16y2 + 4Z2 – 40xy +16yz – 20xz The class will be conducted in Hindi and the notes will be provided in English. (c) any natural number (d) not defined (ii) Degree of polynomial -10 or -10x° is zero, because the exponent of x is zero. Hence, √2 is a polynomial of degree 0, because exponent of x is 0. 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. We have, (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz) At x = -3, p(-3)= 3(—3)3 – 4(-3)2 + 7(-3)- 5 Find the values of a. (i) False Which one of the following is a polynomial? Therefore, (x + 3) is a factor of p(x). (a) -3 Solution: x³ + (-2y)3 + (-6)3 = 3x(-2y)(-6) ⇒ 8m = 8 m = 1 Question 2. (i), Solution: (C) It is not […] Solution: = 3[3x(x – 1) – 1(x – 1)] = 3(3x – 1)(x – 1), (ii) We have, 9x2 – 12x + 4 = 50x2 + 10x = 10x (5x+ 1) NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2. = 8x³ + 2xy2 + 18xz2 + 4x2y + 6xyz – 12x2z – 4x2y – y3 – 9yz2 – 2xy2 – 3y2z + 6xyz + 12x2z + 3y2z + 27z3 + 6xyz + 9yz2 – 18xz2 = 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) = 497 x 1 = 497. Get NCERT Exemplar Solutions for Class 9 Chapter Polynomials here. Solution: The coefficient of x in the expansion of (x + 3)3 is Hence, the value of a is 3/2. (iv) True = (3x – 2) (3x – 2), Question 28. Verify whether the following are true or false. Show that, x3 + y3 + (4)3 = 3xy(4) (d) not defined (iii) The example of trinomial of degree 2 is x2 – 4x + 3. (i) a3 -8b3 -64c3 -2Aabc SOLUTION: (i) The given polynomial is 5x³ + 4x² + 7x The highest power of the variable x is 3. The value of 2492 – 2482 is Give possible expression for the length and breadth of the rectangle whose area is given by 4a2 + 4a – 3. = 50x2 + 10x = 10x (5x + 1) Hence, 0 of x2-3x+2 are land 2. Question 1: (a) 4 (a) Let p (x) = 5x – 4x2 + 3 …(i) For zeroes of p(x), put p(x) = 0=> (2x -1) (x + 4) = 0 Question 8. (i) firstly, determine the factors of quadratic polynomial by splitting middle term. Exercise 2.1: Multiple Choice Questions (MCQs). (a) 5 + x So, x = -1 is zero of x3 + x2 + x + 1 Hence, the value of k is 2. Solution: Question 8: (ix) t² One of the factors of (25x2 – 1) + (1 + 5x)2 is Hence, zero of polynomial is X, (iii) Given, polynomial is q(x) = 2x – 7 For zero of polynomial, put q(x) = 0 (ii) 2√2a3 +8b3 -27c3 +18√2abc (iii) 16x2 + 4)^ + 9z2-^ 6xy – 12yz + 24xz (c) 0 (c) 2 ⇒ 3a = 6 Let p(x) =a5 – 4a2x3 + 2x + 2a + 3 Solving Latest year 2021 Exemplar Problems Solutions for Class 9 Polynomials is the best option to understand the concepts given in NCERT books and do advanced level preparations for Class 9 exams. (iii) p(x) = x3 – 12x2 + 14x -3, g(x)= 2x – 1 – 1 (d) Now, (25x2 -1) + (1 + 5x)2 Polynomials in one variable, zeroes of polynomial, Remainder Theorem, Factorization, and Algebraic Identities. Classify the following polynomials as polynomials in one variable, two variables etc. Solution: Question 23: = (- 5x + 4y + 2z)(- 5x + 4y + 2z), (iii) We have, 16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz NCERT solutions for class 9 Maths is available to download for free from the links below. ⇒ k = 2 (c) Zero of the zero polynomial is any real number. Thus, required polynomial, (ii) If the maximum exponent of a variable is 0, then it is a constant polynomial. (D): In zero polynomial, the coefficient of any power of variable is zero i.e., 0x², 0x5 etc. Solution: Thinking Process On putting x = 0, 1 and – 2, respectively in Eq. Hence, zero of the polynomial p(x) is -5/2. polynomial is divided by the second polynomial x4 + 1 and x-1. (iii) Polynomial 5t – √7 is a linear polynomial, because its degree is 1. (a) -3 (b) 4 (c) 2 (d)-2 = (x + 1)(x2 – 4) Solution: (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. Hence, zero of polynomial is We know that For zero of the polynomial, put p(x) = 0 ∴ 2x + 5 = 0 Write the coefficient of x² in each of the following Given, polynomial is p(x) = (x – 2)2 – (x+ 2)2 (a) -2/5 (d) 1/2 (a) 2 Since, p(x) is divisible by (x+2), then remainder = 0 (c) x4 + x3 + x2 +1 (d) x4 + 3x3 + 3x2 + x +1 (ii) The given polynomial is 4 - y². Hence, zero of polynomial is 4. = (2x + 3) (2x + 1). = (x-1)(x-2)(x-3), (iii) We have, x3 + x2 – 4x – 4 Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. (i) monomial of degree 1. Because a binomial has exactly two terms. = (x + y) (3xy) (b) x3 + x2 + x + 1 = (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 On putting p=1 in Eq. = 25x2 -1 + 1 + 25x2 + 10x [using identity, (a + b)2 = a2 + b2 + 2ab] Solution: Question 22: = 4a2 + 4a – 3 [by splitting middle term] Substituting x = 2 in (2), we get Hence, x – 2 is not a factor of p(x). = 4a2 + 4a – 3 (d) 497 quadratic polynomial. (a) 1 (i) 2x – 1 Solution: = 2(6 – r)(r + 7) or 2(6 – r) (7 + r), Question 24. = (- 5x + 4y + 2z)2 (Hi) True, because a binomial is a polynomial whose degree is a whole number greater than equal to one. Question 10. (ii) p(y) = (y + 2)(y – 2) = 2x2(x – 2) + x(x – 2) – 15(x – 2) Solution: Question 37: Each exponent of the variable x is a whole number. Solution: [using identity, (a + b)2 = a2 + b2 + 2 ab)] Degree of the zero polynomial is NCERT Exemplar Class 9 for Maths Chapter 2 – Polynomials. Solution: So, the degree of the polynomial is 3. = x3 – x2 – 5x2 + 5x + 6x – 6 Solution: Question 2: (iv) x2 – Zxy + y2 + 1 Question 21: = (x + y) (3xy) Which one of the following is a polynomial? Coefficient of x² in 4x³ – 16x² + 17x – 5 is -16. (ii) 3x³ Solution: Solution: NCERT Exemplar Class 9 Maths Chapter 2 Polynomials are part of NCERT Exemplar Class 9 Maths. (c) Zero of the zero polynomial is any real number. Factorise the following: (a) Let p (x) = 5x – 4x2 + 3 …(i) NCERT Exemplar for Class 9 Maths Chapter 5 With Solution | Introduction to Euclid’s Geometry = x2 (x4 + x2 + 1) – 1(x4 + x2 + 1) The value of the polynomial 5x – 4x2 + 3, when x = – 1 is (a) 2 (b)½ (c)-1 (d)-2 L.H.S. ⇒ (-2)3 – 2m(-2)2 + 16 = 0 Solution: Question 6. Hence, the zero of polynomial h(y) is 0. (ii) 101 × 102 p(2) = 4(2)2 + 2 – 2 = 16 ≠ 0 Particular these Exemplar Books Prepare the Students and for Subject … (d) Now, (x + 3)3 = x3 + 33 + 3x (3)(x + 3) (ii) y3 – 5y Firstlyadjust the given number into two number such that one is a multiple of 10 and use the proper identity. Solution: = 27a+ 36+ 9-4= 27a+ 41 (i) Given, polynomial is = (x – 2)(x + 3)(2x – 5), (ii)We have, x3 – 6x2 + 11x – 6 Question 13: = 1000027 + 92700 = 1092727, (ii) We have, 101 × 102 = (100 + 1) (100 + 2) At x= 3, p(3) = 3(3)3 – 4(3)2 + 7(3) – 5 (a) Degree of 4x4 + Ox3 + Ox5 + 5x + 7 is equal to the highest power of variable x. (i) We have, 1 + 64x3 = (1)3 + (4x)3 for p(x) = x² – 4, degree is 2, so it has two zeroes i.e., 2 and -2. (i) p(x)=x – 4 (ii) g(x)= 3 – 6x Solution: Question 24: Question 1: ⇒ a = 2, Question 23. (i) False (iii) The example of trinomial of degree 2 is x2 – 5x+ 4 or 2x2 -x-1. Put t² – 2t = 0 ⇒ t(t – 2) = 0 (i) -3 is a zero of at – 3 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 … (i) Since, x + y + 4 = 0, then ’ (x – 2)2 – (x+ 2)2=0 These handwritten NCERT Mathematics Exemplar Problems solutions are provided absolutely free … Question 5. => x= 4 Question 19. ⇒ t = 0 and t = 2 If a + b + c =0, then a3 + b3 + c3 is equal to Solution: (a) 4 = (4x – 2y + 3z)2 Solution: Classify the following as a constant, linear, quadratic and cubic polynomials Hence, the values of p(0), p(1) and p(-2) are respectively, -3, 3 and – 39. (c) 0 ⇒ -8 – 8m + 16 = 0 Solution: Question 25: It is a polynomial, because each exponent of x is a whole number. By Preparing Exercise wise Exemplar Questions with Solutions to help you to revise complete Syllabus and Score More marks in your exams. (ii) True (i) p(x) = x3 – 5x2 + 4x – 3, g(x) = x – 2. = 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) (ii)Let p(x) = 4x2 + x – 2 … (2) Hence, we cannot exactly determine the degree of variable.

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