P(2√2) = (2√2)2– (2√2)(2√2) + 1 = 8 – 8 + 1 = 1, Question 6. Solution: Hence, the remainder is 50. NCERT Exemplar Polynomials Class 9 (Part - 2) Nov 10, 2020 • 1h . Justify your answer: = 3x2 (x – 1) + 2x(x – 1) -1(x – 1) Question 4: (iii) x3 + x2 – 4x – 4 (iv) 3x3 – x2 – 3x +1 = (y-2)(y + 3) = 0 Hence, one of the factor of given polynomial is 3xy. (iv) 84 – 2r – 2r2 Solution: (i) Since, x + y + 4 = 0, then (i) a3 -8b3 -64c3 -2Aabc Now, p(1) = 2(1)4 – 5(1)3 + 2(1)2 -1 + 2 Solution: = x2 (- z + x – 2y) + 4y2(- z + x – 2y) + z2(- z + x – 2y) + 2xy(- z + x – 2y) + xz(- z + x – 2y) – 2yz (- z + x – 2y) Let p(x) = 8x4 + 4x3 – 16x2 + 10x + m = 2x2(x – 2) + x(x – 2) – 15(x – 2) (a) 0 (b) 1 (c) 4√2 (d) 8 √2 +1 Show that, Here, zero of g(x) is 2. It is not a polynomial, because exponent of x is 1/2 which is not a whole number. (i) 9x2 -12x+ 3 (ii) 9x2 -12xy + 4 (x2 + 4y2 + z2 + 2xy + xz – 2yz)(- z + x – 2y) (a) 2 Solution: Let p(x) =3x3 – 4x2 + 7x – 5 Put 3x + 1 = 0 ⇒ x = -1/3 (iii) Not polynomial (a)-6 (b) 6 (c) 2 (d) -2 p(2) = 4(2)2 + 2 – 2 = 16 ≠ 0 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. Question 18. [using identity, (a + b)3 = a3 + b3 + 3ab (a + b)] (x) √2x – 1 (ii) 2x – 3 is a factor of x + 2x3 -9x2 +12 p(x) = x – 4 Solution: Solution: = x(x-2)-1 (x-2)= (x-1)(x-2) = 8 – 20 + 8 – 3 = – 7 Question 7. Solution: Question 4. Solution: Question 9: p(x) = x4 – 2x3 + 3x2 – 5x + 3(5) – 7 = (4x – 2y + 3z)(4x -2y + 3z), Question 30. Solution: (i), we get p(0) = 10(0)-4(0)2 -3 = 0-0-3= -3 Now, x2-3x + 2 = x2 – 2x – x + 2 Solution: Practice Polynomials questions and become a master of concepts. Solution: Question 2: Factorise (i) We have, NCERT Exemplar Problems Class 9 Maths Solutions are being updated for new academic session 2020-2021. Question 12: we get p(0) = 10(0) – 4(0)2 – 3 = 0 – 0 – 3 = -3 (b) Let assume (x + 1) is a factor of x3 + x2 + x+1. Question 9. ⇒ 3a = 6 Class 9 mathematics is an introduction to various new topics which are not there in previous classes. (ix) Polynomial t² is a quadratic polynomial, because its degree is 2. Let p(x) =3x3 – 4x2 + 7x – 5 Question 3: (i), we get Find the zeroes of the polynomial p(x)= (x – 2)2 – (x+ 2)2. = (2x + 3) (2x + 1). Solution: => 2x-1 = 0 and x+4 = 0 Zero of the polynomial p(x)=2x+5 is (iii) (-x + 2y-3z)2 ⇒ t = 0 and t = 2 (iv) Given polynomial h(y) = 2 y = 2x(x – 5) + 3(x – 5) = (2x + 3)(x – 5) (v) Polynomial 3 = 3x° is a constant polynomial, because its degree is 0. (i) We have, 1033 = (100 + 3)3 Here students are also provided with online learning materials such as NCERT Exemplar Class 9 Maths Solutions. Question 1. (i) The example of monomial of degree 1 is 5y or 10x. = – 27 – 9 – 33 + 69 = 0 = -1 – 2 + 4 – 1 = 0 x3 – y3 = (x – y)(x2 + y2 + xy) and 26a = 26 a = -1. Hence, the zero of polynomial is 0, Question 12: (iii) -4/5 is a zero of 4 – 5y 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. Important questions in Polynomials with video lesson. ⇒ 4a – 1 = 19 ⇒ 4a = 20 Simplify (2x- 5y)3 – (2x+ 5y)3. Question 2: (c) Zero of the zero polynomial is any real number. These Class 9 Maths solutions are solved by subject expert teachers from latest edition books and as per NCERT (CBSE) guidelines. Thinking Process Here, zero of g(x) is 3. Which one of the following is a polynomial? Question 15. Solution: g(1)=110-1= 1-1=0 Hence, p-1 is a factor of g(p). If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is Substituting x = 2 in (1), we get (i) 2x-1 (ii) -10 (i) monomial of degree 1. ⇒ -1 + 1 – 1 + 1 = 0 (iv) Polynomial x2 – 2xy + y2 + 1 is a two variables polynomial, because it contains two variables x and y. p1(3)= p2(3) Since, p(x) is divisible by (x+2), then remainder = 0 27a+41 = 15+a It depends upon the degree of the polynomial NCERT Exemplar Class 9 Maths Solutions Polynomials. Hence, the value of k is 2. (a) x2 + y2 + 2 xy (b) x2 + y2 – xy (c) xy2 (d) 3xy Classify the following polynomials as polynomials in one variable, two variables etc. = (a + b + c – a)[(a + b + c)2 + a2 + (a + b + c)a] – [(b + c) (b2 + c2– be)] (b) -5/2 Verify whether the following are true or false. Solution: ⇒ 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 lf a+b+c=9 and ab + bc + ca = 26, find a2 + b2 + c2. (b) x² + 5 [polynomial and also a binomial]. (iv) Polynomial x2 – Zxy + y2 +1 is a two variables pplynomial, because it contains two variables x and y. = 27-12 + a = 15+a According to’ the question, both the remainders are same. (d) -2 (i) Polynomial x2+ x+ 1 is a one variable polynomial, because it contains only one variable i.e., x. (i) p(x) = x3-2x2-4x-1, g(x)=x + 1 Solution: ⇒ x = ½ and x = -4 ⇒ x = 0 => -5/2 Thus, required polynomial, In this method firstly check the values of a + b+ c, then . [∴ (a – b)3 = a3 – b3 – 3ab(a – b)] Therefore zero of the polynomial is p(x) is 0. (d)½ Given, area of rectangle = 4a2 + 6a-2a-3 = (x+ y)(x2+ y2+ 2xy- x2+ xy- y2) (ii) False (ii) Degree of polynomial – 10 or – 10x° is 0, because the exponent of x is 0. Find the values of a. And to make that base strong students are advised to solve NCERT Exemplar class 9 Maths. (ii) Polynomial 3x³ is a cubic polynomial, because its degree is 3. (a) 4 = [(a + b + c)3 – a3] – (b3+ c3) Solution: Show that p-1 is a factor of p10 -1 and also of p11 -1. = 2x(x+ 4)-1(x + 4) Give an example of a polynomial, which is if a + b+c = Q, now use the identity a3 + b3 + c3 = 3abc. ∴ 3 – 6x = 0 ⇒ x = 3/6 = 1/2. Vivek Patriya. Solution: Solution: Question 23: Thinking Process ∴ p(-1) = (-1)³ – 2(-1)² – 4(-1) -1 ⇒ t = 0 and t – 2 = 0 Here, we see that (x-2y) +(2y-3z)+ (3z-x) = 0 p(- 2) = (- 2)4 – 2(- 2)3 + 3(- 2)2 – 5(- 2) + 8 = (3x + 2y- 4z) (3x + 2y – 4z), (ii)We have, 25X2 + 16y2 + 4z2 – 40xy + 16yz – 20xz Solution: Question 22: Therefore, (x + 3) is a factor of p(x). (D): In zero polynomial, the coefficient of any power of variable is zero i.e., 0x², 0x5 etc. Check whether p(x) is a multiple of g(x) or not We know that, = 2a(2a + 3) -1 (2a + 3) = (2a – 1)(2a + 3) [using identity, (a + b)2 = a2 + b2 + 2 ab)] (ii) 2√2a3 +8b3 -27c3 +18√2abc (i) We have, 1 + 64x3 = (1)3 + (4x)3 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 … Put 4 – 5y = 0 ⇒ y = 4/5 Find the value of the polynomial 3x3 – 4x2 + 7x – 5, when x = 3 and also when x = -3. (i) A Binomial can have atmost two terms. Question 8. (d) 5/2 (b) Now, 4x2 + 8x + 3= 4x2 + 6x + 2x + 3 [by splitting middle term] Expand the following: = x3 – 8y3 – z3 – 6xyz. polynomial is divided by the second polynomial x4 + 1 and x-1. (i) Firstly check the maximum exponent of the variable.. Solution: √2 is a polynomial of degree Question 17: (i) x + 3 is a factor of 69 + 11c – x2 + x3 (ii) p(y) = (y + 2)(y – 2) (i) a3 -8b3 -64c3 -2Aabc Again, putting p = 1 in Eq. Given, polynomial is p(x) = (x – 2)2 – (x+ 2)2 = 3 (b + c)[a(a + b) + c(a + b)] (-1)3 + (-1)2 + (-1) + 1 = 0 => -1+1-1 + 1 = 0 => 0 = 0 Hence, our assumption is true. -[(2x)3 + (5y)3 + 3(2x)(5y)(2x+5y)] (iii) trinomial of degree 2. Hence, one of the zeroes of the polynomial p(x) is ½. (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. Write the degree of each of the following polynomials: (i) 5x³ + 4x² + 7x (ii) 4 - y² (iii) (iv) 3. (iii) 2x2– 7x – 15 Hence, zero of polynomial is X, (iii) Given, polynomial is q(x) = 2x – 7 For zero of polynomial, put q(x) = 0 = 3×27-4×9 + 21-5 = 81-36+21-5 P( 3) =61 Find the value of m, so that 2x -1 be a factor of Let g (p) = p10 -1 …(1) (i), we get Given, polynomial is p(x) = (x – 2)2 – (x + 2)2 (ii) -10 The value of 2492 – 2482 is Practise the solutions to understand how to use the remainder theorem to work out the remainder of a polynomial. = (3x)2 – 2 × 3x × 2 + (2)2 => 2-k = 0 => k= 2 Question 5: Solution: Question 9: HOTS, exemplar, and hard questions in polynomials. (iii) x3-9x+ 3x5 (iv) y3(1-y4) Find the following products: When we divide p1(z) by z – 3, then we get the remainder p,(3). We have to prove that, 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2 i.e., to prove that p (1) =0 and p(2) =0 Question 1: ∴ (x – 2)2 – (x + 2)2 = 0 Which one of the following is a polynomial? = (x – 2) (2x2 + 6x – 5x – 15) It is not a polynomial, because one of the exponents of x is – 2, which is not a whole number. NCERT Exemplar for Class 9 Maths Chapter 2 Polynomials With Solution. (d) -2 Thinking Process (v) If the maximum exponent of a variable is 3, then it is a cubic polynomial. (d) Now, (x + 3)3 = x3 + 33 + 3x (3)(x + 3) P(-2) = 0 (vi) Polynomial 2 + x is a linear polynomial, because its degree is 1. = 2x2 + 8x – x – 4 [by splitting middle term] 2a = 3 Write whether the following statements are True or False. = -1 + 51 = 50 (vi) Polynomial 2 + x is a linear polynomial, because maximum exponent of x is 1. (d) not defined (i), we get Question 15. Solution: m = 1 Which of the following is a factor of (x+ y)3 – (x3 + y3)? (i) Firstly, find the zero of g(x) and then put the value of in p(x) and simplify it. (d) Given p(x) = x + 3, put x = -x in the given equation, we get p(-x) = -x + 3 Justify your answer. Solution: Question 37: Solution: If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. (iv) 0 and 2 are the zeroes of t2 – 2t One of the zeroes of the polynomial 2x2 + 7x – 4 is Zero of the polynomial p(x) = 2x + 5 is ⇒ -2a + 3 = 0 = (x – 2) (2x2 + x – 15) Degree of the zero polynomial is (i) We have, (3a – 2b)3 Solution: Question 5: (vii) Polynomial y3 – y is a cubic polynomial, because maximum exponent of y is 3. Solution: Question 18: Factorise: Here are all questions are solved with a full explanation and available for free to download. We have, a + b + c = 5,ab + bc + ca = 10 = 10000 + 300 + 2 = 10302, (iii) We have, (999)2 = (1000 -1)2 x + 1 is a factor of the polynomial (i) We have, p(x) = x³ – 2x² – 4x – 1 and g(x) = x + 1 = 1000000 + 27 + 900(103) Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. (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). = (x -1) (x2 – 5x + 6) 2y= 0 = (1 + 4x)[(1)2 – (1)(4x) + (4x)2] Solution: Question 28: Hence, 0 of x2-3x+2 are land 2. (b) √2 = -√2x°. Solution: Now, p1(3) = a(3)3 + 4(3)2 + 3(3) – 4 Solution: Hence, x – 2 is a factor of p(x). Hence, one of the factor of given polynomial is 10x. NCERT Exemplar Class 9 Maths Chapter 2 Polynomials are part of NCERT Exemplar Class 9 Maths. = (3x – 2)2 [∴ a2 – 2ab + b2 = (a – b)²] and p(-2) =10 (-2) -4 (-2)2 – 3 (ii) p(x) = x3 – 3x2 + 4x + 50, g(x) = x – 3 (iii) The example of trinomial of degree 2 is x2 – 4x + 3. (i), we get Factorise = 10 – 4 – 3= 10 – 7 = 3 (ii) g(x)= 3 – 6x (b) x3 + x2 + x + 1 At x = -3, p(-3)= 3(-3)3 – 4(-3)2 + 7(-3) – 5 Question 21. (i) 9x2 – 12x + 3 (2), we get Solution: [∴ (a + b)3 = a3 + b3 + 3ab(a + b)] = (4x – 2y + 3z)2 Solution: p(-1)=0 a3 + b3 + c3 = 3abc. (ii) Every polynomial is a binomial Solution: Students can download these NCERT Solutions of class 9 Maths PDFs for free. For zeroes of p(x), put p(x) = 0 (iii) 2x2 -7x.-15 (iv) 84-2r-2r2 ⇒ -8x = 0 (v) -3 is a zero of y2 +y-6 ⇒ a2 + b2 + c2 + 2(26) = 81 [∴ ab + bc + ca = 26] then (5)2 = a2 + b2+ c2 + 2(10) (a) -2/5 Question 1. (i) p(x) = x3 – 5x2 + 4x – 3, g(x) = x – 2. (iv) ‘p(x) = x3-6x2+2x-4, g(x) = 1 -(3/2) x (iv) Further, determine the factor of quadratic polynomial by splitting the middle term. (a) 0 The topic wise list for NCERT Exemplar Class 9 Maths is provided below. Fi nd (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz). Degree of the polynomial 4x4 + 0x3 + 0x5 + 5x + 7 is (iii) Polynomial xy + yz + zx is a three variables polynomial, because it contains three variables x, y and z. (a) -3 Since, (x + 1) is a factor of p(x), then All the solutions in … (ii), we get Hence, zero of the polynomial p(x) is -5/2. 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. (i) 2 – x² + x³ Hence, one of the factor of given polynomial is 10x. Solution: 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) Firstly, determine the factor by using splitting method. (vi) False, because the sum of any two polynomials of same degree is not always same degree. Question 17. (b) 1 When we divide p(x) by g(x) using remainder theorem, we get the remainder p(3) = 2x(2x + 3) + 1 (2x + 3) Therefore, a binomial may have degree 5. Since, x + 2a is a factor of p(x), then put p(-2a) = 0 = x2(x – 1) – 5x (x – 1) + 6(x – 1) (a) Let p (x) = 5x – 4x2 + 3 …(i) ⇒ (x – 2 + x + 2)(x – 2 – x – 2) = 0 (b) 1 Solution: Question 29: ⇒ x3 – 8y3 – 36xy – 216 = 0, Question 40. (i) We have, g(x) = x – 2 Prepare effectively for your Maths exam with our NCERT Solutions for CBSE Class 9 Mathematics Chapter 2 Polynomials. Factorise the following (a) 2 (b)½ (c)-1 (d)-2 Solution: = (x + y) (3xy) Solution: Question 9. (i) Firstly, adjust the given number either in the farm 0f a3 + b3 or a3 -b3 (i) -3 is a zero of at – 3 = 4x³ – 16x² + 17x – 5 = (a – √2b)[a2 + a( √2b) + (√2b)2] Solution: Using suitable identity, evaluate the following (iv) y3(1 – y4) a3+b3 + c3 = 3abc, Exercise 2.2: Short Answer Type Questions. The factorization of 4x2 + 8x+ 3 is (b) 0 (iii) Given, polynomial is q(x) = 2x -7 For zero of polynomial, put q(x) = 2x-7 = 0 (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx p(1) = (1 + 2)(1-2) (i) p(x)=x – 4 (ii) g(x)= 3 – 6x e.g., Let f(x) = x5 + 2 and g(x) = -x5 + 2x2 Solution: = (x – 1) [3x(x + 1) – 1(x + 1)] For zero of polynomial, put g(x) = 0 Which of the following expressions are polynomials? Solution: (ii) We have, (0.2)3 – (0.3)3 + (0.1)3 Solution: e.g., (a) 3x2 + 4x + 5 [polynomial but hot a binomial] The polynomial p{x) = x4 -2x3 + 3x2 -ax+3a-7 when divided by x+1 leaves the remainder 19. NCERT Class 9 Maths Exemplar book has 14 chapters on topics like Rational Numbers, Coordinate Geometry, Triangles, Heron’s formula, Statistics, and Probability. Exercise 2.1: Multiple Choice Questions (MCQs) Question 1: Which one of the following is a polynomial? On putting p = 1 in Eq. Solution: This chapter consists of problems based on polynomial in one variable, Zeroes of a Polynomial, Remainder Theorem, Factorisation of Polynomials and Algebraic Identities. Solution: (ii) -binomial of degree 20. = (- 5x + 4y + 2z)2 9x2 + 4y2 + 16z2 + 12xy – 16yz – 24xz Solution: p(1) = (1 + 2)(1-2) (a)-6 NCERT solutions for class 9 Maths is available to download for free from the links below. Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. (b) √2 = -√2x°. ∴ p(3) = (3)³ – 3(3)² + 4(3) + 50 Factorise 2x4 – Sx3 + 2x2 – x + 2 is divisible by x2 – 3x + 2. It is not a polynomial because it is a rational function. 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 … Because 8 = 8x°, then exponent of the variable x is 0, which is a whole number. Question 5. [∴ a3 + b3 = (a + b)(a2 – ab + b2)] e.g., Let us consider zero polynomial be 0(x-k), where k is a real number For determining the zero, putx-k = 0=>x = k Hence, zero of the zero polynomial be any real number. (B) 1 Let p(x) = x3 -2mx2 +16 (a) Degree of 4x4 + Ox3 + Ox5 + 5x + 7 is equal to the highest power of variable x. Hence, zero of 4 – 5y is 4/5. 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. Solution: = (2x-1)(x+ 4) = 2x2 + 8x-x-4 [by splitting middle term] Degree of the polynomial 4x4 + Ox3 + Ox5 + 5x+ 7 is (ii) True = 4x³ – 6x² + 2x – 10x² + 15x – 5 Expand the following p(x) = x- 4 Solution: Because the sum of any two polynomials of same degree has not always same degree. Question 16. Let p(x) = 2x4 – 5x3 + 2x2 – x + 2 Question 11. Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (-z + x-2y). (i) 1033 (iv) Polynomial 4 – 5y² is a quadratic polynomial, because its degree is 2. = (0.2)3 + (- 0.3)3 + (0.1)3 Solution: Solution: ⇒ -5/2 a = 3/2. ⇒ (2x)(-4) = 0 The coefficient of x in the expansion of (x + 3)3 is (v) Not polynomial 2x= 7 => x =7/2 (c) 18 ⇒ -32a5 + 32a5 – 4a + 2a + 3 = 0 Here we have given NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials. Which of the following is a factor of (x+ y)3 – (x3 + y3)? On putting y =0,1 and -2, respectively in Eq. 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. Solution: Question 32: On putting x = -1 in Eq. Thinking Process [∴ a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2] The value of the polynomial 5x – 4x2 + 3, when x = – 1 is (b) 4 (d) x4 + 3x3 + 3x2 + x + 1 (b) Let p (x) = 2x2 + 7x – 4 Solution: (i) p(x) = x3 – 2x2 – 4x – 1, g(x) = x + 1 (d) 6 Factorise the following = -1 + 51 = 50 (i), we get p(-1) = (-1)51 + 51 (ii) 2x – 3 is a factor of x + 2x3 -9x2 +12 Solution: 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. (C) Any natural number It is a polynomial, because each exponent of x is a whole number. (i) Polynomial x2 + x + 1 is a one variable polynomial, because it contains only one variable i.e., x. (a) 2 (b) 0 (c) 1 (d)½ When we divide p1(z) by z – 3, then we get the remainder p,(3). The value of the polynomial 5x – 4x2 + 3, when x = – 1 is Thinking Process (a) x3 + x2 – x + 1 (i) Given, polynomial is At x= 3, p(3) = 3(3)3 – 4(3)2 + 7(3) – 5 Hence, the zeroes of t² – 2t are 0 and 2. (b) 5 – x Solution: Question 2: = (1 + 4x)(1 – 4x + 16x2), (ii) We have, a3 – 2√2b3 = (a)3 – (√2b)3 (d)-2 If a + b + c =0, then a3 + b3 + c3 is equal to Solution: = a3 + b3 + c3 – 3abc = x(x + 6) + 3(x + 6) = (x + 3) (x + 6) ⇒ (2x -1) (x + 4) = 0 ∴ (0.2)³ + (-0.3)³ + (0.1)3 = 3(0.2) (-0.3) (0.1) Thinking Process p(- 1) = 19 (Given) = 27-12 + a = 15+a = (x-2)(x- 1) Question 39: NCERT Exemplar Solutions For Class 9 Maths. (ii) 25x2 + 16y2 + 4Z2 – 40xy +16yz – 20xz (a) (x +1) (x + 3) (b) (2x+1) (2x + 3) Now, p(1) = 2(1)4 – 5(1)3 + 2(1)2 -1 + 2 =2-5+2-1+2=6-6=0 (c) any real number Hence, we cannot exactly determine the degree of variable. 27a + 41 = 15 + a Solution: Question 36: If p (x) = x2 – 2√2x + 1, then p (2√2) is equal to g(1) = 110 -1 = 1 – 1 = 0 = 3[3x(x – 1) – 1(x – 1)] = 3(3x – 1)(x – 1), (ii) We have, 9x2 – 12x + 4 Hence, zero of polynomial is 4. (i) Let p(x) = 3x2 + 6x – 24 … (1) Hence, the value of a is 3/2. Solution: = 8X3 – y3 + 27z3 + 18xyz. Solution: Question 37. Because every polynomial is not a binomial. (-1)3 + (-1)2 + (-1) + 1 = 0 If x + 1 is a factor of ax3 + x2 – 2x + 4o – 9, find the value of a. We have prepared chapter wise solutions for all characters are given below. ∴ Sum of two polynomials, f(x) + g(x) = x5 + 2 + (-x5 + 2x2) = 2x2 + 2, which is not a polynomial of degree 5. Solution: Question 21. = (a – √2b)(a2 + √2ab + 2b2), Question 35. = (- 5x + 4y + 2z)(- 5x + 4y + 2z), (iii) We have, 16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz and p(-2) =10 (-2)- 4 (-2)2 – 3 (a) x2 + y2 + 2 xy ⇒ 8m = 8 Solution: (i) Polynomial 2 – x2 + x3 is a cubic polynomial, because maximum exponent of x is 3. Question 8: ⇒ 2x – 1 = 0 and x + 4 = 0 Question 14. Solution: Question 20: Factorise p(-1) = 5(-1) -4(-1)2 + 3= -5-4+3 = -6, Question 7: = 32 – 40 + 8 = 40 – 40 = 0 Without actual division, prove that 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2 (c) 1 Question 18: Question 3. Solution: Question 2: 2(-1)2 + k(-1) = 0 => -2a + 3=0 (ii) 25x2 + 16y2 + 4z2 – 40xy + 16yz – 20xz (C) We have, 2x2 – 7x – 15 = 2x2 – 10x + 3x -15 Using long division method (v) A polynomial cannot have more than one zero and p( 2) = 2(2)4 – 5(2)3 + 2(2)2 -2 + 2 (ii) (-x + 2y – 3z)2 = (100)2 + (1 + 2)100 + (1)(2) (b) -1 Solution: Question 34: = 497 x 1 = 497. = 3x(2x + 3) – 1(2x + 3) = (3x – 1)(2x + 3) p(-1) = (-1)4 – 2(-1)3 + 3(-1)2 – a(-1) + 3a – 7 = 1+ 2 + 3 + o + 3a – 7 = 4a – 1 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. Since, remainder ≠ 0, then p(x) is not a multiple of g(x). (a) -2/5 (b) -5/2 (c)2/5 (d)5/2 Solution: (c) 2 Solution: If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p ( ½). (ii) Coefficient of x2 in 3x – 5 is 0. = x3 + 27 + 9x (x + 3) (vi) Not polynomial (ix) t² Solution: Question 25: (c) 0 Solution: and h(p) = p11 -1 …(2) 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 factors of (25x2 – 1) + (1 + 5x)2 is (a) 1 In this class, Vivek Patriya will discuss NCERT Exemplar Polynomials Class 9 (Part - 2) .The class will be helpful for the aspirants of CBSE 9th. Since, x + 1 is a factor of p(x), then p(-1) = 0 One of the zeroes of the polynomial 2x2 + 7x – 4 is Question 13: Solution: (i) 1 + 64x3 = (x – 2)(x + 3)(2x – 5), (ii)We have, x3 – 6x2 + 11x – 6 (d) Now, (25x2 -1) + (1 + 5x)2 (ii) Polynomial y3 – 5y is a one variable polynomial, because it contains only one variable i.e., y. Solution: Question 24: = (x+ y)(x2+ y2+ 2xy – x2+ xy – y2) Also, find the remainder when p(x) is divided by x + 2. (i) False If a + b + c = 0, then a3 + b3 + c3 = 3abc, Question 39. ⇒ a = 2, Question 23. Solution: = 5(5 -10) = 5(-5) = – 25 = R.H.S. Solution: Solution: Therefore, the degree of the given polynomial is 4. = 3(-27)-4×9-21-5 = -81-36-21-5 = -143 p(-3) = -143 When we divide p2(z) by z-3 then we get the remainder p2(3). (iii) trinomial of degree 2. (v) False, because a polynomial can have any number of zeroes. [∴ If a + b + c = 0, then a3 + b3 + c3 3abc] (d) 27 = (x – 1) (3x2 + 3x – x – 1) Simplify (2x- 5y)3 – (2x+ 5y)3. [∴ a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2] According to the question, both the remainders are same. e.g., Let f(x) = x4 + 2 and g(x) = -x4 + 4x3 + 2x. Solution: Question 17: Since, (x + 1) is a factor of p(x), then Question 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. Using suitable identity, evaluate the following: Let g (p) = p10 -1 When we divide p(x) by x+1, we get the remainder p(-1) = 27a3 – 54a2b + 36ab2 – 8b3. Solution: (i) -3 is a zero of at – 3 Without actual division, prove that (iii) We have, p(x) = 4x³ – 12x² + 14x – 3 and g(x) = 2x -1 Solution: Question 6. 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 √2X-1 is a factor of g ( x ) -x4 + 4x3 + +. ) 27a + 41 = 15 – 41 is not a binomial has two... = x³ + x² + x is 0 that it would help students revise! बहुपद ) ( 4x2 + 7x – 5, when x + 2 wise. ) 0 ( d ) 1/2 solution: question 21: find the value of m so. All characters are given below 10 or – 10x° is 0 = x³ + x² is a factor of -1... ) -1 ( c ) 0 ( d ) 1/2 solution: ( ). 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