[∴ a3 + b3 = (a + b)(a2 – ab + b2)] (b) √2 = -√2x°. (c) -1 ⇒ y = 2 and y = -3 (i) We have, 9x2 – 12x + 3 = 3(3x2 – 4x + 1) (d) 27 The factorization of 4x2 + 8x+ 3 is Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (-z + x-2y). (viii) 1 + x + x² Solution: Write the coefficient of x² in each of the following e.g., Let f(x) = x5 + 2 and g(x) = -x5 + 2x2 Hence, zero of polynomial is X (b) 477 = (3x – 2)2 [∴ a2 – 2ab + b2 = (a – b)²] Because given expression is a rational expression, thus, not a polynomial. When we divide p(x) by g(x) using remainder theorem, we get the remainder p(-1) (a) 4 = x3 + 27 + 9x (x + 3) (B) 1 ⇒ 8m = 8 Solution: Question 2: The polynomial p{x) = x4 – 2x3 + 3x2 – ax + 3a – 7 when divided by x + 1 leaves the remainder 19. = 4x3 + 2x + 2 which is not a polynomial of degree 4. ⇒ -1 + 1 – 1 + 1 = 0 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 33: [using identity, a3 + b3 = (a + b)(a2 – ab + b2)] = (x + y)[(x + y)2 -(x2 – xy + y2)] Since, (x + 1) is a factor of p(x), then = (100)2 + (1 + 2)100 + (1)(2) (D): In zero polynomial, the coefficient of any power of variable is zero i.e., 0x², 0x5 etc. If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is 2a =3 Now, p(x) + p(-x) = x + 3 + (-x) + 3 = 6. => x= 4 = 8X3 – y3 + 27z3 + 18xyz. = (x+ y)(x2+ y2+ 2xy – x2+ xy – y2) Since, p(x) is divisible by (x+2), then remainder = 0 p1(3)= p2(3) Now, x2-3x + 2 = x2 – 2x – x + 2 Zero of the polynomial p(x) = 2x + 5 is 9x2 + 4y2 + 16z2 + 12xy – 16yz – 24xz Degree of the zero polynomial is (iii) Polynomial xy + yz + zx is a three variables polynomial, because it contains three variables x, y and z. Solution: Here, we see that (x-2y) +(2y-3z)+ (3z-x) = 0 The coefficient of x in the expansion of (x + 3)3 is 27a+41 = 15+a P(2√2) = (2√2)2– (2√2)(2√2) + 1 = 8 – 8 + 1 = 1, Question 6. Question 8. Solution: (C) It is not […] Solution: (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. iii) 16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz (a) 1 (b) 9 (c) 18 (d) 27 (iii) x3 + x2 – 4x – 4 (iv) 3x3 – x2 – 3x +1 (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx On putting p=1 in Eq. Question 19. We know that, 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 2: => -5/2 On putting x = 2√2 in Eq. = x2(x + 1) – 4(x + 1) p(1) = 10 (1) — 4 (1 )2 -3 (b) abc Thinking Process (ii) x3 – 8y3 – 36xy-216,when x = 2y + 6. Hence, the value of the given polynomial at x = 3 and x = -3 are 61 and -143, respectively. (ii) 25x2 + 16y2 + 4Z2 – 40xy +16yz – 20xz For zero of polynomial, put g(x) = 0 Question 39: Now, p1(3) = a(3)3 + 4(3)2 + 3(3) – 4 (c) 49 Solution: On putting x= -1 in Eq. = 3 x (-1) = -3 (ii) 4x2+ x – 2 If x +1 is a factor of ax3 +x2 -2x+4a-9, then find the value of a. (a) Let p (x) = 5x – 4x2 + 3 …(i) Now, p2(3) = (3)3-4(3)+a Question 15. For zeroes of polynomial, put p(x) = 0 (ii) We have, (x2 – 1) (x4 + x2 + 1) (c) 2 (a) 1 Polynomials in one variable, zeroes of polynomial, Remainder Theorem, Factorization, and Algebraic Identities. (b) 6 (iv) Degree of polynomial y3(1 – y4) or y3 – y7 is 7, because the maximum exponent of y is 7. If a + b + c =0, then a3+b3 + c3 is equal to = (2x + 3) (2x + 1). (iii) xy + yz + zx (v) A polynomial cannot have more than one zero = 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) p(x) = x3 – 3x2 + 4x + 50, g(x) = x – 3 Question 1. (a) 0 (b) 1 (c) 49 (d) 50 Question 16: Solution: Question 8. g(1) = 110 -1 = 1 – 1 = 0 Give an example of a polynomial, which is Classify the following as constant, linear, quadratic and cubic polynomials: If p (x) = x + 3, then p(x) + p(- x) is equal to Solution: [∴ a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2] (i) We have, x2 + 9x +18 = x2 + 6x + 3x +18 Let p(x) = x3 -2mx2 +16 = 2x(2x+ 3) + 1 (2x+ 3) => (-2)3 -2m(-2)2 + 16=0 (d) not defined = (4x – 2y + 3z)(4x -2y + 3z), Question 30. Solution: So, it may have degree 5. (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)). = 50x2 + 10x = 10x (5x + 1) Prove that (a +b +c)3 -a3 -b3 – c3 =3(a +b)(b +c)(c +a). Solution: Question 3: Hence, 0 of x2-3x+2 are land 2. => 2-k = 0 => k= 2 Solution: (c) (2x + 2) (2x + 5) (d) (2x -1) (2x – 3) (iv) 4 – 5y² = (- 5x + 4y + 2z)(- 5x + 4y + 2z), (iii) We have, 16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz [Hint: Factorise x2 – 3x + 2] (d) Now, (25x2 -1) + (1 + 5x)2 (iii) The example of trinomial of degree 2 is x2 – 4x + 3. = 8 – 20 + 8 – 3 = – 7 All solutions are explained using step-by-step approach. It is not a polynomial, because one of the exponents of x is – 2, which is not a whole number. Solution: (a) 0 Factorise (iii) 2x2 -7x.-15 (iv) 84-2r-2r2 Since, 0.2 – 0.3 + 0.1 = 0, (iii) q(x) = 2x – 7 (d) x4 + 3x3 + 3x2 + x + 1 h (1) = (1)11 -1 = 1 – 1 = 0 = 3[3x(x – 1) – 1(x – 1)] = 3(3x – 1)(x – 1), (ii) We have, 9x2 – 12x + 4 = 1000027 + 92700 = 1092727, (ii) We have, 101 × 102 = (100 + 1) (100 + 2) CBSE Maths notes, CBSE physics notes, CBSE chemistry notes. Download for free (or view) PDF file NCERT Class 9 Mathematics Exemplar Problems (Important for UPSC-CSE, CA, UGC-NET) for UPSC-CSE, CA, UGC-NET. If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. Question 17. Solution: It is a polynomial, because each exponent of x is a whole number. Show that p-1 is a factor of p10 -1 and also of p11 -1. Solution: = 4a2 + 4a – 3 [by splitting middle term] g(x) = 3-6x We have, p(x) = x4 – 2x3 + 3x2 – ox + 3a – 7 When we divide p(x) by g(x) using remainder theorem, we get the remainder p(3) Factorise = 3x2 (x – 1) + 2x(x – 1) -1(x – 1) Therefore, we can not exactly determine the highest power of variable, hence cannot define the degree of zero polynomial. NCERT Class 9 New Books for Maths Chapter 2 Polynomials includes all the questions given in CBSE syllabus. On putting p = 1 in Eq. Hence, possible length = 2a -1 and breadth = 2a + 3, Question 1. Factorise the following (i) Firstly, adjust the given number either in the farm 0f a3 + b3 or a3 -b3 (b) x3 + x2 + x + 1 p(2) = 3(2)2 + 6(2) – 24 ∴ a = 5 Zero of the zero polynomial is Hence, p-1 is a factor of g(p). Zero of the polynomial p(x)=2x+5 is Question 4: It is not a polynomial because it is a rational function. With the help of it, candidates can prepare well for the examination. = (a – √2b)[a2 + a( √2b) + (√2b)2] The class will be conducted in Hindi and the notes will be provided in English. Class 9 mathematics is an introduction to various new topics which are not there in previous classes. (iii) True ⇒ 0 = 0 Hence, our assumption is true. (i) The degree of the polynomial is the highest power of the variable x i.e., 6. We have, (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz) (b) (2x + 1) (2x + 3) Solution: Solution: If a + b + c =0, then a3 + b3 + c3 is equal to => -8-8m+16=0 P(-2) = 0 935k watch mins. (a) 2 Question 5. (d) 2abc 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. (viii) Polynomial 1 + x + x² is a quadratic polynomial, because its degree is 2. (a) 0 (b) 1 (c) any real number (d) not defined ⇒ t = 0 and t = 2 (i), we get p(0) = 10(0)-4(0)2 -3 = 0-0-3= -3 Question 21. = 4x³ – 16x² + 17x – 5 Solution: Solution: (v) -3 is a zero of y2 +y-6 (i) False, because a binomial has exactly two terms. Solution: Let p(x) = x3 – 2mx2 + 16 and p(-2) =10 (-2) -4 (-2)2 – 3 (ii) Degree of polynomial – 10 or – 10x° is 0, because the exponent of x is 0. Therefore, the degree of the given polynomial is 4. The value of 2492 – 2482 is Question 14: = x3 + 27 + 9x2 + 27x Hence, the coefficient of x in (x + 3)3 is 27. Since, p(x) is divisible by (x+2), then remainder = 0 (i) Polynomial 2 – x² + x³ is a cubic polynomial, because its degree is 3. Find the value of Hence, p -1 is a factor of h(p). Factorise the following: On putting x = 2√2 in Eq. (i) a3 -8b3 -64c3 -2Aabc = -1 + 51 = 50 NCERT Exemplar Class 9 Maths Solutions Polynomials. Solution: Question 15. Hindi Mathematics. Question 14. (iii) 16x2 + 4)^ + 9z2-^ 6xy – 12yz + 24xz Now, this is divided by x + 2, then remainder is p(-2). Hence, one of the zeroes of the polynomial p(x) is ½. Also, find the remainder when p(x) is divided by x + 2. Hence, zero of polynomial is 4. Solution: Hence, the value of k is 2. (a)-6 Solution: = 10 – 4 – 3= 10 – 7 = 3 = (3x)2 + (2y)2 + (-4z)2 + 2(3x)(2y) + 2(2y)(-4z) + 2(-4z)(3x) ⇒ a2 + b2 + c2 = 81 – 52 = 29, Question 31. (iii) 5t – √7 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. = 2x(2x + 3) + 1 (2x + 3) Hence, the zero of polynomial is 0, Question 12: Given, polynomial is p(x) = (x – 2)2 – (x+ 2)2 Hence, zero of x – 3 is 3. (ii) a3 – 2√2b3 (i) 1 + 64x3 (ii) a3 -2√2b3 ⇒ x = ½ and x = -4 Question 9. (d) -2 = 50x2 + 10x = 10x (5x+ 1) Solution: Question 4: = 3(-27)-4×9-21-5 = -81-36-21-5 = -143 p(-3) = -143 …(i) we see that (x – 2y) +(2y – 3z)+ (3z – x) = 0 (i) a3 -8b3 -64c3 -2Aabc (ii) the coefficient of x3 (iv) Polynomial 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 … (i) We have, p(x) = x³ – 2x² – 4x – 1 and g(x) = x + 1 (i) x2 + x +1 (ii) y3 – 5y = (1 + 4x)(1 – 4x + 16x2), (ii) We have, a3 – 2√2b3 = (a)3 – (√2b)3 ⇒ k = 2 (a) 3 (b) 2x (c) 0 (d) 6 (i) monomial of degree 1. (ii) x3 -8y3 -36xy-216,when x = 2y + 6. 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. If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is NCERT Exemplar Problems Class 9 Maths Solutions are being updated for new academic session 2020-2021. (iii) x3 + x2-4x-4 We have, 84 – 2r – 2r2 = – 2 (r2 + r – 42) NCERT Solutions based on latest NCERT Books as well as NCERT Exemplar Problems chapters are being updated for new academic year 2020-21. (a)-6 (b) 6 (c) 2 (d) -2 Solution: Solution: = 25x2 -1 + 1 + 25x2 + 10x [using identity, (a + b)2 = a2 + b2 + 2ab] p(1) = (1 + 2)(1-2) Since 2x – 1 is afactor of p(x) then p(1/2) = 0, Question 22. For zero of the polynomial, put p(x) = 0 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. e.g., Let f(x) = x4 + 2 and g(x) = -x4 + 4x3 + 2x. Question 3. p(2) = 4(2)2 + 2 – 2 = 16 ≠ 0 These NCERT Exemplar for Class 9 Maths Chapter 2 with Solutions are designed as per the CBSE Class 9th Maths Syllabus. NCERT solutions for class 9 Maths is available to download for free from the links below. [∴ If a + b + c = 0, then a3 + b3 + c3 3abc] We hope the NCERT Exemplar Class 9 Maths Chapter 2 Polynomials will help you. On putting x = -1 in Eq. = 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) ⇒ y(y + 3) – 2(y + 3) = 0 a(- 1)3+ (- 1)2 – 2(-1) + 4a – 9 = 0 Show that, 2a = 3 P(2√2) = (2√2)2– (2√2)(2√2) + 1 =8-8+1=1, Question 6: Because a polynomial can have any number of zeroes. (c) 2/5 Solution: ⇒ a2 + b2 + c2 + 2 (ab + bc + ca) = 81 If x + 1 is a factor of ax3 + x2 – 2x + 4o – 9, find the value of a. = -2 x 125y3 – 30xy(4x) = -250y3 -120x2y. = 3x(2x + 3) – 1(2x + 3) = (3x – 1)(2x + 3) Question 7. (i) firstly, determine the factors of quadratic polynomial by splitting middle term. (b) Let assume (x + 1) is a factor of x3 + x2 + x+1. Hence, the zeroes of y² + y – 6 are 2 and – 3. Factorise: (v) False 2(-1)2 + k(-1) = 0 4x4 + 0x3 + 0x5 + 5x + 7 = 4x4 + 5x + 7 = 3×27-4×9 + 21-5 = 81-36+21-5 P( 3) =61 We have prepared chapter wise solutions for all characters are given below. (iii) (-x + 2y-3z)2 Exercise 2.3: Short Answer Type Questions. (ii) Polynomial y3 – 5y is a one variable polynomial, because it contains only one variable i.e., y. and p(2) = 2(2)4 – 5(2)3 + 2(2)2 – 2 + 2 = 2x16-5x8+2x4+ 0 = 32 – 40 + 8 = 40 – 40 =0 (i), we get p(-1) = (-1)51 + 51 = 497 × 1 = 497. (c) 0 Without actual division, prove that 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2 Hence, one of the factor of given polynomial is 3xy. (a) 4 Question 13. Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (-z + x-2y). Which of the following is a factor of (x+ y)3 – (x3 + y3)? (b) Let p (x) = 2x2 + 7x – 4 (a) 2 (b) 0 (c) 1 (d)½ NCERT Exemplar Class 9 Maths Solutions will give you a thorough understanding of Maths concepts as per the CBSE exam pattern. Here we have given NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials. = x2(x – 1) – 5x (x – 1) + 6(x – 1) Let p(x) = 8x4 + 4x3 – 16x2 + 10x + m = 10000 + 300 + 2 = 10302, (iii) We have, (999)2 = (1000 -1)2 Solution: = x3 +27 + 9x2+27x Hence, the coefficient of x in (x + 3)3 is 27. Factorise the following [using identity, (a + b)2 = a2 + b2 + 2 ab)] (iv) 3x3-x2-3x+1 Solution: Question 1. (ii) Coefficient of x2 in 3x – 5 is 0. = (x -1) (x2 – 5x + 6) Solution: = x(x-2)-1 (x-2)= (x-1)(x-2) Solution: Question 30: NCERT Exemplar for Class 9 Maths Chapter 2 With Solution | Polynomials. Thinking Process Let p(x) =3x3 – 4x2 + 7x – 5 NCERT 9th class Mathematics exemplar book solutions for chapter 2 Polynomials are available in PDF format for free download. (c) 1 Find the zeroes of the polynomial in each of the following, (ii) Further, put the factors equals to zero, then determine the values of x. 2x= 7 => x =7/2 Therefore, remainder is 0. Solution: = (y-2)(y + 3) = 0 (a) 0 (b) 1 ⇒ -5/2 ∴ Sum of two polynomials, lf a+b+c=9 and ab + bc + ca = 26, find a2 + b2 + c2. a = 3/2. Give possible expression for the length and breadth of the rectangle whose area is given by 4a2 + 4a – 3. (ii) The example of binomial of degree 20 is 6x20 + x11 or x20 +1 Here are all questions are solved with a full explanation and available for free to download. (a) 2 (b)½ (c)-1 (d)-2 Solution: (i) 1 + 64x3 Exercise 2.1 Page No: 14. = 8x³ + 2xy2 + 18xz2 + 4x2y + 6xyz – 12x2z – 4x2y – y3 – 9yz2 – 2xy2 – 3y2z + 6xyz + 12x2z + 3y2z + 27z3 + 6xyz + 9yz2 – 18xz2 Solution: (i) We have, 1033 = (100 + 3)3 p(1) = 10 (1) – 4 (1 )2 – 3 (i) We have, g(x) = x – 2 Put t² – 2t = 0 ⇒ t(t – 2) = 0 (vi) Polynomial 2 + x is a linear polynomial, because its degree is 1. NCERT Class 9 Maths Unit 2 is for Polynomials. (iii) the coefficient of x6 ⇒ a2 + b2 + c2 = 5 … (i) (ii) 9x2 – 12x + 4 = 27a+ 36+ 9-4= 27a+ 41 When we divide p2(z) by z-3 then we get the remainder p2(3). Solution: (i) Polynomial x2+ x+ 1 is a one variable polynomial, because it contains only one variable i.e., x. (i) p(x) = 10x – 4x2 – 3 (ii) p(y) = (y + 2)(y – 2) Solution: Degree of the polynomial 4x4 + 0x3 + 0x5 + 5x + 7 is (v) -3 is a zero of y2 + y – 6 Solution: Solution: Question 5: 8x4 + 4x3 – 16x2 + 10x + m p(-1)=0 (iv) We have, (2x – 5) (2x² – 3x + 1) Substituting x = 2 in (1), we get Solution: This chapter consists of problems based on polynomial in one variable, Zeroes of a Polynomial, Remainder Theorem, Factorisation of Polynomials and Algebraic Identities. 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 In this method firstly check the values of a + b+ c, then . => 2x-1 = 0 and x+4 = 0 ∴ (0.2)³ + (-0.3)³ + (0.1)3 = 3(0.2) (-0.3) (0.1) (x2 + 4y2 + z2 + 2xy + xz – 2yz)(- z + x – 2y) Solution: Solution: p(x) = x- 4 ⇒ -8x = 0 Question 6: = (x – 1) (x2 – 3x – 2x + 6) = (3x – 2) (3x – 2), Question 28. Question 13: (i) 9x2 +4y2 + 16z2 +12xy-16yz -24xz Let p1(z) = az3 +4z2 + 3z-4 and p2(z) = z3-4z + o = (3a)3 – (2b)3 – 3(3a)(2b)(3a – 2b) NCERT Class 9 New Books for Maths Chapter 2 Polynomials are given below. 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. (b) 1 (i) x2 + 9x +18 (ii) 6x2 +7x -3 Hence, one of the factor of given polynomial is 3xy. Here, the highest power of x is 4. ⇒ y – 2 = 0 and y + 3 = 0 (i) p(x) = x3-2x2-4x-1, g(x)=x + 1 Using suitable identity, evaluate the following: (a) 4 (b) 5 (c) 3 (d) 7 26a = 26 Simplify (2x- 5y)3 – (2x+ 5y)3. = 5(5 -10) = 5(-5) = – 25 = R.H.S. On putting x = 0,1 and – 2, respectively in Eq. for p(x) = x² – 4, degree is 2, so it has two zeroes i.e., 2 and -2. Visit FlexiPrep for more files and information on Subject-Wise-NCERT-Books-PDF: Mathematics The Class NCERT 9th Math textbook has a total of 15 chapters which are divided into seven units. And to make that base strong students are advised to solve NCERT Exemplar class 9 Maths. ⇒ (x – 2 + x + 2)(x – 2 – x – 2) = 0 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 … (a) 5 + x = 2a(2a + 3) -1 (2a + 3) = (2a – 1)(2a + 3) Hence, possible length = 2a -1 and breadth = 2a + 3, Question 1: Question 32. (iii) If the maximum exponent of a variable is 1, then it is a linear polynomial. Since, remainder ≠ 0, then p(x) is not a multiple of g(x). (c) Let p(x) = 2x2 + kx ’ (ii) p(x) = 2x3 – 11x2 – 4x+ 5, g(x) = 2x + l. Thinking Process Thinking Process (b) x² + 5 [polynomial and also a binomial]. Find the zeroes of the polynomial in each of the following, On putting x = -1 in Eq. Coefficient of x² in 4x³ – 16x² + 17x – 5 is -16. Now, p(x)+ p(-x) = x+ 3+ (-x)+ 3=6. (v) Polynomial 3 is a constant polynomial, because the exponent of variable is 0. Hence, the value of a is 3/2. One of the factors of (25x2 – 1) + (1 + 5x)2 is (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. m = 1 Question 12. Particular these Exemplar Books Prepare the Students and for Subject … Hence, the zeroes of t² – 2t are 0 and 2. 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 … Justify your answer. = a3 + b3 + c3 – 3abc = (1 + 4x)[(1)2 – (1)(4x) + (4x)2] (c) 5x -1 (vi) The degree of the sum of two polynomials each of degree 5 is always 5. (d) Now, (x+ y)3 – (x3 + y3) = (x + y) – (x + y)(x2– xy + y2) (d) 1/2 (i), we get (a) 1 Question 5: Check whether p(x) is a multiple of g(x) or not NCERT Exemplar for Class 9 Maths Chapter 5 With Solution | Introduction to Euclid’s Geometry (i) 9x2 + 4y2+16z2+12xy-16yz-24xz Solution: Question 14: x3 + y3 + (4)3 = 3xy(4) = 2x(x – 5) + 3(x – 5) = (2x + 3)(x – 5) (b) Given, p(x) = x2 – 2√2x + 1 …(i) NCERT Exemplar Class 9 Maths Chapter 2 Polynomials are part of NCERT Exemplar Class 9 Maths. NCERT Class 9 Maths Exemplar book has 14 chapters on topics like Rational Numbers, Coordinate Geometry, Triangles, Heron’s formula, Statistics, and Probability. (iv) h(y) = 2y Solution: Question 29: = x2 + 4y2 + 9z2 – 4xy – 12yz + 6xz, Question 29. If x + 2a is a factor of a5 -4a2x3 +2x + 2a +3, then find the value of a. ⇒ a2 + b2 + c2 = 25 – 20 Students can download these NCERT Solutions of class 9 Maths PDFs for free. (i) -3 is a zero of at – 3 Hence, zero of 4 – 5y is 4/5. (ii) Degree of polynomial -10 or -10x° is zero, because the exponent of x is zero. If p (x) = x + 3, then p(x)+ p (- x) is equal to (ii) 2x – 3 is a factor of x + 2x3 -9x2 +12 = x3 – x2 – 5x2 + 5x + 6x – 6 (a) (x + 1) (x + 3) Solution: Question 33. Prepare effectively for your Maths exam with our NCERT Solutions for CBSE Class 9 Mathematics Chapter 2 Polynomials. The highest power of … Give an example of a polynomial, which is (ii) If the maximum exponent of a variable is 0, then it is a constant polynomial. Solution: = (a – √2b)(a2 + √2ab + 2b2), Question 35. Hence, the values of p(0),p(1) and p(-2) are respectively,-4,-3 and 0. Hence, the zero of polynomial h(y) is 0. = (x + y) (3xy) Check whether p(x) is a multiple of g(x) or not Question 10: Each exponent of the variable x is a whole number. (iv) True It depends upon the degree of the polynomial = (x + 1)(x2 – 4) Since p(x) is divided by x + 1, then remainder is p(-1). 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