According to the question, both the remainders are same. = 4a2 + 4a – 3 (ii) The coefficient of x3 in given polynomial is 1/5. Question 13. = (x – 2) (2x2 + 6x – 5x – 15) (a) 1 Let p(x) =3x3 – 4x2 + 7x – 5 (a) 0                                                   (b) 1 For zeroes of p(x), put p(x) = 0=> (2x -1) (x + 4) = 0 (i) 9x2 -12x+ 3          (ii) 9x2 -12xy + 4 Now, p1(3) = a(3)3 + 4(3)2 + 3(3) – 4 (b) 5 – x (i) p(x)= x – 4 (i) a3 -8b3 -64c3 -2Aabc (a) 3               (b) 2x            (c) 0                   (d) 6 Since, p(x) is divisible by (x+2), then remainder = 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. Question 19: (ii) Every polynomial is a Binomial. Therefore, the degree of the given polynomial is 4. NCERT Exemplar Class 9 Maths. = -x2z + x3 – 2 x2y – 4y2z + 4xy2 – 81y3 – z3 + xz2 – 2yz2– 2xyz + 2x2y – 4xy2 – xz2 + x2z – 2xyz + 2yz2 – 2xyz + 4y2z The polynomial p{x) = x4 -2x3 + 3x2 -ax+3a-7 when divided by x+1 leaves the remainder 19. When we divide p(x) by g(x) using remainder theorem, we get the remainder p(-1) 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. Solution: Question 35: If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is and p( 2) = 2(2)4 – 5(2)3 + 2(2)2 -2 + 2 Fi nd (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz). (b) 0 If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is (d) Let p(x) = x51 + 51 . Because a binomial has exactly two terms. m = 1 If p (x) = x + 3, then p(x) + p(- x) is equal to = -2[r(r + 7) -6(r + 7)] = 2x2 + 8x-x-4 [by splitting middle term] Solution: Solution: (a) 12 Solution: (b) abc (ii) Further, put the factors equals to zero, then determine the values of x. (iii) x3 + x2 – 4x – 4             (iv) 3x3 – x2 – 3x +1 Solution: NCERT 9th class Mathematics exemplar book solutions for chapter 2 Polynomials are available in PDF format for free download. (i) 2x3 – 3x2 – 17x + 30 = x3 + 27 + 9x2 + 27x Hence, the coefficient of x in (x + 3)3 is 27. ⇒ -2a + 3 = 0 (v) -3 is a zero of y2 + y – 6 (iv) the constant term (i) Firstly, adjust the given number either in the farm 0f a3 + b3 or a3 -b3 (a) 2             (b) 0           (c) 1          (d)½ ⇒ 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 Solutions for class 9 Mathematics Polynomials. (i) x + 3 is a factor of 69 + 11x – x2 + x3 = 4a2 + 6a – 2a – 3 = 2x2(x – 2) + x(x – 2) – 15(x – 2) When we divide p(x) by g(x) using remainder theorem, we get the remainder p(3) => x = ½ and x = -4 (ii) a3 – 2√2b3 (iv) Polynomial 4 – 5y² is a quadratic polynomial, because its degree is 2. Question 17. ’ Let p(x) = x3 – 2mx2 + 16 (a) 1     (b) 9     (c) 18     (d) 27 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. Hence, √2 is a polynomial of degree 0, because exponent of x is 0. (c) 49 (c) 5x -1 = (1 + 4x)(1 – 4x + 16x2), (ii)  We have, a3 – 2√2b3 = (a)3 – (√2b)3 Question 36. e.g., Let us consider zero polynomial be 0(x – k), where k is a real number For determining the zero, put x – k = 0 ⇒ x = k 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. When we divide p1(z) by z – 3, then we get the remainder p,(3). (d) 7 Now, p2(3) = (3)3-4(3)+a (iii) Polynomial xy+ yz+ zx is a three variables polynomial, because it contains three variables x, y and z. Because the sum of any two polynomials of same degree has not always same degree. -[(2x)3 + (5y)3 + 3(2x)(5y)(2x+5y)] (i) We have, p(x) = x³ – 2x² – 4x – 1 and g(x) = x + 1 Now, x2-3x + 2 = x2 – 2x – x + 2 Question 16. By actual division, find the quotient and the remainder when the first = -2 (5y)3 – 30xy(2x – 5y + 2x + 5y) It is not a polynomial, because one of the exponents of x is – 2, which is not a whole number. = 8x³ + 2xy2 + 18xz2 + 4x2y + 6xyz – 12x2z – 4x2y – y3 – 9yz2 – 2xy2 – 3y2z + 6xyz + 12x2z + 3y2z + 27z3 + 6xyz + 9yz2 – 18xz2 NCERT Exemplar Solutions For Class 9 Maths. 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. Get NCERT Exemplar Solutions for Class 9 Chapter Polynomials here. On putting p = 1 in Eq. (i) We have, (4a – b + 2c)2 = (4a)2 + (-b)2 + (2c)2 + 2(4a)(-b) + 2(-b)(2c) + 2(2c)(4a) ⇒ k = 2 (i) monomial of degree 1. (ii) Polynomial 3x³ is a cubic polynomial, because its degree is 3. a = 3/2. For zero of polynomial, put g(x) = 0 Put y² + y – 6 = 0 ⇒ y² + 3y – 2y – 6 = 0 (a) 0        (b) 1           (c) any real number               (d) not defined (ii) x3 -8y3 -36xy-216,when x = 2y + 6. dceta.ncert@nic.in 011 2696 2580 NCERT, Sri Aurobindo Marg, New Delhi-110016 011 2696 2580 NCERT, Sri Aurobindo Marg, New Delhi-110016 (i) p(x) = 10x – 4x2 – 3 (ii) p(y) = (y + 2)(y – 2) Substituting x = 2 in (2), we get (v) A polynomial cannot have more than one zero NCERT Exemplar for Class 9 Maths Chapter 2 With Solution | Polynomials. (vi) The degree of the sum of two polynomials each of degree 5 is always 5. Solution: (i) Polynomial 2 – x2 + x3 is a cubic polynomial, because maximum exponent of x is 3. (iv) ‘p(x) = x3-6x2+2x-4, g(x) = 1 -(3/2) x (b) Let assume (x + 1) is a factor of x3 + x2 + x+1. (iv) Polynomial 4- 5y2 is a quadratic polynomial, because maximum exponent of y is 2. Question 3. p(-1)=0 Here, the highest power of x is 4. Solution: Question 2: (i) A Binomial can have atmost two terms. It is not a polynomial, because exponent of x is 1/2 which is not a whole number. Polynomials in one variable, zeroes of polynomial, Remainder Theorem, Factorization, and Algebraic Identities. Show that, On putting p=1 in Eq. Question 10. 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: Solution: Question 40: Find the values of a. Therefore, we can not exactly determine the highest power of variable, hence cannot define the degree of zero polynomial. Solution: Solution: (i) We have, 2X3 – 3x2 – 17x + 30 …(i) Question 6: Find the zeroes of the polynomial p(x)= (x – 2)2 – (x+ 2)2. p(x) = x – 4 (a) Let p (x) = 5x – 4x2 + 3 …(i) (a) 0 (i) firstly, determine the factors of quadratic polynomial by splitting middle term. = (3x)2 – 2 × 3x × 2 + (2)2 Solution: (d) Now, a3+b3 + c3= (a+ b + c) (a2 + b2 + c2 – ab – be – ca) + 3abc (iv) p(x) = x3 – 6x2 + 2x – 4, g(x) = 1 – (3/2) x (vii) Polynomial y³ – y is a cubic polynomial, because its degree is 3. (b) Given, p(x) = 2x+5 = (1000)2 + (1)2 – 2(1000)(1) = -2 x 125y3 – 30xy(4x) = -250y3 -120x2y. So, x = -1 is zero of x3 + x2 + x+1 (i) 9x2 + 4y2+16z2+12xy-16yz-24xz Solution: Question 25: Solution: Using suitable identity, evaluate the following We have, 2x2 – 7x – 15 = 2x2 – 10x + 3x -15 Solution: (c) 487 (ii) True These topics creates the base for higher level of mathematics. ⇒ 2 – k = 0 [using identity, a3 + b3 + c3 – 3 abc = (a + b + c)(a2 + b2 + c2 – ab – be – ca)] (d) -2 (i) We have, (3a – 2b)3 Solution: (i) We have, 9x2 – 12x + 3 = 3(3x2 – 4x + 1) Solution: (c) 2 Solution: (iii) Polynomial xy + yz + zx is a three variables polynomial, because it contains three variables x, y and z. Cbse Class 9 examinations as it will always help you to understand the basic concepts Mathematics! True or False and provide the ncert Exemplar for Class 9 Maths Solutions are solved with a full explanation available... A3 + b3 + c3 = 3abc questions in Polynomials the help of it, candidates can Prepare well the! 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