2a =3 Hence, the degree of a polynomial is 4. Hence, the remainder is 50. Question 20: (c) 0 (d) Now, 2492 – 2482 = (249 + 248) (249 – 248) [using identity, a2 – b2 = (a – b)(a + b)] g(x) = 3 – 6x (ii) Given, polynomial isp(y) = (y+2)(y-2) 27a + 41 = 15 + a (i) We have, 1 + 64x3 = (1)3 + (4x)3 Because a binomial has exactly two terms. The value of 2492 – 2482 is [using identity, a3 + b3 + c3 – 3 abc = (a + b + c)(a2 + b2 + c2 – ab – be – ca)] So, it is a cubic polynomial. Now, p1(3) = a(3)3 + 4(3)2 + 3(3) – 4 Solution: ⇒ t = 0 and t = 2 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. Learn about the degree of a polynomial and the zero of a polynomial with related Maths solutions. ⇒ y – 2 = 0 and y + 3 = 0 Question 5: Solution: = x3 +27 + 9x2+27x Hence, the coefficient of x in (x + 3)3 is 27. ⇒ x = 0 (i) 1033 = 2a(2a + 3) -1 (2a + 3) = (2a – 1)(2a + 3) Hence, possible length = 2a -1 and breadth = 2a + 3, Question 1: Check whether p(x) is a multiple of g(x) or not = (1 + 4x)[(1)2 – (1)(4x) + (4x)2] Now, x2-3x + 2 = x2 – 2x – x + 2 = [(a + b + c)3 – a3] – (b3+ c3) (ii) The coefficient of x3 in given polynomial is 1/5. Solution: Question 9. (c) x4 + x3 + x2 + 1 (iii) xy + yz + zx Factorise the following: ∴ (-2a)5 – 4a2 (-2a)3 + 2(-2a) + 2a + 3 = 0 Without actually calculating the cubes, find the value of 36xy-36xy = 0 Question 7. Solution: (ii) 25x2 + 16y2 + 4Z2 – 40xy +16yz – 20xz = (3x)2 – 2 × 3x × 2 + (2)2 NCERT solutions are really helpful when it comes to a complicated subject like Mathematics. √2 is a polynomial of degree Also, if a + b + c = 0, then a3 + b3 + c3 = 3abc The value of the polynomial 5x – 4x2 + 3, when x = – 1 is The value of 2492 – 2482 is On putting x = -1 in Eq. m = 1 Solution: Number Systems | NCERT Exemplar Solutions | Class 9 Exercise 1.1 Page No: 2 Write the correct answer in each of the following: 1. If x + 2a is a factor of a5 -4a2x3 +2x + 2a +3, then find the value of a. = 2x(x – 5) + 3(x – 5) = (2x + 3)(x – 5) = 12 + 12 – 24 = 0 Solution: ∴ p(2) = (2)3 – 5(2)2 + 4(2) – 3 (a) (x + 1) (x + 3) h (1) = (1)11 —1 = 1 —1 = 0 Hence, p -1 is a factor of h(p). = (4x)2 + (- 2y)2 + (3z)2 + 2(4x)(-2y) + 2(-2y)(3z) + 2(3z)(4x) Solution: = 10000 + 300 + 2 = 10302, (iii) We have, (999)2 = (1000 -1)2 e.g., p(x) = x2 -2, as degree pf p(x) is 2 ,so it has two degree, so it has two zeroes i.e., √2 and —√2. Solution: Question 5: Using suitable identity, evaluate the following: Thinking Process = 2a(2a + 3) -1 (2a + 3) = (2a – 1)(2a + 3) ⇒ x3 + y3 – 12xy + 64 = 0, (ii) Since, x – 2y – 6 = 0, then (ii) y3 – 5y Question 10: = 27a3 – 54a2b + 36ab2 – 8b3. = (x -1) (x2 – 5x + 6) If p (x) = x + 3, then p(x) + p(- x) is equal to P(-2) = 0 Solution: Solution: Solution: These topics creates the base for higher level of mathematics. Question 10: Solution: (ii) Given, polynomial is (ii) Polynomial y3 – 5y is a one variable polynomial, because it contains only one variable i.e., y. NCERT Exemplar Class 9 Maths Chapter 2 Polynomials are part of NCERT Exemplar Class 9 Maths. Solution: (i) 2x3 – 3x2 – 17x + 30 = 2x(4x2 + y2 + 9z2 + 2xy + 3yz – 6xz) – y(4x2 + y2 + 9z2 + 2xy + 3yz – 6xz) + 3z(4x2 + y2 + 9z2 + 2xy + 3yz – 6xz) Therefore, we can not exactly determine the highest power of variable, hence cannot define the degree of zero polynomial. (iii) The coefficient of x6 in given polynomial is -1. Solution: Question 2: (i) Given, polynomial is (ii) Polynomial 3x³ is a cubic polynomial, because its degree is 3. Solution: One of the zeroes of the polynomial 2x2 + 7x – 4 is Fi nd (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz). = (x – 2) [2x(x + 3) – 5(x + 3)] (b) Let p (x) = 2x2 + 7x-4 Now, p2(3) = (3)3-4(3)+a (ii) -1/3 is a zero of 3x+1 = 2x (2x² – 3x + 1) – 5(2x² – 3x + 1) Hence, zero of polynomial is 4. Solution: [ ∴ If a + b + c = 0, then a3 + b3 + c3 = 3abc] = -0.018, Question 38. Solution: Hence, the zero of polynomial is 0, Question 12: Now, p(-3) = (-3)3 – (-3)2 + 11(-3) + 69 (ii) Given, polynomial is Expand the following Here, zero of g(x) is 1/2. (a) Let p (x) = 5x – 4x2 + 3 …(i) Solution: => 2-k = 0 => k= 2 We have, (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz) Degree of the polynomial 4x4 + 0x3 + 0x5 + 5x + 7 is [using identity, (a + b)2 = a2 + b2 + 2 ab)] (ii) We have, (x2 – 1) (x4 + x2 + 1) Question 5. Factorise: NCERT Exemplar for Class 9 Maths Chapter 5 With Solution | Introduction to Euclid’s Geometry (b) -5/2 Hence, the remainder is 50. (i) Polynomial (b) 1 = 2x2 + 8x – x – 4 [by splitting middle term] (i), we get (i) The example of monomial of degree 1 is 5y or 10x. We hope that our NCERT Class 9 New Books for Maths helped with your studies! On putting x = 0, 1 and – 2, respectively in Eq. (iv) h(y) = 2y (2), we get (b) Let assume (x + 1) is a factor of x3 + x2 + x+1. [using identity, a2 – b2 = (a – b)(a + b)] Question 1: (viii) Polynomial 1 + x + x² is a quadratic polynomial, because its degree is 2. (ii) 9x2 – 12x + 4 If a + b + c = 5 and ab + bc + ca = 10, then prove that a3 + b3 + c3 – 3abc = -25. ⇒ 4a – 1 = 19 ⇒ 4a = 20 Question 21: Hence, one of the factor of given polynomial is 3xy. Students can download these NCERT Solutions of class 9 Maths PDFs for free. (iii) q(x) = 2x – 7 -[(2x)3 + (5y)3 + 3(2x)(5y)(2x+5y)] Solution: Since, remainder ≠ 0, then p(x) is not a multiple of g(x). [using identity, (a + b)3 = a3 + b3 + 3ab (a + b)] Solution: Question 35: 27a+41 = 15+a (i), we get p(0) = 10(0)-4(0)2 -3 = 0-0-3= -3 For what value of m is x3 -2mx2 +16 divisible by x + 2? = 27a3 – 8b3 – 18ab(3a – 2b) 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. Question 8: Since, x + 2a is a factor of p(x), then put p(-2a) = 0 (-2a)5 – 4a2 (-2a)3 + 2(-2a) + 2a + 3 = 0 => -32a5 + 32a5 -4a + 2a+ 3 = 0 The class will be conducted in Hindi and the notes will be provided in English. = x(x-2)-1 (x-2)= (x-1)(x-2) Simplify (2x- 5y)3 – (2x+ 5y)3. (b) (2x + 1) (2x + 3) (-1)3 + (-1)2 + (-1) + 1 = 0 => -1+1-1 + 1 = 0 => 0 = 0 Hence, our assumption is true. We have, p(x) = x4 – 2x3 + 3x2 – ox + 3a – 7 (v) -3 is a zero of y2 + y – 6 (i) x2 + 9x +18 (ii) 6x2 +7x -3 (c) 0 (ii) x3 -8y3 -36xy-216,when x = 2y + 6. p(x) = x4 – 2x3 + 3x2 – 5x + 3(5) – 7 Question 10. Solution: (v) A polynomial cannot have more than one zero. 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