Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1. = (3x + y + z)(9x2 + y2 + z2 – 3xy – yz – 3zx). Ex 2.1 Class 9 Maths Question 4. i.e. (viii) We have, p(x) = 2x + 1 = 9a2 + 49b2 + c2 – 42ab + 14bc – 6ac, (v)(- 2x + 5y- 3z)2 = (- 2x)2 + (5y)2 + (- 3z)2 + 2 (- 2x) (5y) + 2 (5y) (- 3z) + 2 (- 3z) (- 2x) = x3 + x2 + 12x2 + 12x + 20x + 20 (ii) 35y2+ 13y -12 = 35y2 + 28y – 15y -12 = (y – 1)(2y2 + 2y + y + 1) ∴ p(0) = (0)3 + 3(0)2 + 3(0) + 1 (i) p (x)= 2x3 + x2 – 2x – 1, g (x) = x + 1 (i) We have, p(x) = x + 5. = (3 – 5a) (3 – 5a) (3 – 5a), (iv) 64a3 -27b3 -144a2b + 108ab2 p( 2) = 2 + 2 + 2(2)2 – (2)3 (ii) We have y2 + √2 = y2 + √2y0 Thus, zero of 2x + 5 is \(\frac { -5 }{ 2 }\) . Telanagana State Board Class 9 Mathematics Textbook Solution Chapter-wise Telanagana SCERT Class IX Math Solution. Ex 2.1 Class 9 Maths Question 2. (iv) We have, p(x) = 3x – 2. (iii) 6x2 + 5x – 6 (i) We have, (i) We know that Hindi Medium and English Medium both are available to free download. For example, if you are weak in class 9 maths, you can’t make a great career in the field of engineering and mathematics. ∴ 3x3 + 7x is not divisib1e by 7 + 3x. = 2 + k + √2 =0 Solution: (ii) Here, p (x) = 2x2 + kx + √2 (iv) x + π (ii) x = – 1 These assignments will be available in updated form along with new assignments and chapter wise tests with solutions. (i) x = 0 = 10000 + (-900) + 20 = 9120, (iii) We have 104 x 96 = (100 + 4) (100 – 4) (i) ∵ (x + y)3 = x3 + y3 + 3xy(x + y) Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Maths for Class 9 so that you can refer them as and when required. Solution: CBSE Class 11 Maths Worksheet for students has been used by teachers & students to develop logical, lingual, analytical, and … = -14 + 13 Thus, x3 – 3x2 – 9x – 5 = (x + 1)(x – 5)(x +1), (iii) We have, x3 + 13x2 + 32x + 20 (iv) 3 = (3)3 – (5a)3 – 3(3)(5a)(3 – 5a) (v) p (x) = x2, x = 0 So, it is a linear polynomial. ⇒ p (-1) ≠ 0 We have, 27y3 + 125z3 = (3y)3 + (5z)3 (iv) p (x) = (x-1) (x+1) = 8x3 + 1 + 6x(2x + 1) = 4k[(3y + 5) x (y – 1)] (iii) (- 2x + 3y + 2z)2 Chapter 14 Probability. (iv) Given that p(x) = (x – 1)(x + 1) = 1000000000 – 8 – 6000(1000 – 2) ∴ p(1) = (1)2 – 1 = 1 – 1=0 = ( 100)2 + [(- 5) + (- 4)] 100 + (- 5 x – 4) = (1000)3– (2)3 – 3(1000)(2)(1000 – 2) Thus, 12x2 -7x + 3 = (2x – 1) (x + 3), (ii) We have, 2x2 + 7x + 3 = 2x2 + x + 6x + 3 (iv) 3x2 – x – 4 = [(x)2 – (1)2](x – 2) (i) We have, (x+ 4) (x + 10) (iii) p (x) = kx2 – √2 x + 1 The coefficient of x2 is \(\frac { \pi }{ 2 }\). Solution: Extra questions along with questions of NCERT book complete the topic . ∴ p(a) = (a)3 – a(a)2 + 6(a) – a We have, 64m3 – 343n3 = (4m)3 – (7n)3 Terms of Service. Solution: Question 13. ⇒ x3 + y3 + 3xy(-z) = -z3 [∵ x + y = -z] Homework Help with Chapter-wise solutions and Video explanations. Thus, 3x2 – x – 4 = (3x – 4)(x + 1), Question 5. = 4k[3y2 – 3y + 5y – 5] Thus, zero of 3x is 0. = (3y + 5z)[(3y)2 – (3y)(5z) + (5z)2] Question 3. Login to view more pages. ∴ p (- 1) =(- 1)3- (-1)2 – (2 + √2)(-1) + √2 Extra questions based on the topic Number System. = 1000000 -1 – 30000 + 300 sin θ and tan θ) without evaluating θ. = 2 + 2 + 8 – 8 = 4 (iii) The zero of x is 0. 1et p(x) = 5x – 4x2 + 3 [Using (a – b)3 = a3 – b3 – 3ab (a – b)] Thus, the required remainder is -π3 + 3π2 – 3π+1. It is a polynomial in one variable i.e., x (ii) p (t) = 2 +1 + 2t2 -t3 (i) The given polynomial is 5x3 + 4x2 + 7x. ⇒ x3 + y3 + 3xy(x + y) = -z3 Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given ⇒ p(1) = k + 2 = 0 (v) (3 – 2x) (3 + 2x) 27x3 + y3 + z3 – 9xyz = (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z) (iii) 5t – √7 Find p (0), p (1) and p (2) for each of the following polynomials. Chapter-wise NCERT Solutions for Class 9 Maths Chapter 2 Polynomials solved by Expert Teachers as per NCERT (CBSE) Book guidelines. Since, p(2) = 0, so, x = 2 is also a zero of (x + 1)(x – 2). = (2a)3 + (b)3 + 3(2a)(b)(2a + b) ∴ p(0) = (0 – 1)(0 + 1) = (-1)(1) = -1 Download NCERT Solutions for Class 9 Maths in Hindi Medium for CBSE Board, UP Board (High School), MP Board, Gujrat Board and other board’s students who are following NCERT Books. 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. NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1 are part of NCERT Solutions for Class 9 Maths. ⇒ x3 + y3 + z3 = 3xyz However, it is possible to avoid such a scenario by taking authentic NCERT solutions for class 9 maths from a reliable source. Thus, the required remainder is \(-\frac { 27 }{ 8 }\) . (iii) 27-125a3 -135a+225a2 Solution: (vii) We have, p(x) = cx + d. Since, p(x) = 0 Thus, zero of x – 5 is 5. = 2y3 – 2y2 + 3y2 – 3y + y – 1 = (4a – 3b)3 = x2 + 4y2 + 16z2 + 4xy + 16yz + 8 zx, (ii) (2x – y + z)2 = (2x)2 + (- y)2 + z2 + 2 (2x) (- y)+ 2 (- y) (z) + 2 (z) (2x) The coefficient of x2 is 0. They are in a list with arrows. = (y – 1)[2y(y + 1) + 1(y + 1)] (v) We have x10+  y3 + t50 = 10000 – 16 = 9984, Question 3. These NCERT solutions are created by the BYJU’S expert faculties to help students in the preparation of their board exams. p(2) = (2)3 = 8 (iii) We have 3 √t + t√2 = 3 √t1/2 + √2.t ⇒ p(3) = 0, so g(x) is a factor of p(x). [Hint See question 9] Solution: [Using (a – b)3 = a3 – b3 – 3ab (a – b)] In this NCERT Solutions for Class 9 Maths Chapter 2, Polynomials, you learn about the definition of Polynomials which comes from the word “poly” which means “many” and the word “nominal” which means “term”. (ii) Given that p(t) = 2 + t + 2t2 – t3 Rational Numbers, irrational Numbers, rationalize irrational numbres, operation on real numbers, laws of … = (x + 1)(x – 5)(x + 1) ∴ p (-1) = (-1)3 + (-1)2 + (-1) + 1 . (x + a) (x + b) = x2 + (a + b) x + ab [Using a3 – b3 – 3 ab(a – b) = (a – b)3] FREE Downloadable! = 2y2(y – 1) + 3y(y – 1) + 1(y – 1) NCERT Solutions for Class 9 Maths: The Class 9 Maths textbook solutions of exercises for all 15 chapters are included in this article. Find the remainder when x3 + 3x2 + 3x + 1 is divided by = 0 + 0 + 0 + 1 = 1 (ii) The given polynomial is 4- y2. Mathematics Part - II Solutions Textbook Solutions for Class 9 Math. = -8 + 12 – 6 + 1 (iii) We have, p(x) = 2x + 5. It is not a polynomial, because one of the exponents of y is -1, Write the following cubes in expanded form (iii) x3 + 13x2 + 32x + 20 (i) 25a2 – 35a + 12 = 25a2 – 20a – 15a + 12 = 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3) Thus, x3 – 2x2 – x + 2 = (x – 1)(x + 1)(x – 2), (ii) We have, x3 – 3x2 – 9x – 5 Since, p(x) = 0 = (2a + b)(2a + b)(2a + b), (ii) 8a3 – b3 – 12o2b + 6ab2 (ii) y2 + √2 (iii) Let p (x) = x4 + 3x3 + 3x2 + x + 1 . = (2a – b)3 = 2(-1) + 1 + 2 – 1 (i) x + 1 (ii) We have, p(x) x3 + 3x2 + 3x + 1 and g(x) = x + 2 = (3x + y)2 Give one example each of a binomial of degree 35, and of a monomial of degree 100. (i) Let p (x) = x3 + x2 + x + 1 Question from very important topics are covered by NCERT Exemplar Class 9.. You also get idea about the type of questions and method to answer in your Class 9th … Question 12. [Using a3 + b3 + 3 ab(a + b) = (a + b)3] Hence, verified. Using identity, If x + y + z = 0, show that x3 + y3 + z3 = 3 xyz. [Using a2 + 2ab + b2 = (a + b)2] = 10000 + (-9) + 20 = 9120 Solution: Factorise ∴ p(3) = (3)3 – 4(3)2 + 3 + 6 Question 1. (iv) Since, 3 = 3x° [∵ x°=1] (v) p (x) = 3x ∴ p(0) = (0)3 = 0, p(1) = (1)3 = 1 (ii) 2x2 + y2 + 8z2 – 2√2xy + 4√2yz – 8xz (iv) (y2+ \(\frac { 3 }{ 2 }\)) (y2– \(\frac { 3 }{ 2 }\)) These solutions are also applicable for UP board (High School) NCERT Books 2020 – 2021 onward. = (3x -1) (4x -1) ⇒ p(-2) ≠ 0, so g(x) is not a factor of p(x). Thus, x3 + 13x2 + 32x + 20 CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. So, it is a linear polynomial. = (2 a + b)3 = (2x)2 + (3y)2 + (- 4z)2 + 2 (2x) (3y) + 2 (3y) (- 4z) + 2 (- 4z) (2x) = (2y -1)2 Solution: R.H.S CBSE Class 9 Maths Worksheet for students has been used by teachers & students to develop logical, lingual, … ∴ (998)3 = (1000-2)3 (i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz (i) (x + 4)(x + 10) = k – 3 + k NCERT Book NCERT Sol. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1, drop a comment below and we will get back to you at the earliest. (iii) p(2) = 5(2) – 4(2)2 + 3 = 10 – 4(4) + 3 Solution: = x2(x + 1) + 12x(x +1) + 20(x + 1) Hence, verified. (iii) p (x) = x2 – 1, x = x – 1 So, it is a quadratic polynomial. ⇒ x = \(\frac { 2 }{ 3 }\) = (2a – b) (2a – b) (2a – b), (iii) 27 – 125a3 – 135a + 225a2 = 1000000 + 8 + 600(100 + 2) Class 9 maths printable worksheets, online practice and online tests. Using identity, They give a detailed and stepwise explanation of each answer to the problems given in the exercises in the NCERT … (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx, (i) (x + 2y + 4z)2 = 3 x x x (x – 4) which is not a whole number. NCERT Solutions Class 9 Maths Chapter 2 Polynomials. NCERT solutions for session 2019-20 is now available to download in PDF form. We have, (x + y)3 = x3 + y3 + 3xy(x + y) …(1) = 2 + 1 + 2 – 1 = 4 = (y – 1)(y + 1)(2y + 1) Find the remainder when x3 – ax2 + 6x – a is divided by x – a. [Using a3 + b3 + 3 ab(a + b) = (a + b)3] Class 9 Maths Chapter 2 Polynomials This chapter guides you through algebraic expressions called polynomial and various terminologies related to it. ⇒ (x + y)(x2 + y2 – xy) = x3 + y3 The zero of x + 1 is -1. (i) We have 4x2 – 3x + 7 = 4x2 – 3x + 7x0 (iv) x3 – x2 – (2 +√2 )x + √2 (i) We have, 9x2 + 6xy + y2 = 2 x \(\frac { 1 }{ 2 }\) x (x + y + z)(x2 + y2 + z2 – xy – yz – zx) Since, p(1) = 2(1)2 + k(1) + √2 Expand each of the following, using suitable identity (iii) p (x) = 2x + 5 We have, (3x + 4) (3x – 5) = (3x)2 + (4 – 5) x + (4) (- 5) (i) We have, p (x)= 2x3 + x2 – 2x – 1 and g (x) = x + 1 Download free printable assignments for CBSE Class 9 Polynomials with important chapter wise questions, students must practice NCERT Class 9 Polynomials assignments, question booklets, workbooks and topic wise test papers with solutions as it will help them in revision of important and difficult concepts Class 9 Polynomials.Class Assignments for Grade 9 Polynomials, … So, it is a quadratic polynomial. (iii) 104 x 96 Solution: Evaluate the following using suitable identities (v) 5 + 2x = 4x2 + y2 + z2 – 4xy – 2yz + 4zx, (iii) (- 2x + 3y + 2z)2 = (- 2x)2 + (3y)2 + (2z)2 + 2 (- 2x) (3y)+ 2 (3y) (2z) + 2 (2z) (- 2x) ∴ p (-1)= (-1)4 + 3 (-1)3 + 3 (-1)2 + (- 1) + 1 (i) 12x2 – 7x +1 (ii) Volume 12ky2 + 8ky – 20k ∴ P(0) = (0)2 – 0 + 1 = 0 – 0 + 1 = 1 (ii) 4 – y2 [∵ (a2 – b2) = (a + b)(a-b)] = a3 – a3 + 6a – a = 5a Solution: Find the zero of the polynomial in each of the following cases We have, p(x) = 3x3+7x. Solution: Practise the solutions to understand how to use the remainder theorem to work out the remainder of a polynomial. (i) The given polynomial is 2 + x2 + x. [Using (a + b)(a -b) = a2– b2] = (3x + y + z)[(3x)3 + y3 + z3 – (3x × y) – (y × 2) – (z × 3x)] So, the degree of the polynomial is 3. Question 2. ⇒ (x – y)3 + 3xy(x – y) = x3 – y3 = k – √2 + 1 = 0 (v) The degree of 3t is 1. = \(\frac { 1 }{ 2 }\) (x + y + 2)(x2 + y2 + y2 + z2 + z2 + x2 – 2xy – 2yz – 2zx) Factorise the following using appropriate identities Also, p(-1) = (-1)2 -1 = 1 – 1 = 0 (ii) (2x – y + z)2 Find the value of the polynomial 5x – 4x2 + 3 at Chapter 13 Geometrical Constructions. = x2 (x + 1) – 4x(x + 1) – 5(x + 1) Since, \(p(\frac { 1 }{ 2 } )\) ≠ 0, so, x = \(\frac { 1 }{ 2 }\) is not a zero of 2x + 1. Use the Factor Theorem to determine whether g (x) is a factor of p (x) in each of the following cases (i) 8a3 +b3 +12a2b+6ab2 (i) We have, 99 = (100 -1) = (x + 1)(x2 – 5x + x – 5) Class 9 Mathematics Notes for FBISE. Teachoo is free. ⇒ x = \(\frac { -5 }{ 2 }\) Solutions to all NCERT Exercise Questions and Examples of Chapter 2 Class 9 Polynomials are provided free at Teachoo. = (x + 1)(x + 2)(x + 10), (iv) We have, 2y3 + y2 – 2y – 1 [Using (x + a)(x + b) = x2 + (a + b)x + ab] Chapter 2: Polynomials. = (4m – 7n)[(4m)2 + (4m)(7n) + (7n)2] p(1) = (1 – 1)(1 +1) = (0)(2) = 0 Thus, the value of 5x – 4x2 + 3 at x = 0 is 3. (i) (- 12)3 + (7)3 + (5)3 Solution: = – π3 + 3π2 – 3π +1 (ii) x4 + x3 + x2 + x + 1 (iii) 3 √t + t√2 = 10000 + 1000 + 21=11021, (ii) We have, 95 x 96 = (100 – 5) (100 – 4) = (x + 1)(x2 + 2x + 10x + 20) [Using a2 – 2ab + b2 = (a- b)2] Since, p(1) = 0, so x = 1 is a zero of x2 -1. = (100)2-42 Exercise 13.2 Solution. Here, exponent of every variable is a whole number, but x10 + y3 + t50 is a polynomial in x, y and t, i.e., in three variables. Click on exercise or topic link below to get started. = 10 – 16 + 3 = -3 ⇒ 2x + 5 =0 The highest power of the variable x is 3. 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