Thus, the required remainder is -π3 + 3π2 – 3π+1. Thus, x3 + 13x2 + 32x + 20 Chapter-wise NCERT Solutions for Class 9 Maths Chapter 2 Polynomials solved by Expert Teachers as per NCERT (CBSE) Book guidelines. (i) Given that p(y) = y2 – y + 1. = (2a – b) (2a – b) (2a – b), (iii) 27 – 125a3 – 135a + 225a2 Then, x + y + z = -12 + 7 + 5 = 0 (iii) x4 + 3x3 + 3x2 + x + 1 We have, (3x + 4) (3x – 5) = (3x)2 + (4 – 5) x + (4) (- 5) = 0 + 0 + 0 + 1 = 1 The zero of x + 1 is -1. = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) (iii) x2 – \(\frac { { y }^{ 2 } }{ 100 }\) (ii) x3 – y3 = (x – y) (x2 + xy + y2) ⇒ x3 + y3 + 3xy(x + y) = -z3 Thus, 12x2 -7x + 3 = (2x – 1) (x + 3), (ii) We have, 2x2 + 7x + 3 = 2x2 + x + 6x + 3 They give a detailed and stepwise explanation of each answer to the problems given in the exercises in the NCERT … (ii) p(-1) = 5(-1) – 4(-1)2 + 3 = 4k x (3y + 5) x (y – 1) Thus, zero of 3x – 2 is \(\frac { 2 }{ 3 }\). = – π3 + 3π2 – 3π +1 Factorise each of the following (i) Area 25a2 – 35a + 12 Use the Factor Theorem to determine whether g (x) is a factor of p (x) in each of the following cases Teachoo provides the best content available! (i) (99)3 However, it is possible to avoid such a scenario by taking authentic NCERT solutions for class 9 maths from a reliable source. the remainder is not 0. = k – √2 + 1 = 0 ∴ The possible dimensions of the cuboid are 3, x and (x – 4). (i) p(x)=x+5 = 2 + k + √2 =0 = 4[3ky2 + 2ky – 5k] = 4[k(3y2 + 2y – 5)] CBSE Class 9 Maths Worksheet for students has been used by teachers & students to develop logical, lingual, … (iv) p (x) = (x + 1) (x – 2), x = – 1,2 (i) 25a2 – 35a + 12 = 25a2 – 20a – 15a + 12 = 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3) (iii) We have, = x3 – 4x2 + x + 6 and g (x) = x – 3 Extra questions for class 9 maths chapter 1 with solution. Solution: = (3x + y + z)(9x2 + y2 + z2 – 3xy – yz – 3zx). ∴ (998)3 = (1000-2)3 ∴ P(0) = (0)2 – 0 + 1 = 0 – 0 + 1 = 1 ⇒ p (-1) ≠ 0 ⇒ p (- 1) = 0 ∴ 993 = (100 – 1)3 = 2 + 2 + 8 – 8 = 4 It is a polynomial in one variable i.e., x (i) 8a3 +b3 +12a2b+6ab2 = 8x3 + 1 + 6x(2x + 1) (ii) 2x2 + y2 + 8z2 – 2√2xy + 4√2yz – 8xz = 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) = 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) = k – 3 + k So, the degree of the polynomial is 1. = -1 + 1 – 1 + 1 Class-IX CHAPTER – 1 Number System (Maths Assignment) 1. = -14 + 13 = (2x + 3y – 4z)2 = (2x + 3y + 4z) (2x + 3y – 4z), (ii) 2x2 + y2 + 8z2 – 2√2xy + 4√2yz – 8xz Question 2. (iii) The given polynomial is 5t – √7 . (vi) r2 = (2a)3 + (b)3 + 6ab(2a + b) The highest power of the variable x is 3. (x + a) (x + b) = x2 + (a + b) x + ab Students first revise all the topics from NCERT book and then Solve the sums in this worksheet. (ii) x4 + x3 + x2 + x + 1 Chapter-9 Chapter-2 Sol. = x2(x + 1) + 12x(x +1) + 20(x + 1) Write the following numbers in p/q form (i) 2.015 (ii) 0.235 Ans (399 235 ' 198 999) 4. and zero of 7 + 3x is \(-\frac { 7 }{ 3 }\). R.H.S 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. Evaluate the following products without multiplying directly Find p (0), p (1) and p (2) for each of the following polynomials. = 4x2 + 9y2 + 4z2 – 12xy + 12yz – 8zx, (iv) (3a -7b- c)2 = (3a)2 + (- 7b)2 + (- c)2 + 2 (3a) (- 7b) + 2 (- 7b) (- c) + 2 (- c) (3a) So, (x+ 1) is a factor of x3 + x2 + x + 1. Find the remainder when x3 + 3x2 + 3x + 1 is divided by Since, p(x) = 0 ⇒ x – 5 = 0 ⇒ x = -5 1et p(x) = 5x – 4x2 + 3 (i) Abmomial of degree 35 can be 3x35 -4. Chapter 14 Probability. Solution: [Using (a + b)(a -b) = a2– b2] Expand each of the following, using suitable identity ⇒ p(3) = 0, so g(x) is a factor of p(x). These expert faculties solve and provide the NCERT Solution for class 9 so that it would help students to solve the problems comfortably. Mathematics Part - II Solutions Textbook Solutions for Class 9 Math. = x2 + 14x+40, (ii) We have, (x+ 8) (x -10) Classify the following as linear, quadratic and cubic polynomials. Ex 2.1 Class 9 Maths Question 2. (v) We have, p(x) = x2 ⇒ 2x + 5 =0 Since, p(x) = 0 = (x + 1)[x(x – 5) + 1(x – 5)] = (y – 1)(2y2 + 2y + y + 1) = \(\frac { 1 }{ 2 }\)(x + y + z)[(x – y)2+(y – z)2+(z – x)2] = 2 + 0 + 0 – 0=2 = (2a + b)(2a + b)(2a + b), (ii) 8a3 – b3 – 12o2b + 6ab2 We have, (x + y)3 = x3 + y3 + 3xy(x + y) …(1) ∴ P(-1) = (-1)4 + (-1)3 + (-1)2 + (-1)+1 Find the value of k, if x – 1 is a factor of p (x) in each of the following cases It is a polynomial in one variable i.e., y = (4a)3 – (3b)3 – 3(4a)(3b)(4a – 3b) = 4k[3y2 – 3y + 5y – 5] (iv) Since, 3 = 3x° [∵ x°=1] If x + y + z = 0, show that x3 + y3 + z3 = 3 xyz. (iv) p (x) = 3x – 2 = (4a – 3b)3 = 1000000 + 8 + 60000 + 1200 = 1061208, (iii) We have, 998 = 1000 – 2 Thus, 2y3 + y2 – 2y – 1 Question 3. Since, p(x) = 0 (i) The given polynomial is 2 + x2 + x. 10 Questions. (i) p(y) = y2 – y +1 (iv) (3a -7b – c)z Solution: = 2(-1) + 1 + 2 – 1 Since, p(1) = 2(1)2 + k(1) + √2 (iv) p (x) = (x-1) (x+1) Class 9 maths printable worksheets, online practice and online tests. ⇒ (x + y)3 – 3(x + y)(xy) = x3 + y3 (i) We have 4x2 – 3x + 7 = 4x2 – 3x + 7x0 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. Solution: (iv) √2 x – 1 ⇒ 2x = -5 The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.. For a better understanding of this chapter… (iii) Given that p(x) = x3 (iii) Let p (x) = x4 + 3x3 + 3x2 + x + 1 . Solution: Solution: Login to view more pages. Hence, verified. Thus, x3 – 3x2 – 9x – 5 = (x + 1)(x – 5)(x +1), (iii) We have, x3 + 13x2 + 32x + 20 (ii) (2x – y + z)2 For (x – 1) to be a factor of p(x), p(1) should be equal to 0. (ii) p (t) = 2 +1 + 2t2 -t3 All the chapterwise questions with solutions to help you to revise complete CBSE syllabus and score more marks in Your board examinations. (ii) (28)3 + (- 15)3 + (- 13)3 (ii) y2 + √2 Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1. = (2x + 1)(x + 3) = 3[-420] = -1260, (ii) We have, (28)3 + (-15)3 + (-13)3 (iii) p (x) = x3 – 4x2 + x + 6, g (x) = x – 3 = (y – 1)(2y2 + 3y + 1) = (x + 1)(x2 + 12x + 20) (i) (x + 4)(x + 10) So, the degree of the polynomial is 3. (v) (3 – 2x) (3 + 2x) = [(x)2 – (1)2](x – 2) = 1 – 1 + 1 – 1 + 1 Thus, the required remainder is 5a. (ii) We have, 12ky2 + 8ky – 20k = 8x3 + 12x2 + 6x + 1, (ii) (2a – 3b)3 = (2a)3 – (3b)3 – 3(2a)(3b)(2a – 3b) [By (2)] Factorise = (3y + 5z)[(3y)2 – (3y)(5z) + (5z)2] Click on exercise or topic link below to get started. (i) 4 x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz (iv) 1 + x [Using a3 + b3 + 3 ab(a + b) = (a + b)3] Factorise each of the following 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”. Using identity, We have, p(x) = 3x3+7x. ∴ p(o) = (0)2 = 0 = ( 100)2 + [(- 5) + (- 4)] 100 + (- 5 x – 4) Ex 2.1 Class 9 Maths Question 1. (i) (x+2y+ 4z)2 (v) 3t (i) x + 1 ⇒ 3x = 2 Since, p(x) = 0 => ax = 0 => x-0 Question 4. To help students in making easy preparations, we are providing the MCQs for class 9 NCERT Maths. (iii) x3 + 13x2 + 32x + 20 power of the variable y is 2. Thus, the value of 5x – 4x2 + 3 at x = 0 is 3. [Using a3 – b3 – 3 ab(a – b) = (a – b)3] ∴ \(p(-\frac { 1 }{ 3 } )\quad =\quad 3(-\frac { 1 }{ 3 } )\quad +\quad 1\quad =\quad -1\quad +\quad 1\quad =\quad 0\) = (2x)2 + (3y)2 + (- 4z)2 + 2 (2x) (3y) + 2 (3y) (- 4z) + 2 (- 4z) (2x) (iii) The zero of x is 0. (iv) x3 – x2 – (2 +√2 )x + √2 Solution: (viii) p (x) = 2x + 1, x = \(\frac { 1 }{ 2 }\) = x2 – 2x – 80, (iii) We have, (3x + 4) (3x – 5) Solutions to all NCERT Exercise Questions and Examples of Chapter 2 Class 9 Polynomials are provided free at Teachoo. Thus, the required remainder = 0, (ii) The zero of \(x-\frac { 1 }{ 2 }\) is \(\frac { 1 }{ 2 }\) (iii) p(2) = 5(2) – 4(2)2 + 3 = 10 – 4(4) + 3 (i) p(0) = 5(0) – 4(0)2 + 3 = 0 – 0 + 3 = 3 = 1000000 – 1 – 300(100 – 1) Let x = 28, y = -15 and z = -13. [Using (x + a)(x + b) = x2 + (a + b)x + ab] = (3x + y + z)[(3x)3 + y3 + z3 – (3x × y) – (y × 2) – (z × 3x)] Chapter -1 Sol. = (3x -1) (4x -1) In this article, you will get the MCQs on Class 9 Maths Chapter 4: Linear Equations in Two Variables. Using the identity, (i) The zero of x + 1 is -1. NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1 are part of NCERT Solutions for Class 9 Maths. Telanagana State Board Class 9 Mathematics Textbook Solution Chapter-wise Telanagana SCERT Class IX Math Solution. Let x = -12, y = 7 and z = 5. There is plenty to learn in this chapter about the definition and examples of polynomials, coefficient, degrees, and terms in a polynomial. = 4x2 + 25y2 + 9z2 – 20xy – 30yz + 12zx, Question 5. Volume of a cuboid = (Length) x (Breadth) x (Height) (iv) x + π We hope the NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1, help you. Since, p(x) = 0 Solution: Along with recalling the knowledge of linear … Chapter 4 Linear Equations in Two Variables. After an in-depth analysis, our expert panel has drafted the solutions so that students of class 9 can easily refer to them during their exams or to complete their homework. = (3 – 5a) (3 – 5a) (3 – 5a), (iv) 64a3 -27b3 -144a2b + 108ab2 = (x + 1)(x + 2)(x + 10) = (3x + y)2 = x3 + x2 – 4x2 – 4x – 5x – 5 , [∵ (a2 – b2) = (a + b)(a-b)] (ii) x3 – 3x2 – 9x – 5 (i) Let p (x) = x3 + x2 + x + 1 So, it is a quadratic polynomial. ∴ p(a) = (a)3 – a(a)2 + 6(a) – a Since, p(1) = k(1)2 – (1) + 1 = 4k[(3y + 5) x (y – 1)] (ii) (102)3 You have these advantages of browsing notes from our website. (ii) 8a3 -b3-12a2b+6ab2 Write the coefficients of x2 in each of the following NCERT Solutions for Class 9 Maths are a set of solutions in the form of chapter-wise solutions made specifically for Class 9th students. We have, p(x) = x3 – ax2 + 6x – a and zero of x – a is a. = 1000000000 – 8 – 6000(1000 – 2) (vii) p (x) = cx + d, c ≠ 0 where c and d are real numbers. Thus, the possible length and breadth are (7y – 3) and (5y + 4). Chapter-10 Chapter-3 Sol. Thus, 2×2 + 7x + 3 = (2x + 1)(x + 3), (iii) We have, 6x2 + 5x – 6 = 6x2 + 9x – 4x – 6 (vi) p (x) = 1x + m, x = – \(\frac { m }{ 1 }\) [Using (a + b)3 = a3 + b3 + 3ab (a + b)] ∴ 1023 = (100 + 2)3 = (x + 1)(x2 – 5x + x – 5) (ii) A monomial of degree 100 can be √2y100. = 2y2(y – 1) + 3y(y – 1) + 1(y – 1) Verify 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. Thus, zero of 2x + 5 is \(\frac { -5 }{ 2 }\) . = 8a3 – 27b3 – 36a2b + 54ab2, Question 7. = 10000 + 1000 + 21=11021, (ii) We have, 95 x 96 = (100 – 5) (100 – 4) k = -2 – √2 = -(2 + √2), (iii) Here, p (x) = kx2 – √2 x + 1 Evaluate the following using suitable identities NCERT Solutions for Class 9 Maths: The Class 9 Maths textbook solutions of exercises for all 15 chapters are included in this article. The coefficient of x2 is 1. (iii) We have, p(x) = 2x + 5. NCERT solutions for session 2019-20 is now available to download in PDF form. Thus, 3x2 – x – 4 = (3x – 4)(x + 1), Question 5. 3. Since, p(2) = 0, so, x = 2 is also a zero of (x + 1)(x – 2). Important questions in Number systems with video lesson. 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Thus, the value of 5x – 4x2 + 3 at x = 2 is – 3. NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4; NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3; NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2; NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Ex 9.2 ⇒ x = -5. Note: Important questions have also been marked for your reference. Thus, the possible length and breadth are (5a – 3) and (5a – 4). P(2) = (2 – 1)(2 + 1) = (1)(3) = 3. = (- √2x)2 + (y)2 + (2 √2z)2y + 2(- √2x) (y)+ 2 (y) (2√2z) + 2 (2√2z) (- √2x) 12x2 – 7x + 1 = 12x2 – 4x- 3x + 1 (i) We have, ∴ p(3) = (3)3 – 4(3)2 + 3 + 6 Ex 2.1 Class 9 Maths Question 5. Thus, zero of cx + d is \(-\frac { d }{ c }\), Question 1. (iii) (- 2x + 3y + 2z)2 Solution: (i) p (x)= 2x3 + x2 – 2x – 1, g (x) = x + 1 (iii) y + y2+4 Using identity, Ex 2.1 Class 9 Maths Question 3. (ii) The given polynomial is 4- y2. x3 – y3 = (x – y)(x2 + xy + y2) ⇒ x3 + y3 – 3xyz = -z3 (v) 5 + 2x GoyalAssignments.com provides Free Downloads of Study materials (Assignments, Model Test Paper, Sample Paper, Previous Paper) of all subjects for CBSE Class 9 & 10 students. = 27 – 36 + 3 + 6 = 0 State reasons for your answer. = 10000 + (10) x 100 + 21 (iv) p (x) = kx2 – 3x + k Solution: = 10000 + (-9) + 20 = 9120 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, … Since, p(x) = 0 (viii) We have, p(x) = 2x + 1 Represent geometrically 8.1 on number line. 27x3 + y3 + z3 – 9xyz = (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z) (v) The zero of 5 + 2x is \(-\frac { 5 }{ 2 }\) . = (2y -1)2 = (x + 1)(x – 5)(x + 1) = (100)2-42 Extra questions along with questions of NCERT book complete the topic . All these questions are based on the important fundamental concepts given in NCERT Class 9 Maths. CBSE Worksheets for Class 11 Maths: One of the best teaching strategies employed in most classrooms today is Worksheets. Prepare effectively for your Maths exam with our NCERT Solutions for CBSE Class 9 Mathematics Chapter 2 Polynomials. ⇒ (x – y)3 + 3xy(x – y) = x3 – y3 Ex 2.1 Class 9 Maths Question 4. = (2y – 1)(2y – 1 ), Question 4. (ii) 64m3 – 343n3 (v) p (x) = 3x (iv) 3 ⇒ p(-1) = 0, so g(x) is a factor of p(x). (iii) 6x2 + 5x – 6 (i) We have, p (x)= 2x3 + x2 – 2x – 1 and g (x) = x + 1 Middle school is one of the most crucial periods a student undergoes in the course of their schooling years. ⇒ 3x = 0 ⇒ x = 0 (iii) p (x) = 2x + 5 (iii) The given polynomial is \(\frac { \pi }{ 2 } { x }^{ 2 }\) + x. Since,( \(-\frac { 490 }{ 9 }\)) ≠ 0 = 3 x x x (x – 4) = x3 + x2 + 12x2 + 12x + 20x + 20 = ( 100)2 + (3 + 7) (100)+ (3 x 7) So, it is a quadratic polynomial. p(1) = (1 – 1)(1 +1) = (0)(2) = 0 = 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) = 2√2 = (4m – 7n)[(4m)2 + (4m)(7n) + (7n)2] = 9x2 – x – 20, Question 2. p(-π) = (-π)3 + 3(- π)22 + 3(- π) +1 = 1 – 3 + 3 – 1 + 1 = 1 NCERT Solutions Class 9 Maths Chapter 2 Polynomials are provided here. 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. (iv) Given that p(x) = (x – 1)(x + 1) These ncert book chapter wise questions and answers are very helpful for CBSE board exam. (iii) x ⇒ (x + y)(x2 + y2 – xy) = x3 + y3 DronStudy provides you Chapter wise Solutions for Class 9th Maths, Science and Social Studies. Thus, x3 – 2x2 – x + 2 = (x – 1)(x + 1)(x – 2), (ii) We have, x3 – 3x2 – 9x – 5 (ii) x – \(\frac { 1 }{ 2 }\) We know that if x + y + z = 0, then, x3 + y3 + z3 = 3xyz (ii) p (x) = 2x2 + kx + √2 We have, (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z) Hence, verified. (i) x3 + y3 = (x + y)-(x2 – xy + y2) Solution: (v) x10+ y3+t50 (i)We have, 103 x 107 = (100 + 3) (100 + 7) ⇒ x + 5 = 0 Chapter 9 Trigonometric Ratios – Mathematics Easter Holiday Assignment Chapter 9.2 Finding Trigonometric Ratios by Constructing Right-Angled Triangles Key Points If one of the trigonometric ratios of an acute angle θ = 4 1 cos e.g. Verify whether the following are zeroes of the polynomial, indicated against them. Since, \(p(\frac { 1 }{ 2 } )\) ≠ 0, so, x = \(\frac { 1 }{ 2 }\) is not a zero of 2x + 1. So, it is a cubic polynomial. (i) 4x2 – 3x + 7 = 1000000000 – 8 – 6000000 +12000 sin θ and tan θ) without evaluating θ. Since, p(-1) = 0, so, x = -1 is a zero of (x + 1)(x – 2). CLASS 9 IX Sample Papers X Sample Papers CLASS 8 CLASS 07 Class 06 ... Chapter - 2: Polynomials. The coefficient of x2 is -1. (ii) Volume 12ky2 + 8ky – 20k Since p(-1) = 0, so, x = -1, is also a zero of x2 – 1. We have, (x + 4) (x + 10) = x2+(4 + 10) x + (4 x 10) (ii) Given that p(t) = 2 + t + 2t2 – t3 (i) ∵ (x + y)3 = x3 + y3 + 3xy(x + y) (ii) p (x) = 5x – π, x = \(\frac { 4 }{ 5 }\) The highest power of variable t is 1. (i) x2+ x (i) 9x2 + 6xy + y2 (iv) (y2+ \(\frac { 3 }{ 2 }\)) (y2– \(\frac { 3 }{ 2 }\)) (iii) The degree of y + y2 + 4 is 2. Solution: = (4a – 3b)(4a – 3b)(4a – 3b). We have, 27y3 + 125z3 = (3y)3 + (5z)3 CBSE Class 11 Maths Worksheet for students has been used by teachers & students to develop logical, lingual, analytical, and … (iii) (998)3 Answers to each and every question is explained in an easy to understand way, with videos of all the questions. (i) Here, p(x) = x2 + x + k because each exponent of x is a whole number. [Using (x + a)(x + b) = x2 + (a + b)x + ab] Question 2. Factorise CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. ⇒ (x – y)[(x – y)2 + 3xy)] = x3 – y3 = (2y)2 + 2(2y)(1) + (1)2 (i) x3+x2+x +1 [Using a3 + b3 + 3 ab(a + b) = (a + b)3] It is not a polynomial, because one of the exponents of t is \(\frac { 1 }{ 2 }\), Question 10. = 2 x \(\frac { 1 }{ 2 }\) x (x + y + z)(x2 + y2 + z2 – xy – yz – zx) (v) p (x) = x2, x = 0 These solutions are also applicable for UP board (High School) NCERT Books 2020 – 2021 onward. (vi) We have, p(x) = ax, a ≠ 0. Learn about the degree of a polynomial and the zero of a polynomial with related Maths solutions. (ii) Area 35y2 + 13y – 12 = -8 + 12 – 6 + 1 (i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz CBSETuts.com provides you Free PDF download of NCERT Exemplar of Class 9 Maths chapter 2 Polynomials solved by expert teachers as per NCERT (CBSE) book guidelines. Hence, if x + y + z = 0, then Contains solved exercises, review questions, MCQs, important board questions and chapter overview. (iii) x = 2 (v) We have x10+  y3 + t50 = a3 – a3 + 6a – a = 5a [Using a2 + 2ab + b2 = (a + b)2] Thus, the value of 5x – 4x2 + 3 at x = -1 is -6. (iii) 27-125a3 -135a+225a2 Solution: Solution: (x+ a) (x+ b) = x2 + (a + b) x+ ab. 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. = (x – 1)(x + 1)(x – 2) (i) We have, 3x2 – 12x = 3(x2 – 4x) = x2 (x + 1) – 4x(x + 1) – 5(x + 1) (ii) 4 – y2 Chapter-2 Chapter-10 Sol. (vii) The degree of 7x3 is 3. (i) 2 + x2 + x = (3x + y)(3x + y), (ii) We have, 4y2 – 4y + 12 ⇒ k = -2. Download NCERT Solutions For Class 9 Maths in PDF based on latest pattern of CBSE in 2020 - 2021. Factorise the following using appropriate identities which is not a whole number. (ii) p (x) = x – 5 These assignments will be available in updated form along with new assignments and chapter wise tests with solutions. Solution: (iii) 104 x 96 Question 16. Thus, the possible dimensions of the cuboid are 4k, (3y + 5) and (y -1). [Using a3 + b3 + 3 ab(a + b) = (a + b)3] Use suitable identities to find the following products (ii) 2x2 + 7x + 3 which is not a whole number. (iv) The given polynomial is √2 x – 1. = (1000)3– (2)3 – 3(1000)(2)(1000 – 2) Get here MCQs on Class 9 Maths Chapter 2- Polynomials with answers. Solution: Give one example each of a binomial of degree 35, and of a monomial of degree 100. So, it is a cubic polynomial. Class 9 Chapter 2 Polynomials Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks. Write the following cubes in expanded form Download File. Solution: Homework Help with Chapter-wise solutions and Video explanations. Then, x + y + z = 28 – 15 – 13 = 0 P(1) = 2 + 1 + 2(1)2 – (1)3 (i) p(x) = 3x + 1,x = –\(\frac { 1 }{ 3 }\) (ii) We have, p(x) = x – 5. (i) The degree of x2 + x is 2. = x(3x – 4) + 1(3x – 4) = (3x – 4)(x + 1) Using identity, (iv) 2y3 + y2 – 2y – 1 = (- √2x + y + 2 √2z) (- √2x + y + 2 √2z), Question 6. (iii) P (x) = x3 (iii) p (x) = kx2 – √2 x + 1 = 1000000 -1 – 30000 + 300 = -1 Answers to each and every question is explained in an easy to understand way, with videos of all the questions. FREE Downloadable! Exercise 14.1 Solution. = 4k[3y(y – 1) + 5(y – 1)] = -2 + 1 + 2 -1 = 0 We have, (iv) We have, p(x) = 3x – 2. x3 + y3 + z3 = 3xyz, Question 14. Download free printable assignments for CBSE Class 9 Mathematics with important chapter wise questions, students must practice NCERT Class 9 Mathematics 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 Mathematics.Class Assignments for Grade 9 Mathematics, … = x2 + (2y)2 + (4z)2 + 2 (x) (2y) + 2 (2y) (4z) + 2(4z) (x) ∴ p(-1) = 2(-1)3 + (-1)2 – 2(-1) – 1 (iii) 3 √t + t√2 Since, p(0) = 0, so, x = 0 is a zero of x2. ⇒ x = \(\frac { 2 }{ 3 }\) Without actually calculating the cubes, find the value of each of the following = (3 – 5a)3 = 1000300 – 30001 = 970299, (ii) We have, 102 =100 + 2 = 8a3 – 27b3 – 18ab(2a – 3b) p( 2) = 2 + 2 + 2(2)2 – (2)3 (vii) 7x3 p(1) = (1)2 – 1 + 1 = 1 – 1 + 1 = 1 In this … ⇒ p(-2) ≠ 0, so g(x) is not a factor of p(x). = 10000 + (-900) + 20 = 9120, (iii) We have 104 x 96 = (100 + 4) (100 – 4) ⇒ p(1) = k + 2 = 0 because each exponent of y is a whole number. x3 + y3 = (x + y)(x2 – xy + y2) ⇒ (x – y)(x2 + y2 + xy) = x3 – y3 Factorise 27x3 +y3 +z3 -9xyz. (ii) 35y2+ 13y -12 = 35y2 + 28y – 15y -12 (i) We have , p(x) = 3x + 1 (ii) The degree of x – x3 is 3. Thus, 6x2 + 5x – 6 = (2x + 3)(3x – 2), (iv) We have, 3x2 – x – 4 = 3x2 – 4x + 3x – 4 (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx, (i) (x + 2y + 4z)2 ∴ (28)3 + (-15)3 + (-13)3 = 3(28)(-15)(-13) ∴ p (- 1) =(- 1)3- (-1)2 – (2 + √2)(-1) + √2 Hindi Medium and English Medium both are available to free download. Solution: ∴ p(0) = (0)3 = 0, p(1) = (1)3 = 1 ∴ p(-1) = (-1)3 + 3(-1)2 + 3(-1) +1 ⇒ 3x – 2 = 0 We know that Question 1. (ii) Let p (x) = x4 + x3 + x2 + x + 1 Get NCERT solutions for Class 9 Maths free with videos of each and every exercise question and examples. = 27 – 4(9) + 3 + 6 After the completion of chapters from NCERT Books and Exemplar Books, students should go for Assignments. (ii) 2 – x2 + x3 Thus, the required remainder is \(-\frac { 27 }{ 8 }\) . 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