(iii) x (iv) 64a3 -27b3 -144a2b + 108ab2 (iv) The degree of 1 + x is 1. Question 1. Thus, 3x2 – x – 4 = (3x – 4)(x + 1), Question 5. It is not a polynomial, because one of the exponents of y is -1, Prepare effectively for your Maths exam with our NCERT Solutions for CBSE Class 9 Mathematics Chapter 2 Polynomials. = (x + 1)(x2 + 12x + 20) (i) 9x2 + 6xy + y2 Since, p(1) = (1)2 +1 + k = x(2x + 1) + 3(2x + 1) Solution: So, it is a quadratic polynomial. (ii) The given polynomial is 4- y2. = 994011992, Question 8. (iii) P (x) = x3 (i) We have , p(x) = 3x + 1 = -1 – 1 + 2 + √2 + √2 (ii) x3 – 3x2 – 9x – 5 Represent geometrically 8.1 on number line. (iv) The zero of x + π is -π. 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 … = 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) Hence, verified. = 1000000000 – 8 – 6000000 +12000 We have, (3x + 4) (3x – 5) = (3x)2 + (4 – 5) x + (4) (- 5) Get here MCQs on Class 9 Maths Chapter 2- Polynomials with answers. Verify It is not a polynomial, because one of the exponents of t is \(\frac { 1 }{ 2 }\), So, it is a linear polynomial. [∵ (a2 – b2) = (a + b)(a-b)] Find the zero of the polynomial in each of the following cases Solution: (ii) Here, p (x) = 2x2 + kx + √2 = (y – 1)(2y2 + 3y + 1) Since, p(x) = 0 ⇒ x – 5 = 0 ⇒ x = -5 = – 5x – 4x2 + 3 = -9 + 3 = -6 NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12. Factorise So, it is a quadratic polynomial. (iii) (3x + 4) (3x – 5) = x2(x + 1) + 12x(x +1) + 20(x + 1) Download NCERT Solutions For Class 9 Maths in PDF based on latest pattern of CBSE in 2020 - 2021. Middle school is one of the most crucial periods a student undergoes in the course of their schooling years. which is not a whole number. Chapter -1 Sol. What are the possible expressions for the dimensions of the cuboids whose volumes are given below? = 10000 + (-900) + 20 = 9120, (iii) We have 104 x 96 = (100 + 4) (100 – 4) We have, (x + 8) (x – 10) = x2 + [8 + (-10)] x + (8) (- 10) (i) p(y) = y2 – y +1 (i) x = 0 = 3(5460) = 16380. ⇒ k = \(\frac { 3 }{ 4 }\), Question 4. ⇒ k = √2 -1, (iv) Here, p(x) = kx2 – 3x + k Extra questions for class 9 maths chapter 1 with solution. (ii) y2 + √2 = 2 + 0 + 0 – 0=2 So, (x + 1) is not a factor of x4 + 3x3 + 3x2 + x+ 1. 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. Thus, the required remainder is \(-\frac { 27 }{ 8 }\) . (ii) p (t) = 2 +1 + 2t2 -t3 NCERT Solutions for Class 9 Maths: The Class 9 Maths textbook solutions of exercises for all 15 chapters are included in this article. Solution: = 1 – 3 + 3 – 1 + 1 = 1 Evaluate the following products without multiplying directly Then, x + y + z = -12 + 7 + 5 = 0 Using identity, = 27 – 4(9) + 3 + 6 = (1000)3– (2)3 – 3(1000)(2)(1000 – 2) = x2 – 2x – 80, (iii) We have, (3x + 4) (3x – 5) (iii) We have, = x3 – 4x2 + x + 6 and g (x) = x – 3 Question 9. (ii) We have, p(x) x3 + 3x2 + 3x + 1 and g(x) = x + 2 Also, p(-1) = (-1)2 -1 = 1 – 1 = 0 (ii) We have, 12ky2 + 8ky – 20k Solution: p(2) = (2)2 – 2 + 1 = 4 – 2 + 1 = 3 (ii) x – \(\frac { 1 }{ 2 }\) (iv) y+ \(\frac { 2 }{ y }\) We know that if x + y + z = 0, then, x3 + y3 + z3 = 3xyz NCERT Solutions Class 9 Maths Chapter 2 Polynomials are provided here. CLASS 9 IX Sample Papers X Sample Papers CLASS 8 CLASS 07 Class 06 ... Chapter - 2: Polynomials. = 4k[3y2 – 3y + 5y – 5] The coefficient of x2 is -1. (ii) x = – 1 and zero of 7 + 3x is \(-\frac { 7 }{ 3 }\). (i) (99)3 Chapter-2 Chapter-10 Sol. Check whether 7 + 3x is a factor of 3x3+7x. (ii) p (x) = x – 5 = (x + 1)(x2 – 4x – 5) [Using (x + a)(x + b) = x2 + (a + b)x + ab] is given, we can find the other two trigonometric ratios (i.e. (i) 8a3 +b3 + 12a2b+6ab2 So, (x + 1) is not a factor of x3 – x2 – (2 + √2) x + √2. = x3 + x2 – 4x2 – 4x – 5x – 5 , Factorise each of the following Answers to each and every question is explained in an easy to understand way, with videos of all the questions. You have these advantages of browsing notes from our website. (iv) We have, p(x) = (x + 1)(x – 2) NCERT Solutions for Class 9 Maths are a set of solutions in the form of chapter-wise solutions made specifically for Class 9th students. = 4[3ky2 + 2ky – 5k] = 4[k(3y2 + 2y – 5)] = ( 100)2 + (3 + 7) (100)+ (3 x 7) Thus, the required remainder is 5a. (iii) p (x) = x2 – 1, x = x – 1 ∴ 3x3 + 7x is not divisib1e by 7 + 3x. = 2√2 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. = 8x3 + 1 + 6x(2x + 1) Question 10. x3 + y3 + z3 = 3xyz, Question 14. the remainder is not 0. = 4k[(3y + 5) x (y – 1)] Solution: (i) Abmomial of degree 35 can be 3x35 -4. Thus, zero of x + 5 is -5. = 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) Area of a rectangle = (Length) x (Breadth) After the completion of chapters from NCERT Books and Exemplar Books, students should go for Assignments. (i) x3 – 2x2 – x + 2 Solution: = (3x + y + z)[(3x)3 + y3 + z3 – (3x × y) – (y × 2) – (z × 3x)] Terms of Service. [Hint See question 9] = (2y – 1)(2y – 1 ), Question 4. (i) 4x2 – 3x + 7 Chapter-wise NCERT Solutions for Class 9 Maths Chapter 2 Polynomials solved by Expert Teachers as per NCERT (CBSE) Book guidelines. (x + a) (x + b) = x2 + (a + b) x + ab which is not a whole number. (ii) x – x3 ⇒ (x – y)(x2 + y2 + xy) = x3 – y3 They give a detailed and stepwise explanation of each answer to the problems given in the exercises in the NCERT … [Using (a – b)3 = a3 – b3 – 3ab (a – b)] So, the degree of the polynomial is 3. = (4a)3 – (3b)3 – 3(4a)(3b)(4a – 3b) (ii) 4y2-4y + 1 Solutions to all NCERT Exercise Questions and Examples of Chapter 2 Class 9 Polynomials are provided free at Teachoo. = a3 – a3 + 6a – a = 5a = 1000300 – 30001 = 970299, (ii) We have, 102 =100 + 2 Let x = 28, y = -15 and z = -13. = k – 3 + k = (3x -1) (4x -1) This solution is strictly revised in accordance … CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. = 7y(5y + 4) – 3(5y + 4) = (5 y + 4)(7y – 3) (i) (x+2y+ 4z)2 = 2(-1) + 1 + 2 – 1 We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz [Using a3 + b3 + 3 ab(a + b) = (a + b)3] CBSE Class 11 Maths Worksheet for students has been used by teachers & students to develop logical, lingual, analytical, and … ∴ p (- 1) =(- 1)3- (-1)2 – (2 + √2)(-1) + √2 (ii) The given polynomial is 2 – x2 + x3. = x2 (x + 1) – 4x(x + 1) – 5(x + 1) = 10000 – 16 = 9984, Question 3. ∴ (28)3 + (-15)3 + (-13)3 = 3(28)(-15)(-13) = 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) (vii) We have, p(x) = cx + d. Since, p(x) = 0 Expand each of the following, using suitable identity = (2y -1)2 Question 3. (i) We have, x3 – 2x2 – x + 2 Also, p( 2) = (2 + 1)(2 – 2) = (3)(0) = 0 Verify whether the following are zeroes of the polynomial, indicated against them. = (4a – 3b)3 = -1 + 1 – 1 + 1 DronStudy provides you Chapter wise Solutions for Class 9th Maths, Science and Social Studies. = 3[-420] = -1260, (ii) We have, (28)3 + (-15)3 + (-13)3 = (4a – 3b)(4a – 3b)(4a – 3b). = (- √2x)2 + (y)2 + (2 √2z)2y + 2(- √2x) (y)+ 2 (y) (2√2z) + 2 (2√2z) (- √2x) (i) x + 1 27x3 + y3 + z3 – 9xyz = (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z) ⇒ 3x – 2 = 0 Which of the following expressions are polynomials in one variable and which are not? (iv) p (x) = 3x – 2 NCERT Solutions Class 9 Maths Chapter 2 Polynomials. [Using (a + b)3 = a3 + b3 + 3ab (a + b)] (i) (x + 4)(x + 10) Download File. = 10000 + 1000 + 21=11021, (ii) We have, 95 x 96 = (100 – 5) (100 – 4) Solution: ∴ p(3) = (3)3 – 4(3)2 + 3 + 6 (i) p (x)= 2x3 + x2 – 2x – 1, g (x) = x + 1 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. (viii) We have, p(x) = 2x + 1 Ex 2.1 Class 9 Maths Question 5. = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) ⇒ x = -5. (i) Area 25a2 – 35a + 12 Chapter-10 Chapter-3 Sol. Hence, verified. Thus, 2y3 + y2 – 2y – 1 = (3y + 5z)(9y2 – 15yz + 25z2), (ii) We know that Without actually calculating the cubes, find the value of each of the following Extra questions based on the topic Number System. = 1000000 + 8 + 60000 + 1200 = 1061208, (iii) We have, 998 = 1000 – 2 p(1) = (1 – 1)(1 +1) = (0)(2) = 0 The coefficient of x2 is 1. Thus, the required remainder is -π3 + 3π2 – 3π+1. R.H.S = (3y + 5z)[(3y)2 – (3y)(5z) + (5z)2] Unit 1 - Matrices & Determinants. Chapter 4 Linear Equations in Two Variables. = 10000 + (-9) + 20 = 9120 Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 Exercise 1.5 Exercise 1.6 ... Class 9 Mathematics Notes are free and will always remain free. Solution: = 8a3 – 27b3 – 36a2b + 54ab2, Question 7. = 10 – 16 + 3 = -3 Solution: Find p (0), p (1) and p (2) for each of the following polynomials. Question 3. Chapter 13 Geometrical Constructions. p(1) = k(1)2 – 3(1) + k The highest power of the variable x is 3. = \(\frac { 1 }{ 2 }\) (x + y + 2)[(x2 + y2 – 2xy) + (y2 + z2 – 2yz) + (z2 + x2 – 2zx)] Find the remainder when x3 + 3x2 + 3x + 1 is divided by NCERT Book NCERT Sol. Thus, x3 – 2x2 – x + 2 = (x – 1)(x + 1)(x – 2), (ii) We have, x3 – 3x2 – 9x – 5 (iii) (998)3 (iv) (y2+ \(\frac { 3 }{ 2 }\)) (y2– \(\frac { 3 }{ 2 }\)) = 9x2 – x – 20, Question 2. (ii) (102)3 = (y – 1)(y + 1)(2y +1), Question 1. ∴ P(0) = (0)2 – 0 + 1 = 0 – 0 + 1 = 1 Along with recalling the knowledge of linear … = \(\frac { 1 }{ 2 }\) (x + y + 2)(x2 + y2 + y2 + z2 + z2 + x2 – 2xy – 2yz – 2zx) (iii) x2 – \(\frac { { y }^{ 2 } }{ 100 }\) ∴ p(-1) = (-1 +1) (-1 – 2) = (0)(- 3) = 0 For (x – 1) to be a factor of p(x), p(1) should be equal to 0. (i) We have, (-12)3 + (7)3 + (5)3 (iii) 27-125a3 -135a+225a2 So, it is a linear polynomial. Question 2. So, it is a quadratic polynomial. = (3x)2 + 2(3x)(y) + (y)2 Question 15. (iv) 1 + x Chapter wise assignments for class 9 Maths are given below updated for new academic session 2020-2021. Chapter - 3 Pair of Linear Equations. = (2y)2 + 2(2y)(1) + (1)2 So, the degree of the polynomial is 0. Exercise 13.1 Solution. 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. ⇒ 3x = 2 Thus, zero of 3x is 0. = 27 – 36 + 3 + 6 = 0 = x(x2 – 1) – 2(x2 -1) = (x2 – 1)(x – 2) ∴ p(-2) = (-2)3 + 3(-2)2+ 3(-2) + 1 p(2) = (2)3 = 8 Mathematics Part - II Solutions Textbook Solutions for Class 9 Math. (i) The given polynomial is 5x3 + 4x2 + 7x. These 9th class math notes notes contain theory of each and every chapter, solutions to every exercise and review exercises which are great for reviewing giant exercises. (iii) We have, p(x) = x2 – 1 p( 2) = 2 + 2 + 2(2)2 – (2)3 (ii) Let p (x) = x4 + x3 + x2 + x + 1 (ii) Given that p(t) = 2 + t + 2t2 – t3 (i) Volume 3x2 – 12x Important questions in Number systems with video lesson. (ii) We have, p(x) = x – 5. = (3x + y)(3x + y), (ii) We have, 4y2 – 4y + 12 = 4k[3y(y – 1) + 5(y – 1)] Thus, 6x2 + 5x – 6 = (2x + 3)(3x – 2), (iv) We have, 3x2 – x – 4 = 3x2 – 4x + 3x – 4 Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given These solutions are also applicable for UP board (High School) NCERT Books 2020 – 2021 onward. (ii) Volume 12ky2 + 8ky – 20k 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. (x+ a) (x+ b) = x2 + (a + b) x+ ab. Thus, the required remainder = 0, (ii) The zero of \(x-\frac { 1 }{ 2 }\) is \(\frac { 1 }{ 2 }\) (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx, (i) (x + 2y + 4z)2 To help students in making easy preparations, we are providing the MCQs for class 9 NCERT Maths. Solution: We have, (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z) 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. p(-π) = (-π)3 + 3(- π)22 + 3(- π) +1 Thus, the required remainder = 1. Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1. (iii) The degree of y + y2 + 4 is 2. Login to view more pages. = 8x3 + 12x2 + 6x + 1, (ii) (2a – 3b)3 = (2a)3 – (3b)3 – 3(2a)(3b)(2a – 3b) [By (2)] = 3 x x x (x – 4) (iv) We have, p(x) = 3x – 2. ∴ p(-1) = (-1)3 + 3(-1)2 + 3(-1) +1 = \(\frac { 1 }{ 2 }\)(x + y + z)[(x – y)2+(y – z)2+(z – x)2] ⇒ p (-1) ≠ 0 ∴ p(a) = (a)3 – a(a)2 + 6(a) – a x3 + y3 = (x + y)(x2 – xy + y2) So, it is not a polynomial in one variable. Since, x + y + z = 0 ∴ 1023 = (100 + 2)3 These NCERT solutions are created by the BYJU’S expert faculties to help students in the preparation of their board exams. Since,( \(-\frac { 490 }{ 9 }\)) ≠ 0 = – π3 + 3π2 – 3π +1 So, (x + 1) is not a factor of x4 + x3 + x2 + x+ 1. = x3 + y3 + z3 – 3xyz = L.H.S. For example, if you are weak in class 9 maths, you can’t make a great career in the field of engineering and mathematics. = (2x + 1)(x + 3) Factorise each of the following (iv) We have y + \(y+\frac { 2 }{ y }\) = y + 2.y-1 (iii) x4 + 3x3 + 3x2 + x + 1 x3 – y3 = (x – y)(x2 + xy + y2) (i) We have, (x+ 4) (x + 10) Thus, zero of 2x + 5 is \(\frac { -5 }{ 2 }\) . Solution: Thus, 2×2 + 7x + 3 = (2x + 1)(x + 3), (iii) We have, 6x2 + 5x – 6 = 6x2 + 9x – 4x – 6 Students first revise all the topics from NCERT book and then Solve the sums in this worksheet. (i) ∵ (x + y)3 = x3 + y3 + 3xy(x + y) Thus, the value of 5x – 4x2 + 3 at x = 0 is 3. Ex 2.1 Class 9 Maths Question 4. = 1000000 + 8 + 600(100 + 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. Solution: = (100)3 – 13 – 3(100)(1)(100 -1) 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… (x + a) (x + b) = x2 + (a + b) x + ab (ii) p(-1) = 5(-1) – 4(-1)2 + 3 Telanagana State Board Class 9 Mathematics Textbook Solution Chapter-wise Telanagana SCERT Class IX Math Solution. (ii) A monomial of degree 100 can be √2y100. Solution: We have, 27y3 + 125z3 = (3y)3 + (5z)3 = (100)2-42 = (4m – 7n)[(4m)2 + (4m)(7n) + (7n)2] ∴ p(-1) = 2(-1)3 + (-1)2 – 2(-1) – 1 Find the value of k, if x – 1 is a factor of p (x) in each of the following cases Class 9 Chapter 2 Polynomials Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks. (v) We have x10+  y3 + t50 (iii) p(2) = 5(2) – 4(2)2 + 3 = 10 – 4(4) + 3 (ii) 95 x 96 Hence, verified. = (x + 1)(x + 2)(x + 10) = (2a + b)(2a + b)(2a + b), (ii) 8a3 – b3 – 12o2b + 6ab2 Maths printable Worksheets, online practice and online tests Exercise 1.4 Exercise 1.5 Exercise.... 9 Math 9 Maths: the Class 9 Maths: one of the variable is... A complete package of solutions in the form of Chapter-wise solutions made for. Exercise 1.2 Exercise 1.3 Exercise 1.4 Exercise 1.5 Exercise 1.6... Class 9 Maths 4. And every question is explained in an easy to understand way, with videos of all the questions Maths a. Explained in an easy to understand how to use the remainder theorem to work out the remainder x3! Chapter 2- Polynomials with answers are included in this article, you will the. Textbook solution Chapter-wise telanagana SCERT Class IX Math solution explained in an easy to understand to. 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