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”. = (4m – 7n)(16m2 + 28mn + 49n2), Question 11. Question 1. For example, if you are weak in class 9 maths, you can’t make a great career in the field of engineering and mathematics. Class 9 maths printable worksheets, online practice and online tests. Check whether 7 + 3x is a factor of 3x3+7x. (x+ a) (x+ b) = x2 + (a + b) x+ ab. Teachoo provides the best content available! = (3y + 5z)(9y2 – 15yz + 25z2), (ii) We know that Solution: Chapter 2: Polynomials. (iv) p (x) = (x-1) (x+1) The coefficient of x2 is \(\frac { \pi }{ 2 }\). (i) 9x2 + 6xy + y2 = (y – 1)(y + 1)(2y +1), Question 1. ⇒ 2x + 5 =0 Solution: = x2(x + 1) + 12x(x +1) + 20(x + 1) Thus, zero of ax is 0. 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. 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 … (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx, (i) (x + 2y + 4z)2 (vii) 7x3 p(1) = (1)2 – 1 + 1 = 1 – 1 + 1 = 1 Question 13. (iii) x4 + 3x3 + 3x2 + x + 1 ⇒ x = \(\frac { -5 }{ 2 }\) [Using a3 + b3 + 3 ab(a + b) = (a + b)3] We have, (v) p (x) = 3x Hence, verified. (ii) The given polynomial is 2 – x2 + x3. Use suitable identities to find the following products ∴ p(3) = (3)3 – 4(3)2 + 3 + 6 So, (x + 1) is not a factor of x3 – x2 – (2 + √2) x + √2. = 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) (i) p(y) = y2 – y +1 = (2y -1)2 (i) We have, 9x2 + 6xy + y2 Let p(x) = x3 + 3x2 + 3x +1 Prepare effectively for your Maths exam with our NCERT Solutions for CBSE Class 9 Mathematics Chapter 2 Polynomials. = 2 + 2 + 8 – 8 = 4 because each exponent of y is a whole number. ∴ p(0) = (0 – 1)(0 + 1) = (-1)(1) = -1 Class 9 Maths Chapter 2 Polynomials This chapter guides you through algebraic expressions called polynomial and various terminologies related to it. = x(x2 – 1) – 2(x2 -1) = (x2 – 1)(x – 2) (vi) [ \(\frac { 1 }{ 4 }\)a –\(\frac { 1 }{ 4 }\)b + 1] 2 (iv) 1 + x = (x + 1)(x2 – 5x + x – 5) Thus, zero of 3x – 2 is \(\frac { 2 }{ 3 }\). NCERT Solutions Class 9 Maths Chapter 2 Polynomials are provided here. Verify that = 2y3 – 2y2 + 3y2 – 3y + y – 1 The coefficient of x2 is -1. Chapter 14 Probability. (i) The degree of x2 + x is 2. = 27 – 4(9) + 3 + 6 ∴ p(o) = (0)2 = 0 (i) We have, (x+ 4) (x + 10) Use the Factor Theorem to determine whether g (x) is a factor of p (x) in each of the following cases Since p(-1) = 0, so, x = -1, is also a zero of x2 – 1. 1et p(x) = 5x – 4x2 + 3 Solution: Since, p(1) = 2(1)2 + k(1) + √2 = (x + 1)(x – 5)(x + 1) Verify = (4a – 3b)(4a – 3b)(4a – 3b). = ( 100)2 + (3 + 7) (100)+ (3 x 7) The zero of x + 1 is -1. Ex 2.1 Class 9 Maths Question 4. Extra questions based on the topic Number System. NCERT Solutions Class 9 Maths Chapter 2 Polynomials. Question 3. = 7y(5y + 4) – 3(5y + 4) = (5 y + 4)(7y – 3) (i) 5x3+4x2 + 7x 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. To help students in making easy preparations, we are providing the MCQs for class 9 NCERT Maths. P(2) = (2 – 1)(2 + 1) = (1)(3) = 3. Middle school is one of the most crucial periods a student undergoes in the course of their schooling years. These solutions are also applicable for UP board (High School) NCERT Books 2020 – 2021 onward. = 0 + 0 + 0 + 1 = 1 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. Question 9. Thus, zero of cx + d is \(-\frac { d }{ c }\), Question 1. (iv) (y2+ \(\frac { 3 }{ 2 }\)) (y2– \(\frac { 3 }{ 2 }\)) Give one example each of a binomial of degree 35, and of a monomial of degree 100. = (x + 1)(x2 – 4x – 5) = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) (iii) Given that p(x) = x3 = (100)3 + (2)3 + 3(100)(2)(100 + 2) (iv) x + π Thus, zero of x + 5 is -5. Learn about the degree of a polynomial and the zero of a polynomial with related Maths solutions. 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. (i) x3+x2+x +1 (i) Here, p(x) = x2 + x + k ∴ p (- 1) =(- 1)3- (-1)2 – (2 + √2)(-1) + √2 which is not a whole number. = 1000000 – 1 – 300(100 – 1) For (x – 1) to be a factor of p(x), p(1) should be equal to 0. (iii) (3x + 4) (3x – 5) Write the coefficients of x2 in each of the following = 2 + k + √2 =0 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. = (3 – 5a)3 ⇒ x3 + y3 – 3xyz = -z3 Find the remainder when x3 + 3x2 + 3x + 1 is divided by (i) p(x)=x+5 ⇒ p (-1) ≠ 0 Write the following cubes in expanded form Chapter-wise NCERT Solutions for Class 9 Maths Chapter 2 Polynomials solved by Expert Teachers as per NCERT (CBSE) Book guidelines. = 2(-1) + 1 + 2 – 1 (iv) We have, p(x) = (x + 1)(x – 2) ∴ p(0) = (0)3 + 3(0)2 + 3(0) + 1 ∴ 1023 = (100 + 2)3 Practise the solutions to understand how to use the remainder theorem to work out the remainder of a polynomial. (vi) r2 x3 + y3 + z3 = 3xyz, Question 14. Let x = 28, y = -15 and z = -13. = (2a)3 – (b)3 – 3(2a)(b)(2a – b) Using identity, Using identity, Solution: ∴ 3x3 + 7x is not divisib1e by 7 + 3x. Unit 1 - Matrices & Determinants. The highest power of variable t is 1. (i)We have, 103 x 107 = (100 + 3) (100 + 7) We know that Note: Important questions have also been marked for your reference. k = -2 – √2 = -(2 + √2), (iii) Here, p (x) = kx2 – √2 x + 1 Represent geometrically 8.1 on number line. Factorise 27x3 +y3 +z3 -9xyz. (iii) p (x) = x3 – 4x2 + x + 6, g (x) = x – 3 Solution: = (2a – b) (2a – b) (2a – b), (iii) 27 – 125a3 – 135a + 225a2 = 3[-420] = -1260, (ii) We have, (28)3 + (-15)3 + (-13)3 (ii) 2x2 + y2 + 8z2 – 2√2xy + 4√2yz – 8xz Thus, zero of x – 5 is 5. = (x + 1)[x(x – 5) + 1(x – 5)] = 1 – 1 + 1 – 1 + 1 ∴ 993 = (100 – 1)3 Thus, 2×2 + 7x + 3 = (2x + 1)(x + 3), (iii) We have, 6x2 + 5x – 6 = 6x2 + 9x – 4x – 6 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. It is a complete package of solutions to problems of your really tough book. The highest = 2 + 0 + 0 – 0=2 Students first revise all the topics from NCERT book and then Solve the sums in this worksheet. Chapter-9 Chapter-2 Sol. ⇒ p(-1) = 0, so g(x) is a factor of p(x). Since, p(x) = 0 Solutions to all NCERT Exercise Questions and Examples of Chapter 2 Class 9 Polynomials are provided free at Teachoo. Rearranging the terms, we have x3 – x – 2x2 + 2 (ii) (x+8) (x -10) Since, p(0) = 0, so, x = 0 is a zero of x2. 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. = 4k[(3y + 5) x (y – 1)] Thus, x3 – 2x2 – x + 2 = (x – 1)(x + 1)(x – 2), (ii) We have, x3 – 3x2 – 9x – 5 = 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) = (100)2-42 Important questions in Number systems with video lesson. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. = x2 – 2x – 80, (iii) We have, (3x + 4) (3x – 5) = 1 – 3 + 3 – 1 + 1 = 1 (iv) 3 We have, p(x) = 3x3+7x. ⇒ k = -2. So, it is a quadratic polynomial. After the completion of chapters from NCERT Books and Exemplar Books, students should go for Assignments. = -1 Question 2. Thus, the possible length and breadth are (5a – 3) and (5a – 4). = 1000000 + 8 + 60000 + 1200 = 1061208, (iii) We have, 998 = 1000 – 2 = (2y – 1)(2y – 1 ), Question 4. Using identity, ∴ \(p(-\frac { 1 }{ 3 } )\quad =\quad 3(-\frac { 1 }{ 3 } )\quad +\quad 1\quad =\quad -1\quad +\quad 1\quad =\quad 0\) Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given Solution: Thus, 6x2 + 5x – 6 = (2x + 3)(3x – 2), (iv) We have, 3x2 – x – 4 = 3x2 – 4x + 3x – 4 ⇒ 3x – 2 = 0 (iii) The given polynomial is 5t – √7 . 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, … (iii) 104 x 96 = 994011992, Question 8. (vi) p (x) = 1x + m, x = – \(\frac { m }{ 1 }\) (ii) 64m3 – 343n3 (iii) x3 + 13x2 + 32x + 20 (i) p (x)= 2x3 + x2 – 2x – 1, g (x) = x + 1 (i) x3 – 2x2 – x + 2 However, it is possible to avoid such a scenario by taking authentic NCERT solutions for class 9 maths from a reliable source. Thus, the possible length and breadth are (7y – 3) and (5y + 4). (ii) x – \(\frac { 1 }{ 2 }\) Verify whether the following are zeroes of the polynomial, indicated against them. In this article, you will get the MCQs on Class 9 Maths Chapter 4: Linear Equations in Two Variables. = (2x + 1)(x + 3) = 4[3ky2 + 2ky – 5k] = 4[k(3y2 + 2y – 5)] (i) (x + 4)(x + 10) (iii) The degree of y + y2 + 4 is 2. Write the following numbers in p/q form (i) 2.015 (ii) 0.235 Ans (399 235 ' 198 999) 4. = x(3x – 4) + 1(3x – 4) = (3x – 4)(x + 1) Contains solved exercises, review questions, MCQs, important board questions and chapter overview. = -1 + 3- 3 + 1 = 0 (ii) The degree of x – x3 is 3. (v) We have, p(x) = 3x. = (1000)3– (2)3 – 3(1000)(2)(1000 – 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. (iv) x3 – x2 – (2 +√2 )x + √2 (i) (- 12)3 + (7)3 + (5)3 = -π3 + 3π2 + (-3π) + 1 which is not a whole number. = 10 – 16 + 3 = -3 and zero of 7 + 3x is \(-\frac { 7 }{ 3 }\). (i) Volume 3x2 – 12x (ii) p (x) = 5x – π, x = \(\frac { 4 }{ 5 }\) Solution: x3 + y3 = (x + y)(x2 – xy + y2) = 2 + 1 + 2 – 1 = 4 = (3 – 5a) (3 – 5a) (3 – 5a), (iv) 64a3 -27b3 -144a2b + 108ab2 [Using (a + b)(a -b) = a2– b2] These NCERT solutions are created by the BYJU’S expert faculties to help students in the preparation of their board exams. We have, 64m3 – 343n3 = (4m)3 – (7n)3 CBSE Worksheets for Class 11 Maths: One of the best teaching strategies employed in most classrooms today is Worksheets. (i) (x+2y+ 4z)2 We have, (x + 8) (x – 10) = x2 + [8 + (-10)] x + (8) (- 10) x3 +y3 +z3 – 3xyz = \(\frac { 1 }{ 2 }\) (x + y+z)[(x-y)2 + (y – z)2 +(z – x)2] Solution: Since, p(1) = (1)2 +1 + k [Using a2 + 2ab + b2 = (a + b)2] = (x + 1)(x2 + 2x + 10x + 20) (ii) p (x) = 2x2 + kx + √2 = -2 + 1 + 2 -1 = 0 = (y – 1)(y + 1)(2y + 1) = 4x2 + 25y2 + 9z2 – 20xy – 30yz + 12zx, Question 5. Thus, the required remainder is \(-\frac { 27 }{ 8 }\) . So, it is a quadratic polynomial. (iii) p (x) = kx2 – √2 x + 1 (iii) x = 2 (iv) p (x) = (x + 1) (x – 2), x = – 1,2 Thus, the possible dimensions of the cuboid are 4k, (3y + 5) and (y -1). (i) 4 x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz Thus, zero of 3x is 0. = k – √2 + 1 = 0 [Using a2 – 2ab + b2 = (a- b)2] Represent the following irrational numbers on number line. = (- √2x + y + 2 √2z)2 (iii) The given polynomial is \(\frac { \pi }{ 2 } { x }^{ 2 }\) + x. Solution: ⇒ (x – y)3 + 3xy(x – y) = x3 – y3 (iii) y + y2+4 Chapterwise basic concepts & formulas for classes IX & X, Assignment pdf for math for classes IX & X NCERT Mathematics Book[10] with solutions NCERT Book NCERT Sol. In this … [Using a3 – b3 – 3 ab(a – b) = (a – b)3] = 4 x k x (3y2 + 2y – 5) (iii) 3 √t + t√2 [Using a3 + b3 + 3 ab(a + b) = (a + b)3] Exercise 13.1 Solution. (ii) We have, 12ky2 + 8ky – 20k 27x3 + y3 + z3 – 9xyz = (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z) [Using (a – b)3 = a3 – b3 – 3ab (a – b)] p(1) = k(1)2 – 3(1) + k = (4a – 3b)3 = (3x)2 + 2(3x)(y) + (y)2 p(-π) = (-π)3 + 3(- π)22 + 3(- π) +1 Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1. = (2x + 3)(3x – 2) = (x – 1)(x + 1)(x – 2) = (2a + b)(2a + b)(2a + b), (ii) 8a3 – b3 – 12o2b + 6ab2 (v) We have x10+  y3 + t50 = (x + 1)(x + 2)(x + 10), (iv) We have, 2y3 + y2 – 2y – 1 All answers are solved step by step with videos of every question.Topics includeChapter 1 Number systems- What are Rational, Irrational, Real numbers, Law of Exponents, Expressing numbers in p/q Find p (0), p (1) and p (2) for each of the following polynomials. Let x = -12, y = 7 and z = 5. ∴ p(-2) = (-2)3 + 3(-2)2+ 3(-2) + 1 (ii) p (x) = x – 5 = (- √2x + y + 2 √2z) (- √2x + y + 2 √2z), Question 6. = 10000 + (10) x 100 + 21 Answers to each and every question is explained in an easy to understand way, with videos of all the questions. = k – 3 + k (iii) x2 – \(\frac { { y }^{ 2 } }{ 100 }\) Evaluate the following products without multiplying directly i.e. = (3x + y)2 Variables and expression are called as indeterminate and coefficients. Teachoo is free. (i) p (x) = x2 + x + k ⇒ 3x = 0 ⇒ x = 0 Determine which of the following polynomials has (x +1) a factor. Hence, if x + y + z = 0, then [Using (a – b)3 = a3 – b3 – 3ab (a – b)] Question 1. because each exponent of x is a whole number. (iii) p (x) = 2x + 5 It is a polynomial in one variable i.e., y Expand each of the following, using suitable identity NCERT Exemplar Class 9 Maths is very important resource for students preparing for IX Board Examination.Here we have provided NCERT Exemplar Problems Solutions along with NCERT Exemplar Problems Class 9. Question 2. = (3x -1) (4x -1) We have, (3x + 4) (3x – 5) = (3x)2 + (4 – 5) x + (4) (- 5) = x2 (x + 1) – 4x(x + 1) – 5(x + 1) = 8a3 – 27b3 – 18ab(2a – 3b) = a3 – a3 + 6a – a = 5a Solution: = 8x3 + 12x2 + 6x + 1, (ii) (2a – 3b)3 = (2a)3 – (3b)3 – 3(2a)(3b)(2a – 3b) [By (2)] Factorise each of the following ⇒ x = -5. (i) x + 1 = (3x + y)(3x + y), (ii) We have, 4y2 – 4y + 12 Since, p(1) = k(1)2 – (1) + 1 Chapter 4 Linear Equations in Two Variables. = – 5x – 4x2 + 3 = -9 + 3 = -6 = – π3 + 3π2 – 3π +1 Solution: (ii) 35y2+ 13y -12 = 35y2 + 28y – 15y -12 the remainder is not 0. (iii) p(2) = 5(2) – 4(2)2 + 3 = 10 – 4(4) + 3 Solution: (iv) 3x2 – x – 4 (v) x10+ y3+t50 sin θ and tan θ) without evaluating θ. Since, x + y + z = 0 = \(\frac { 1 }{ 2 }\) (x + y + z)[2(x2 + y2 + z2 – xy – yz – zx)] = 3x(2x + 3) – 2(2x + 3) Since, p(-1) = 0, so, x = -1 is a zero of (x + 1)(x – 2). So, the degree of the polynomial is 3. We have, (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z) (i) Given that p(y) = y2 – y + 1. = (x + 1)(x2 + 12x + 20) (ii) Let p (x) = x4 + x3 + x2 + x + 1 ∴ p(a) = (a)3 – a(a)2 + 6(a) – a Since, p(x) = 0 => ax = 0 => x-0 = 1000000 -1 – 30000 + 300 (i) ∵ (x + y)3 = x3 + y3 + 3xy(x + y) ∴ p (-1)= (-1)4 + 3 (-1)3 + 3 (-1)2 + (- 1) + 1 (i) Abmomial of degree 35 can be 3x35 -4. Question 2. = 4x (3x – 1 ) -1 (3x – 1) = \(\frac { 1 }{ 2 }\) (x + y + 2)[(x2 + y2 – 2xy) + (y2 + z2 – 2yz) + (z2 + x2 – 2zx)] = (x + 1)(x + 2)(x + 10) (i) We have, (-12)3 + (7)3 + (5)3 NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12. (iv) y+ \(\frac { 2 }{ y }\) ⇒ x = \(\frac { 2 }{ 3 }\) = (100)3 – 13 – 3(100)(1)(100 -1) This solution is strictly revised in accordance … (vi) The degree of r2 is 2. (i) We have, 3x2 – 12x = 3(x2 – 4x) ∴ p(-1) = 2(-1)3 + (-1)2 – 2(-1) – 1 All the chapterwise questions with solutions to help you to revise complete CBSE syllabus and score more marks in Your board examinations. (i) Area 25a2 – 35a + 12 Question 3. (i) p(0) = 5(0) – 4(0)2 + 3 = 0 – 0 + 3 = 3 Thus, x3 + 13x2 + 32x + 20 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, Chapter 4 Linear Equations in Two Variables, Chapter 5 Introduction to Euclid Geometry, Chapter 9 Areas of Parallelograms and Triangles, 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. [Using (a + b)3 = a3 + b3 + 3ab (a + b)] NCERT Solutions for Class 9 Maths: The Class 9 Maths textbook solutions of exercises for all 15 chapters are included in this article. If x + y + z = 0, show that x3 + y3 + z3 = 3 xyz. 10 Questions. Thus, the value of 5x – 4x2 + 3 at x = 0 is 3. (ii) (2x – y + z)2 12x2 – 7x + 1 = 12x2 – 4x- 3x + 1 (vi) We have, p(x) = ax, a ≠ 0. They are in a list with arrows. ∴ \(p(\frac { 1 }{ 2 } )\quad =\quad 2(\frac { 1 }{ 2 } )+1=\quad 1+1\quad =\quad 2\) CBSE Class 11 Maths Worksheet for students has been used by teachers & students to develop logical, lingual, analytical, and … (vii) p (x) = cx + d, c ≠ 0 where c and d are real numbers. ⇒ (x – y)(x2 + y2 + xy) = x3 – y3 (i) 103 x 107 Mathematics Part - II Solutions Textbook Solutions for Class 9 Math. The highest power of the variable x is 3. ∴ P(0) = (0)2 – 0 + 1 = 0 – 0 + 1 = 1 Solution: Factorise (i) We have, 99 = (100 -1) (iii) 27-125a3 -135a+225a2 Using the identity, Since, p(1) = 0, so x = 1 is a zero of x2 -1. They give a detailed and stepwise explanation of each answer to the problems given in the exercises in the NCERT … = (y – 1)(2y2 + 3y + 1) = x3 + x2 – 4x2 – 4x – 5x – 5 , ⇒ p(1) = k + 2 = 0 Thus, the required remainder = \(\frac { 27 }{ 8 }\). These expert faculties solve and provide the NCERT Solution for class 9 so that it would help students to solve the problems comfortably. (i) We have, (v) 3t (i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz (iv) 2y3 + y2 – 2y – 1 Hence, verified. = 8x3 + 1 + 6x(2x + 1) Solution: Thus, 2y3 + y2 – 2y – 1 ⇒ (x + y)[(x + y)2-3xy] = x3 + y3 Also, p(-1) = (-1)2 -1 = 1 – 1 = 0 Solution: It is not a polynomial, because one of the exponents of y is -1, (ii) p(-1) = 5(-1) – 4(-1)2 + 3 Solution: ∴ (28)3 + (-15)3 + (-13)3 = 3(28)(-15)(-13) (ii) x – x3 (iv) Let p (x) = x3 – x2 – (2 + √2) x + √2 Area of a rectangle = (Length) x (Breadth) (ii) 4 – y2 (ii) ∵ (x – y)3 = x3 – y3 – 3xy(x – y) ⇒ p(3) = 0, so g(x) is a factor of p(x). [Using (x + a)(x + b) = x2 + (a + b)x + ab] = x(2x + 1) + 3(2x + 1) (ii) We have, p(x) = x – 5. (ii) Here, p (x) = 2x2 + kx + √2 (ii) x = – 1 (ii) x4 + x3 + x2 + x + 1 You have these advantages of browsing notes from our website. Ex 2.1 Class 9 Maths Question 5. p(2) = (2)3 = 8 Since, p(2) = 0, so, x = 2 is also a zero of (x + 1)(x – 2). = 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) Login to view more pages. (ii) p(x)= x3 + 3x2 + 3x + 1, g (x) = x + 2 CLASS 9 IX Sample Papers X Sample Papers CLASS 8 CLASS 07 Class 06 ... Chapter - 2: Polynomials. (viii) We have, p(x) = 2x + 1 Chapter-10 Chapter-3 Sol. Factorise Maths Assignment Class 9th Chapter 1 Important questions based on chapter 1 class 9. (iv) The given polynomial is √2 x – 1. (viii) p (x) = 2x + 1, x = \(\frac { 1 }{ 2 }\) = 2√2 Chapter 13 Geometrical Constructions. x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) Find the remainder when x3 – ax2 + 6x – a is divided by x – a. Class 9 Chapter 2 Polynomials Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks. Solution: = (3x + y + z)[(3x)3 + y3 + z3 – (3x × y) – (y × 2) – (z × 3x)] = 9x2 – x – 20, Question 2. and (x – y)3 = x3 – y3 – 3xy(x – y) …(2), (i) (2x + 1)3 = (2x)3 + (1)3 + 3(2x)(1)(2x + 1) [By (1)] Chapter-1 Chapter-9 Sol. Hindi Medium and English Medium both are available to free download. (vii) P (x) = 3x2 – 1, x = – \(\frac { 1 }{ \sqrt { 3 } }\),\(\frac { 2 }{ \sqrt { 3 } }\) (iii) 5t – √7 All these questions are based on the important fundamental concepts given in NCERT Class 9 Maths. Telanagana State Board Class 9 Mathematics Textbook Solution Chapter-wise Telanagana SCERT Class IX Math Solution. = 4k x (3y + 5) x (y – 1) Solution: (i) x2+ x (iii) We have 3 √t + t√2 = 3 √t1/2 + √2.t = x2 + (2y)2 + (4z)2 + 2 (x) (2y) + 2 (2y) (4z) + 2(4z) (x) ⇒ x3 + y3 + 3xy(x + y) = -z3 (ii) 2 – x2 + x3 Write the degree of each of the following polynomials. Solution: = (4m – 7n)[(4m)2 + (4m)(7n) + (7n)2] is given, we can find the other two trigonometric ratios (i.e. Click on exercise or topic link below to get started. Question 12. Thus, the required remainder = 0, (ii) The zero of \(x-\frac { 1 }{ 2 }\) is \(\frac { 1 }{ 2 }\) Solution: Since, \(p(\frac { 1 }{ 2 } )\) ≠ 0, so, x = \(\frac { 1 }{ 2 }\) is not a zero of 2x + 1. (iv) (3a -7b – c)z (iii) 6x2 + 5x – 6 So, (x+ 1) is a factor of x3 + x2 + x + 1. (v) We have, p(x) = x2 Exercise 14.1 Solution. (i) Let p (x) = x3 + x2 + x + 1 = 2 x \(\frac { 1 }{ 2 }\) x (x + y + z)(x2 + y2 + z2 – xy – yz – zx) Thus, x3 – 3x2 – 9x – 5 = (x + 1)(x – 5)(x +1), (iii) We have, x3 + 13x2 + 32x + 20 Solution: (i) 25a2 – 35a + 12 = 25a2 – 20a – 15a + 12 = 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3) Thus, zero of 2x + 5 is \(\frac { -5 }{ 2 }\) . 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. = 3 x x x (x – 4) = (2x)2 + (3y)2 + (- 4z)2 + 2 (2x) (3y) + 2 (3y) (- 4z) + 2 (- 4z) (2x) ⇒ k = \(\frac { 3 }{ 4 }\), Question 4. (i) 27y3 + 125z3 (ii) 8a3 -b3-12a2b+6ab2 (i) The given polynomial is 2 + x2 + x. We have, p(x) = x3 – ax2 + 6x – a and zero of x – a is a. Ex 2.1 Class 9 Maths Question 3. ⇒ cx + d = 0 ⇒ cx = -d ⇒ \( x =-\frac { d }{ c }\) The coefficient of x2 is 1. (ii) 95 x 96 Solution: CBSE Worksheets for Class 9 Maths: One of the best teaching strategies employed in most classrooms today is Worksheets. = 2y2(y – 1) + 3y(y – 1) + 1(y – 1) (ii) Area 35y2 + 13y – 12 (v) The degree of 3t is 1. Class 9 Mathematics Notes for FBISE. Chapter -1 Sol. = 2k – 3 = 0 = 3(5460) = 16380. (iv) Since, 3 = 3x° [∵ x°=1] (vi) p (x)= ax, a≠0 (v) (- 2x + 5y – 3z)2 ⇒ x + 5 = 0 ⇒ k = √2 -1, (iv) Here, p(x) = kx2 – 3x + k = -8 + 12 – 6 + 1 ∴ p(0) = (0)3 = 0, p(1) = (1)3 = 1 = -1 + 1 – 1 + 1 Thus, 7 + 3x is not a factor of 3x3 + 7x. The coefficient of x2 is 0. ⇒ 2x = -5 So, the degree of the polynomial is 2. Since,( \(-\frac { 490 }{ 9 }\)) ≠ 0 ∴ p(-1) = (-1 +1) (-1 – 2) = (0)(- 3) = 0 Since, p(x) = 0 Get here MCQs on Class 9 Maths Chapter 2- Polynomials with answers. P(1) = 2 + 1 + 2(1)2 – (1)3 Thus, the required remainder = 1. (vii) The degree of 7x3 is 3. (iii) (- 2x + 3y + 2z)2 Solution: 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… (ii) Volume 12ky2 + 8ky – 20k = [(x)2 – (1)2](x – 2) 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. Evaluate the following using suitable identities = (2a)3 + (b)3 + 6ab(2a + b) (i) We have 4x2 – 3x + 7 = 4x2 – 3x + 7x0 These assignments will be available in updated form along with new assignments and chapter wise tests with solutions. So, it is a quadratic polynomial. State reasons for your answer. There is plenty to learn in this chapter about the definition and examples of polynomials, coefficient, degrees, and terms in a polynomial. (iii) We have, p(x) = 2x + 5. ⇒ p(-2) ≠ 0, so g(x) is not a factor of p(x). FREE Downloadable! Chapter wise assignments for class 9 Maths are given below updated for new academic session 2020-2021. ⇒ (x + y)(x2 + y2 – xy) = x3 + y3 Exercise 13.2 Solution. = 1000000 + 8 + 600(100 + 2) Question 15. ⇒ x3 + y3 + 3xy(-z) = -z3 [∵ x + y = -z] Thus, 12x2 -7x + 3 = (2x – 1) (x + 3), (ii) We have, 2x2 + 7x + 3 = 2x2 + x + 6x + 3 (ii) We have, p(x) x3 + 3x2 + 3x + 1 and g(x) = x + 2 (ii) A monomial of degree 100 can be √2y100. (iii) \(\frac { \pi }{ 2 }\) x2 + x = x3 + x2 + 12x2 + 12x + 20x + 20 (iv) We have, p(x) = 3x – 2. Solution: = 10000 + (-900) + 20 = 9120, (iii) We have 104 x 96 = (100 + 4) (100 – 4) Solution: Solution: Without actually calculating the cubes, find the value of each of the following We have, 27y3 + 125z3 = (3y)3 + (5z)3 (i) p(x) = 3x + 1,x = –\(\frac { 1 }{ 3 }\) Question 16. (i) (99)3 ∴ p(-1) = (-1)3 + 3(-1)2 + 3(-1) +1 = 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) (iii) p (x) = x2 – 1, x = x – 1 Download File. So, it is a cubic polynomial. (i) The zero of x + 1 is -1. = 1000000000 – 8 – 6000000 +12000 = \(\frac { 1 }{ 2 }\)(x + y + z)[(x – y)2+(y – z)2+(z – x)2] Which of the following expressions are polynomials in one variable and which are not? Since, p(x) = 0 p(1) = (1 – 1)(1 +1) = (0)(2) = 0 (x + a) (x + b) = x2 + (a + b) x + ab x3 – y3 = (x – y)(x2 + xy + y2) Since, p(x) = 0 ⇒ x – 5 = 0 ⇒ x = -5 So, the degree of the polynomial is 1. = -1 – 1 + 2 + √2 + √2 So, it is a linear polynomial. 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