Find p (0), p (1) and p (2) for each of the following polynomials. There is plenty to learn in this chapter about the definition and examples of polynomials, coefficient, degrees, and terms in a polynomial. Homework Help with Chapter-wise solutions and Video explanations. Chapter -1 Sol. Chapter-9 Chapter-2 Sol. = (x + 1)(x2 – 5x + x – 5) Chapter wise assignments for class 9 Maths are given below updated for new academic session 2020-2021. Also, p( 2) = (2 + 1)(2 – 2) = (3)(0) = 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. = 1000000 -1 – 30000 + 300 3. (ii) p (x) = 5x – π, x = \(\frac { 4 }{ 5 }\) = 2 + k + √2 =0 = (x + 1)(x2 – 4x – 5) ⇒ k = \(\frac { 3 }{ 4 }\), Question 4. ∴ p(1) = (1)2 – 1 = 1 – 1=0 ∴ p(3) = (3)3 – 4(3)2 + 3 + 6 [Using (a + b)3 = a3 + b3 + 3ab (a + b)] 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. (v) We have x10+  y3 + t50 = (2a)3 – (b)3 – 3(2a)(b)(2a – b) Chapter-2 Chapter-10 Sol. = (2x + 3)(3x – 2) = x2 – 2x – 80, (iii) We have, (3x + 4) (3x – 5) ∴ p(0) = (0)3 + 3(0)2 + 3(0) + 1 Prepare effectively for your Maths exam with our NCERT Solutions for CBSE Class 9 Mathematics Chapter 2 Polynomials. Thus, zero of ax is 0. x3 + y3 = (x + y)(x2 – xy + y2) = (3x)2 + 2(3x)(y) + (y)2 Using identity, = a3 – a3 + 6a – a = 5a p(2) = (2)2 – 2 + 1 = 4 – 2 + 1 = 3 Login to view more pages. Solution: = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) Important questions in Number systems with video lesson. Evaluate the following products without multiplying directly (iv) 64a3 -27b3 -144a2b + 108ab2 We have, 64m3 – 343n3 = (4m)3 – (7n)3 (ii) 95 x 96 Factorise the following using appropriate identities Chapter 13 Geometrical Constructions. (v) The zero of 5 + 2x is \(-\frac { 5 }{ 2 }\) . So, it is a linear polynomial. (iii) p (x) = x2 – 1, x = x – 1 = (3x + y + z)[(3x)3 + y3 + z3 – (3x × y) – (y × 2) – (z × 3x)] ∴ p(0) = (0 – 1)(0 + 1) = (-1)(1) = -1 (i) 2 + x2 + x (iv) 3 x3 +y3 +z3 – 3xyz = \(\frac { 1 }{ 2 }\) (x + y+z)[(x-y)2 + (y – z)2 +(z – x)2] (ii) The given polynomial is 4- y2. So, it is a linear polynomial. ⇒ x3 + y3 – 3xyz = -z3 Solution: = 4[3ky2 + 2ky – 5k] = 4[k(3y2 + 2y – 5)] (vi) The degree of r2 is 2. Factorise Solution: p(1) = (1)2 – 1 + 1 = 1 – 1 + 1 = 1 and zero of 7 + 3x is \(-\frac { 7 }{ 3 }\). (ii) p (t) = 2 +1 + 2t2 -t3 So, the degree of the polynomial is 2. ∴ p(-1) = (-1)3 + 3(-1)2 + 3(-1) +1 ⇒ p (-1) ≠ 0 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 4. (i) We have, p (x)= 2x3 + x2 – 2x – 1 and g (x) = x + 1 (iii) The given polynomial is \(\frac { \pi }{ 2 } { x }^{ 2 }\) + x. DronStudy provides you Chapter wise Solutions for Class 9th Maths, Science and Social Studies. (iv) The given polynomial is √2 x – 1. (i) 27y3 + 125z3 Class 9 Maths Chapter 2 Polynomials This chapter guides you through algebraic expressions called polynomial and various terminologies related to it. Hence, verified. Solution: (i) x3 + y3 = (x + y)-(x2 – xy + y2) So, (x + 1) is not a factor of x4 + 3x3 + 3x2 + x+ 1. is given, we can find the other two trigonometric ratios (i.e. Since, p(x) = 0 ⇒ x – 5 = 0 ⇒ x = -5 State reasons for your answer. Find the remainder when x3 – ax2 + 6x – a is divided by x – a. Write the following cubes in expanded form (ii) x – x3 = 2y3 – 2y2 + 3y2 – 3y + y – 1 Then, x + y + z = -12 + 7 + 5 = 0 (vi) We have, p(x) = ax, a ≠ 0. CLASS 9 IX Sample Papers X Sample Papers CLASS 8 CLASS 07 Class 06 ... Chapter - 2: Polynomials. So, the degree of the polynomial is 0. Represent the following irrational numbers on number line. Question 9. ⇒ (x – y)3 + 3xy(x – y) = x3 – y3 (iii) (998)3 Exercise 13.2 Solution. ∴ \(p(\frac { 1 }{ 2 } )\quad =\quad 2(\frac { 1 }{ 2 } )+1=\quad 1+1\quad =\quad 2\) (vii) P (x) = 3x2 – 1, x = – \(\frac { 1 }{ \sqrt { 3 } }\),\(\frac { 2 }{ \sqrt { 3 } }\) = (2x + 1)(x + 3) We have, (x + 8) (x – 10) = x2 + [8 + (-10)] x + (8) (- 10) = 8a3 – 27b3 – 18ab(2a – 3b) (iii) 104 x 96 Middle school is one of the most crucial periods a student undergoes in the course of their schooling years. = 27 – 4(9) + 3 + 6 (iii) 3 √t + t√2 Class 9 Mathematics Notes for FBISE. (i) 5x3+4x2 + 7x = ( 100)2 + (3 + 7) (100)+ (3 x 7) = x2 + (2y)2 + (4z)2 + 2 (x) (2y) + 2 (2y) (4z) + 2(4z) (x) = x2(x + 1) + 12x(x +1) + 20(x + 1) NCERT Solutions Class 9 Maths Chapter 2 Polynomials. They give a detailed and stepwise explanation of each answer to the problems given in the exercises in the NCERT … (ii) 8a3 -b3-12a2b+6ab2 (i) p(y) = y2 – y +1 (iii) (3x + 4) (3x – 5) (ii) 2 – x2 + x3 (iv) Since, 3 = 3x° [∵ x°=1] Since, p(0) = 0, so, x = 0 is a zero of x2. Solution: = (2y -1)2 (ii) 35y2+ 13y -12 = 35y2 + 28y – 15y -12 (x + a) (x + b) = x2 + (a + b) x + ab (ii) ∵ (x – y)3 = x3 – y3 – 3xy(x – y) ⇒ x = \(\frac { -5 }{ 2 }\) Chapter-1 Chapter-9 Sol. (vi) p (x) = 1x + m, x = – \(\frac { m }{ 1 }\) = (3x + y)2 These NCERT solutions are created by the BYJU’S expert faculties to help students in the preparation of their board exams. So, (x+ 1) is a factor of x3 + x2 + x + 1. (ii) We have, p(x) x3 + 3x2 + 3x + 1 and g(x) = x + 2 We have, (iii) y + y2+4 (iv) x3 – x2 – (2 +√2 )x + √2 = (4a – 3b)(4a – 3b)(4a – 3b). Thus, the required remainder = \(\frac { 27 }{ 8 }\). ⇒ 2x = -5 ∴ p (-1) = (-1)3 + (-1)2 + (-1) + 1 . Solution: Using the identity, (ii) Let p (x) = x4 + x3 + x2 + x + 1 = 1 – 1 + 1 – 1 + 1 ∴ 1023 = (100 + 2)3 12x2 – 7x + 1 = 12x2 – 4x- 3x + 1 = (x + 1)(x + 2)(x + 10), (iv) We have, 2y3 + y2 – 2y – 1 So, (x + 1) is not a factor of x3 – x2 – (2 + √2) x + √2. (iv) p (x) = (x + 1) (x – 2), x = – 1,2 Since, p(-1) = 0, so, x = -1 is a zero of (x + 1)(x – 2). = 8x3 + 12x2 + 6x + 1, (ii) (2a – 3b)3 = (2a)3 – (3b)3 – 3(2a)(3b)(2a – 3b) [By (2)] 1et p(x) = 5x – 4x2 + 3 So, (x + 1) is not a factor of x4 + x3 + x2 + x+ 1. Thus, zero of 3x is 0. ⇒ k = -2. (i) Area 25a2 – 35a + 12 Write the degree of each of the following polynomials. ⇒ x3 + y3 + 3xy(-z) = -z3 [∵ x + y = -z] (viii) p (x) = 2x + 1, x = \(\frac { 1 }{ 2 }\) So, the degree of the polynomial is 1. (i) x3 – 2x2 – x + 2 Thus, zero of cx + d is \(-\frac { d }{ c }\), Question 1. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. [Using a2 – 2ab + b2 = (a- b)2] They are in a list with arrows. Thus, zero of 2x + 5 is \(\frac { -5 }{ 2 }\) . ∴ P(-1) = (-1)4 + (-1)3 + (-1)2 + (-1)+1 ⇒ p (-1) ≠ 0 (iv) p (x) = 3x – 2 This solution is strictly revised in accordance … (v) x10+ y3+t50 (i) p (x) = x2 + x + k = (2a)3 + (b)3 + 3(2a)(b)(2a + b) Thus, the required remainder = 0, (ii) The zero of \(x-\frac { 1 }{ 2 }\) is \(\frac { 1 }{ 2 }\) = (1000)3– (2)3 – 3(1000)(2)(1000 – 2) NCERT solutions for session 2019-20 is now available to download in PDF form. Thus, 2×2 + 7x + 3 = (2x + 1)(x + 3), (iii) We have, 6x2 + 5x – 6 = 6x2 + 9x – 4x – 6 = (x + 1)[x(x – 5) + 1(x – 5)] (ii) p(x)= x3 + 3x2 + 3x + 1, g (x) = x + 2 (i) Abmomial of degree 35 can be 3x35 -4. = (2y – 1)(2y – 1 ), Question 4. (iii) The zero of x is 0. = [(x)2 – (1)2](x – 2) (i) We have, x3 – 2x2 – x + 2 (vii) We have, p(x) = cx + d. Since, p(x) = 0 On signing up you are confirming that you have read and agree to Evaluate the following using suitable identities Teachoo is free. All the chapterwise questions with solutions to help you to revise complete CBSE syllabus and score more marks in Your board examinations. Solutions to all NCERT Exercise Questions and Examples of Chapter 2 Class 9 Polynomials are provided free at Teachoo. (ii) Area 35y2 + 13y – 12 = (- √2x)2 + (y)2 + (2 √2z)2y + 2(- √2x) (y)+ 2 (y) (2√2z) + 2 (2√2z) (- √2x) Then, x + y + z = 28 – 15 – 13 = 0 (i) We have, (-12)3 + (7)3 + (5)3 Since, p(x) = 0 (iv) Given that p(x) = (x – 1)(x + 1) Get here MCQs on Class 9 Maths Chapter 2- Polynomials with answers. Since, p(x) = 0 ! = (3x + y)(3x + y), (ii) We have, 4y2 – 4y + 12 The coefficient of x2 is \(\frac { \pi }{ 2 }\). (v) p (x) = 3x (i) We have 4x2 – 3x + 7 = 4x2 – 3x + 7x0 Along with recalling the knowledge of linear … So, it is not a polynomial in one variable. (v) p (x) = x2, x = 0 Solution: Chapter 4 Linear Equations in Two Variables. Ex 2.1 Class 9 Maths Question 2. Solution: = (2a – b)3 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)] It is a polynomial in one variable i.e., x Hence, if x + y + z = 0, then Thus, 3x2 – x – 4 = (3x – 4)(x + 1), Question 5. Since, p(x) = 0 = 3[-420] = -1260, (ii) We have, (28)3 + (-15)3 + (-13)3 = (100)3 – 13 – 3(100)(1)(100 -1) (ii) We have, p(x) = 5x – π x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) Let x = 28, y = -15 and z = -13. (i) We know that Practise the solutions to understand how to use the remainder theorem to work out the remainder of a polynomial. 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. Ex 2.1 Class 9 Maths Question 1. Solution: (iv) We have, p(x) = (x + 1)(x – 2) Question 10. 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. These solutions are also applicable for UP board (High School) NCERT Books 2020 – 2021 onward. (i) We have, 3x2 – 12x = 3(x2 – 4x) (ii) p (x) = x – 5 Solution: ⇒ x3 + y3 + z3 = 3xyz Give one example each of a binomial of degree 35, and of a monomial of degree 100. x3 – y3 = (x – y)(x2 + xy + y2) Unit 1 - Matrices & Determinants. It is a polynomial in one variable i.e., y ⇒ 3x – 2 = 0 ⇒ (x + y)[(x + y)2-3xy] = x3 + y3 10 Questions. Let p(x) = x3 + 3x2 + 3x +1 (iv) The degree of 1 + x is 1. (iv) x + π Thus, 2y3 + y2 – 2y – 1 (ii) p (x) = 2x2 + kx + √2 = (2a – b) (2a – b) (2a – b), (iii) 27 – 125a3 – 135a + 225a2 = 2 + 2 + 8 – 8 = 4 Solution: These ncert book chapter wise questions and answers are very helpful for CBSE board exam. ⇒ (x + y)3 – 3(x + y)(xy) = x3 + y3 Question 12. (i) We have , p(x) = 3x + 1 ⇒ (x – y)(x2 + y2 + xy) = x3 – y3 = 4k[3y2 – 3y + 5y – 5] The coefficient of x2 is 0. = -1 + 1 – 1 + 1 P(1) = 2 + 1 + 2(1)2 – (1)3 Question 2. = 1 – 3 + 3 – 1 + 1 = 1 Variables and expression are called as indeterminate and coefficients. (i) 25a2 – 35a + 12 = 25a2 – 20a – 15a + 12 = 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3) = 1000000000 – 8 – 6000(1000 – 2) Chapter 14 Probability. = (- √2x + y + 2 √2z)2 (i) Here, p(x) = x2 + x + k p(1) = k(1)2 – 3(1) + k Solution: (iv) (3a -7b – c)z Solution: (i) (- 12)3 + (7)3 + (5)3 because each exponent of x is a whole number. = (3x + y + z)(9x2 + y2 + z2 – 3xy – yz – 3zx). Chapter-10 Chapter-3 Sol. Using identity, (i)We have, 103 x 107 = (100 + 3) (100 + 7) ∴ \(p(-\frac { 1 }{ 3 } )\quad =\quad 3(-\frac { 1 }{ 3 } )\quad +\quad 1\quad =\quad -1\quad +\quad 1\quad =\quad 0\) ⇒ (x + y)(x2 + y2 – xy) = x3 + y3 ∴ 993 = (100 – 1)3 Determine which of the following polynomials has (x +1) a factor. Question 1. (v) 5 + 2x = (y – 1)(2y2 + 2y + y + 1) [Using a3 + b3 + 3 ab(a + b) = (a + b)3] (ii) We have, p(x) = x – 5. Which of the following expressions are polynomials in one variable and which are not? (ii) 2x2 + 7x + 3 = 4 x k x (3y2 + 2y – 5) Question 1. Telanagana State Board Class 9 Mathematics Textbook Solution Chapter-wise Telanagana SCERT Class IX Math Solution. Since, p(x) = 0 = -1 + 3- 3 + 1 = 0 x3 + y3 + z3 = 3xyz, Question 14. Chapter 2: Polynomials. Mathematics Part - II Solutions Textbook Solutions for Class 9 Math. (iv) √2 x – 1 which is not a whole number. = x3 + x2 – 4x2 – 4x – 5x – 5 , = (3)3 – (5a)3 – 3(3)(5a)(3 – 5a) [Using a3 + b3 + 3 ab(a + b) = (a + b)3] = 2√2 (iv) p (x) = kx2 – 3x + k = (x + 1)(x + 2)(x + 10) ⇒ x + y = -z (x + y)3 = (-z)3 ⇒ p(-2) ≠ 0, so g(x) is not a factor of p(x). CBSE Worksheets for Class 11 Maths: One of the best teaching strategies employed in most classrooms today is Worksheets. Factorise each of the following (iii) p(2) = 5(2) – 4(2)2 + 3 = 10 – 4(4) + 3 Since, p(1) = (1)2 +1 + k = x3 + y3 + z3 – 3xyz = L.H.S. Ex 2.1 Class 9 Maths Question 3. Thus, the possible length and breadth are (5a – 3) and (5a – 4). Write the following numbers in p/q form (i) 2.015 (ii) 0.235 Ans (399 235 ' 198 999) 4. ∴ P(0) = (0)2 – 0 + 1 = 0 – 0 + 1 = 1 NCERT Solutions Class 9 Maths Chapter 2 Polynomials are provided here. = (y – 1)(y + 1)(2y +1), Question 1. Since, p(1) = k(1)2 – (1) + 1 To help students in making easy preparations, we are providing the MCQs for class 9 NCERT Maths. = (3x -1) (4x -1) = (2a)3 + (b)3 + 6ab(2a + b) NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1 are part of NCERT Solutions for Class 9 Maths. We have, p(x) = x3 – ax2 + 6x – a and zero of x – a is a. (iv) We have, p(x) = 3x – 2. = 2 + 0 + 0 – 0=2 Solution: = 9x2 – x – 20, Question 2. Thus, zero of x + 5 is -5. = -8 + 12 – 6 + 1 Write the coefficients of x2 in each of the following (i) x3+x2+x +1 Classify the following as linear, quadratic and cubic polynomials. Chapter-3 Chapter-11 Sol. So, the degree of the polynomial is 3. 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. = (3 – 5a)3 ∴ p(-1) = 2(-1)3 + (-1)2 – 2(-1) – 1 (iv) 3x2 – x – 4 = 8x3 + 1 + 6x(2x + 1) Thus, the value of 5x – 4x2 + 3 at x = -1 is -6. Thus, the value of 5x – 4x2 + 3 at x = 2 is – 3. Solution: Thus, the value of 5x – 4x2 + 3 at x = 0 is 3. (iii) P (x) = x3 = 1000300 – 30001 = 970299, (ii) We have, 102 =100 + 2 Maths Assignment Class 9th Chapter 1 Important questions based on chapter 1 class 9. = \(\frac { 1 }{ 2 }\) (x + y + 2)(x2 + y2 + y2 + z2 + z2 + x2 – 2xy – 2yz – 2zx) 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 ⇒ 3x = 2 Thus, the required remainder is \(-\frac { 27 }{ 8 }\) . = 3x° [ ∵ x°=1 ] so, the degree of 1 + x 0. Degree of 7x3 is 3 is 4- y2 topics from NCERT book Chapter wise questions and answers are very for. Θ and tan θ ) class 9 maths chapter 2 assignment evaluating θ Polynomials Exercise questions with solutions to help students to the! 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