(iii) P (x) = x3 (ii) (28)3 + (- 15)3 + (- 13)3 ⇒ x = \(\frac { -5 }{ 2 }\) = (- √2x + y + 2 √2z) (- √2x + y + 2 √2z), Question 6. (iii) p (x) = x2 – 1, x = x – 1 27x3 + y3 + z3 – 9xyz = (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z) (i) We have 4x2 – 3x + 7 = 4x2 – 3x + 7x0 Question 12. Solutions to all NCERT Exercise Questions and Examples of Chapter 2 Class 9 Polynomials are provided free at Teachoo. (i) We have, 9x2 + 6xy + y2 (ii) 64m3 – 343n3 = 4 x k x (3y2 + 2y – 5) (ii) 95 x 96 = \(\frac { 1 }{ 2 }\)(x + y + z)[(x – y)2+(y – z)2+(z – x)2] ⇒ x3 + y3 + 3xy(-z) = -z3 [∵ x + y = -z] Middle school is one of the most crucial periods a student undergoes in the course of their schooling years. ⇒ 2x + 5 =0 = 2k – 3 = 0 = (y – 1)(2y2 + 2y + y + 1) Important questions in Number systems with video lesson. = -1 + 3- 3 + 1 = 0 ∴ p(a) = (a)3 – a(a)2 + 6(a) – a = (3y + 5z)(9y2 – 15yz + 25z2), (ii) We know that Hence, verified. Thus, the required remainder = 1. The coefficient of x2 is -1. (i) p(0) = 5(0) – 4(0)2 + 3 = 0 – 0 + 3 = 3 Chapter-2 Chapter-10 Sol. (i) p(x)=x+5 Thus, zero of cx + d is \(-\frac { d }{ c }\), Question 1. (iv) p (x) = (x + 1) (x – 2), x = – 1,2 Solution: (iii) The zero of x is 0. is given, we can find the other two trigonometric ratios (i.e. (i) (99)3 Since, p(x) = 0 ⇒ x3 + y3 – 3xyz = -z3 Thus, x3 – 2x2 – x + 2 = (x – 1)(x + 1)(x – 2), (ii) We have, x3 – 3x2 – 9x – 5 ∴ (998)3 = (1000-2)3 = 4k[3y2 – 3y + 5y – 5] [Using a3 + b3 + 3 ab(a + b) = (a + b)3] (iii) x = 2 P(2) = (2 – 1)(2 + 1) = (1)(3) = 3. ∴ 3x3 + 7x is not divisib1e by 7 + 3x. (i) 8a3 +b3 + 12a2b+6ab2 = (x + 1)(x – 5)(x + 1) NCERT Solutions for Class 9 Maths are a set of solutions in the form of chapter-wise solutions made specifically for Class 9th students. So, it is a linear polynomial. Solution: Thus, 2y3 + y2 – 2y – 1 Solution: (iii) The degree of y + y2 + 4 is 2. (iii) We have, p(x) = x2 – 1 Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1. (i) We have , p(x) = 3x + 1 So, the degree of the polynomial is 0. Find p (0), p (1) and p (2) for each of the following polynomials. It is a polynomial in one variable i.e., x = (4m – 7n)[(4m)2 + (4m)(7n) + (7n)2] Solution: We have, p(x) = x3 – ax2 + 6x – a and zero of x – a is a. Question 4. [Using (x + a)(x + b) = x2 + (a + b)x + ab] (iii) We have 3 √t + t√2 = 3 √t1/2 + √2.t (i) x3 – 2x2 – x + 2 Ex 2.1 Class 9 Maths Question 1. Let x = -12, y = 7 and z = 5. Thus, the value of 5x – 4x2 + 3 at x = 0 is 3. = x2 + 14x+40, (ii) We have, (x+ 8) (x -10) (iv) 3x2 – x – 4 = 27 – 4(9) + 3 + 6 Using identity, Verify whether the following are zeroes of the polynomial, indicated against them. (vii) 7x3 Thus, zero of ax is 0. (i) Abmomial of degree 35 can be 3x35 -4. Represent geometrically 8.1 on number line. Question 2. (iv) The zero of x + π is -π. Classify the following as linear, quadratic and cubic polynomials. Factorise 27x3 +y3 +z3 -9xyz. ⇒ (x – y)(x2 + y2 + xy) = x3 – y3 We have, (3x + 4) (3x – 5) = (3x)2 + (4 – 5) x + (4) (- 5) (i) p (x)= 2x3 + x2 – 2x – 1, g (x) = x + 1 ∴p(0) = 2 + 0 + 2(0)2 – (0)3 (iv) 2y3 + y2 – 2y – 1 Learn about the degree of a polynomial and the zero of a polynomial with related Maths solutions. ∴ 993 = (100 – 1)3 CLASS 9 IX Sample Papers X Sample Papers CLASS 8 CLASS 07 Class 06 ... Chapter - 2: Polynomials. Class 9 Mathematics Notes for FBISE. All these questions are based on the important fundamental concepts given in NCERT Class 9 Maths. (iii) The given polynomial is \(\frac { \pi }{ 2 } { x }^{ 2 }\) + x. ∴ \(p(-\frac { 1 }{ 3 } )\quad =\quad 3(-\frac { 1 }{ 3 } )\quad +\quad 1\quad =\quad -1\quad +\quad 1\quad =\quad 0\) (i) 4x2 – 3x + 7 (i) We know that Thus, 6x2 + 5x – 6 = (2x + 3)(3x – 2), (iv) We have, 3x2 – x – 4 = 3x2 – 4x + 3x – 4 Class 9 Chapter 2 Polynomials Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks. = 10000 + (-9) + 20 = 9120 = (3 – 5a)3 = (2a – b) (2a – b) (2a – b), (iii) 27 – 125a3 – 135a + 225a2 Solution: CBSE Worksheets for Class 11 Maths: One of the best teaching strategies employed in most classrooms today is Worksheets. = (4m – 7n)(16m2 + 28mn + 49n2), Question 11. ∴ (-12)3 + (7)3 + (5)3 = 3[(-12)(7)(5)] = ( 100)2 + [(- 5) + (- 4)] 100 + (- 5 x – 4) = 2y3 – 2y2 + 3y2 – 3y + y – 1 Thus, the value of 5x – 4x2 + 3 at x = 2 is – 3. Thus, the possible length and breadth are (5a – 3) and (5a – 4). = (x + 1)(x2 – 4x – 5) (i) x + 1 = 4[3ky2 + 2ky – 5k] = 4[k(3y2 + 2y – 5)] (i) We have, p (x)= 2x3 + x2 – 2x – 1 and g (x) = x + 1 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. In this … = 3 x x x (x – 4) (ii) x = – 1 ∴ p(-2) = (-2)3 + 3(-2)2+ 3(-2) + 1 = 3(5460) = 16380. However, it is possible to avoid such a scenario by taking authentic NCERT solutions for class 9 maths from a reliable source. Thus, 2×2 + 7x + 3 = (2x + 1)(x + 3), (iii) We have, 6x2 + 5x – 6 = 6x2 + 9x – 4x – 6 = (x + 1)[x(x + 2) + 10(x + 2)] ∴ P(0) = (0)2 – 0 + 1 = 0 – 0 + 1 = 1 So, it is a cubic polynomial. (iii) x4 + 3x3 + 3x2 + x + 1 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, So, the degree of the polynomial is 1. (v) We have x10+  y3 + t50 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)] = (2a)3 + (b)3 + 6ab(2a + b) (i) 25a2 – 35a + 12 = 25a2 – 20a – 15a + 12 = 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3) Solution: (ii) Let p (x) = x4 + x3 + x2 + x + 1 Solution: (ii) p (t) = 2 +1 + 2t2 -t3 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… [Using (a – b)3 = a3 – b3 – 3ab (a – b)] = 1 – 1 + 1 – 1 + 1 Hindi Medium and English Medium both are available to free download. So, it is a quadratic polynomial. ∴ p(0) = (0 – 1)(0 + 1) = (-1)(1) = -1 Since, p(x) = 0 ∴ p (-1) = (-1)3 + (-1)2 + (-1) + 1 . = (2x + 3y – 4z)2 = (2x + 3y + 4z) (2x + 3y – 4z), (ii) 2x2 + y2 + 8z2 – 2√2xy + 4√2yz – 8xz For (x – 1) to be a factor of p(x), p(1) should be equal to 0. Solution: DronStudy provides you Chapter wise Solutions for Class 9th Maths, Science and Social Studies. Let x = 28, y = -15 and z = -13. = (x + 1)(x2 + 12x + 20) Since, p(x) = 0 (ii) A monomial of degree 100 can be √2y100. Factorise the following using appropriate identities Expand each of the following, using suitable identity It is a complete package of solutions to problems of your really tough book. = (1000)3– (2)3 – 3(1000)(2)(1000 – 2) = 8x3 + 12x2 + 6x + 1, (ii) (2a – 3b)3 = (2a)3 – (3b)3 – 3(2a)(3b)(2a – 3b) [By (2)] All the chapterwise questions with solutions to help you to revise complete CBSE syllabus and score more marks in Your board examinations. (iii) y + y2+4 [Using (a + b)(a -b) = a2– b2] p(1) = (1)2 – 1 + 1 = 1 – 1 + 1 = 1 3. Question 1. Write the coefficients of x2 in each of the following (ii) 8a3 -b3-12a2b+6ab2 = \(\frac { 1 }{ 2 }\) (x + y + 2)(x2 + y2 + y2 + z2 + z2 + x2 – 2xy – 2yz – 2zx) = ( 100)2 + (3 + 7) (100)+ (3 x 7) (i) We have, (x+ 4) (x + 10) = (y – 1)(y + 1)(2y +1), Question 1. = 2 + 0 + 0 – 0=2 (i) Area 25a2 – 35a + 12 Solution: because each exponent of x is a whole number. (i) We have, x3 – 2x2 – x + 2 Question 2. Solution: (iv) We have, p(x) = 3x – 2. (iv) We have y + \(y+\frac { 2 }{ y }\) = y + 2.y-1 (i) x3 + y3 = (x + y)-(x2 – xy + y2) Without actually calculating the cubes, find the value of each of the following 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. (iv) p (x) = kx2 – 3x + k 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. State reasons for your answer. (ii) The degree of x – x3 is 3. 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. Rational Numbers, irrational Numbers, rationalize irrational numbres, operation on real numbers, laws of … 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, … (v) The zero of 5 + 2x is \(-\frac { 5 }{ 2 }\) . = (x + 1)(x2 + 2x + 10x + 20) (ii) p(-1) = 5(-1) – 4(-1)2 + 3 (iii) We have, p(x) = 2x + 5. (iii) x2 – \(\frac { { y }^{ 2 } }{ 100 }\) Thus, the required remainder is \(-\frac { 27 }{ 8 }\) . = -1 CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. (iv) The degree of 1 + x is 1. Write the degree of each of the following polynomials. Question 15. Use the Factor Theorem to determine whether g (x) is a factor of p (x) in each of the following cases [Using a2 – 2ab + b2 = (a- b)2] = x(3x – 4) + 1(3x – 4) = (3x – 4)(x + 1) Ex 2.1 Class 9 Maths Question 5. = 1000000000 – 8 – 6000(1000 – 2) (i) x = 0 Class-IX CHAPTER – 1 Number System (Maths Assignment) 1. Thus, zero of 3x – 2 is \(\frac { 2 }{ 3 }\). Hence, verified. = (100)3 + (2)3 + 3(100)(2)(100 + 2) = (3x + y + z)(9x2 + y2 + z2 – 3xy – yz – 3zx). = (3x)2 + 2(3x)(y) + (y)2 Solution: = (100)3 – 13 – 3(100)(1)(100 -1) (ii) The given polynomial is 2 – x2 + x3. 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. (iv) x + π ⇒ x = -5. (iii) (998)3 = 4k x (3y + 5) x (y – 1) ⇒ (x – y)[(x – y)2 + 3xy)] = x3 – y3 These expert faculties solve and provide the NCERT Solution for class 9 so that it would help students to solve the problems comfortably. ∴ p(-1) = (-1 +1) (-1 – 2) = (0)(- 3) = 0 (iv) y+ \(\frac { 2 }{ y }\) Since, p(-1) = 0, so, x = -1 is a zero of (x + 1)(x – 2). = 1000300 – 30001 = 970299, (ii) We have, 102 =100 + 2 So, it is a linear polynomial. Since, x + y + z = 0 = 8x3 + 1 + 6x(2x + 1) Solution: = -1 + 1 – 1 + 1 (i) Here, p(x) = x2 + x + k = -8 + 12 – 6 + 1 Solution: NCERT Solutions for Class 9 Maths: The Class 9 Maths textbook solutions of exercises for all 15 chapters are included in this article. p(2) = (2)2 – 2 + 1 = 4 – 2 + 1 = 3 1et p(x) = 5x – 4x2 + 3 This solution is strictly revised in accordance … Since, \(p(\frac { 1 }{ 2 } )\) ≠ 0, so, x = \(\frac { 1 }{ 2 }\) is not a zero of 2x + 1. ∴ The possible dimensions of the cuboid are 3, x and (x – 4). Write the following numbers in p/q form (i) 2.015 (ii) 0.235 Ans (399 235 ' 198 999) 4. (v) p (x) = 3x (iii) p (x) = 2x + 5 Practise the solutions to understand how to use the remainder theorem to work out the remainder of a polynomial. (i) Let p (x) = x3 + x2 + x + 1 (i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz Thus, 12x2 -7x + 3 = (2x – 1) (x + 3), (ii) We have, 2x2 + 7x + 3 = 2x2 + x + 6x + 3 ⇒ (x + y)[(x + y)2-3xy] = x3 + y3 NCERT Solutions Class 9 Maths Chapter 2 Polynomials. (ii) p (x) = 5x – π, x = \(\frac { 4 }{ 5 }\) Question 10. [Using (a – b)3 = a3 – b3 – 3ab (a – b)] Thus, 7 + 3x is not a factor of 3x3 + 7x. (iii) x3 + 13x2 + 32x + 20 CBSE Class 11 Maths Worksheet for students has been used by teachers & students to develop logical, lingual, analytical, and … Solution: (i) p(x) = 3x + 1,x = –\(\frac { 1 }{ 3 }\) (iv) 3 The highest power of variable t is 1. We have, (x + 4) (x + 10) = x2+(4 + 10) x + (4 x 10) We know that = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) Teachoo is free. (v) p (x) = x2, x = 0 = 9x2 – x – 20, Question 2. = [(x)2 – (1)2](x – 2) FREE Downloadable! (vii) We have, p(x) = cx + d. Since, p(x) = 0 (v) (- 2x + 5y – 3z)2 NCERT solutions for session 2019-20 is now available to download in PDF form. Get here MCQs on Class 9 Maths Chapter 2- Polynomials with answers. (i) p(y) = y2 – y +1 = x3 + y3 + z3 – 3xyz = L.H.S. 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, … (i) x2+ x (i)We have, 103 x 107 = (100 + 3) (100 + 7) = 2(-1) + 1 + 2 – 1 Solution: CBSE Class 9 Maths Worksheet for students has been used by teachers & students to develop logical, lingual, … (i) The degree of x2 + x is 2. (ii) (x+8) (x -10) ⇒ 3x = 2 Thus, the required remainder is 5a. Solution: (iv) x3 – x2 – (2 +√2 )x + √2 Solution: (vii) p (x) = cx + d, c ≠ 0 where c and d are real numbers. x3 +y3 +z3 – 3xyz = \(\frac { 1 }{ 2 }\) (x + y+z)[(x-y)2 + (y – z)2 +(z – x)2] (vi) p (x) = 1x + m, x = – \(\frac { m }{ 1 }\) = (3x -1) (4x -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. = 8a3 – 27b3 – 18ab(2a – 3b) (i) Volume 3x2 – 12x (iii) \(\frac { \pi }{ 2 }\) x2 + x Telanagana State Board Class 9 Mathematics Textbook Solution Chapter-wise Telanagana SCERT Class IX Math Solution. We have, 27y3 + 125z3 = (3y)3 + (5z)3 Since, p(1) = k(1)2 – (1) + 1 = 1 – 3 + 3 – 1 + 1 = 1 (iv) Let p (x) = x3 – x2 – (2 + √2) x + √2 Since p(-1) = 0, so, x = -1, is also a zero of x2 – 1. Since,( \(-\frac { 490 }{ 9 }\)) ≠ 0 Solution: Thus, zero of 3x is 0. There is plenty to learn in this chapter about the definition and examples of polynomials, coefficient, degrees, and terms in a polynomial. Mathematics Part - II Solutions Textbook Solutions for Class 9 Math. (i) 8a3 +b3 +12a2b+6ab2 = 2 + 2 + 8 – 8 = 4 ∴ p(0) = (0)3 + 3(0)2 + 3(0) + 1 because each exponent of y is a whole number. = (4a – 3b)(4a – 3b)(4a – 3b). x3 + y3 = (x + y)(x2 – xy + y2) Find the value of k, if x – 1 is a factor of p (x) in each of the following cases (iii) p (x) = kx2 – √2 x + 1 = 2y2(y – 1) + 3y(y – 1) + 1(y – 1) Solution: = (100)2-42 So, it is not a polynomial in one variable. ∴ p(o) = (0)2 = 0 Solution: (iii) x (iii) 3 √t + t√2 = \(\frac { 1 }{ 2 }\) (x + y + 2)[(x2 + y2 – 2xy) + (y2 + z2 – 2yz) + (z2 + x2 – 2zx)] ∴ \(p(\frac { 1 }{ 2 } )\quad =\quad 2(\frac { 1 }{ 2 } )+1=\quad 1+1\quad =\quad 2\) Exercise 13.2 Solution. The coefficient of x2 is 1. On signing up you are confirming that you have read and agree to 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. (ii) 35y2+ 13y -12 = 35y2 + 28y – 15y -12 (i) (x + 4)(x + 10) (ii) Volume 12ky2 + 8ky – 20k which is not a whole number. = (2 a + b)3 ∴ p(-1) = (-1)3 + 3(-1)2 + 3(-1) +1 What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (i) 12x2 – 7x +1 NCERT Solutions Class 9 Maths Chapter 2 Polynomials are provided here. (i) 2 + x2 + x The highest (ii) Given that p(t) = 2 + t + 2t2 – t3 (viii) p (x) = 2x + 1, x = \(\frac { 1 }{ 2 }\) (iv) Given that p(x) = (x – 1)(x + 1) We have, 64m3 – 343n3 = (4m)3 – (7n)3 ⇒ cx + d = 0 ⇒ cx = -d ⇒ \( x =-\frac { d }{ c }\) = (y – 1)[2y(y + 1) + 1(y + 1)] (vi) r2 = (2y -1)2 To help students in making easy preparations, we are providing the MCQs for class 9 NCERT Maths. = a3 – a3 + 6a – a = 5a Question 2. Chapter-3 Chapter-11 Sol. R.H.S = 4k[3y(y – 1) + 5(y – 1)] = (x + 1)(x + 2)(x + 10) Variables and expression are called as indeterminate and coefficients. Terms of Service. = (- √2x + y + 2 √2z)2 Using identity, p(1) = (1 – 1)(1 +1) = (0)(2) = 0 (iii) (- 2x + 3y + 2z)2 (iii) 104 x 96 (ii) We have, p(x) = 5x – π Question 13. (iii) 6x2 + 5x – 6 ⇒ P (-1) ≠ 1 NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1 are part of NCERT Solutions for Class 9 Maths. = 10 – 16 + 3 = -3 Question 3. (iii) p(2) = 5(2) – 4(2)2 + 3 = 10 – 4(4) + 3 = (x + 1)(x2 – 5x + x – 5) (i) p (x) = x2 + x + k 12x2 – 7x + 1 = 12x2 – 4x- 3x + 1 p( 2) = 2 + 2 + 2(2)2 – (2)3 ⇒ 3x – 2 = 0 Question 3. Solution: It is not a polynomial, because one of the exponents of t is \(\frac { 1 }{ 2 }\), [Using (x + a)(x + b) = x2 + (a + b)x + ab] (iii) Given that p(x) = x3 So, it is a quadratic polynomial. = 10000 – 16 = 9984, Question 3. ⇒ p (-1) ≠ 0 ⇒ (x – y)3 + 3xy(x – y) = x3 – y3 (ii) (2x – y + z)2 After the completion of chapters from NCERT Books and Exemplar Books, students should go for Assignments. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. 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 (ii) 2 – x2 + x3 ∴ P(-1) = (-1)4 + (-1)3 + (-1)2 + (-1)+1 ⇒ x + 5 = 0 Thus, zero of x + 5 is -5. = (3x + y)2 = 10000 + (10) x 100 + 21 So, (x + 1) is not a factor of x4 + x3 + x2 + x+ 1. = 1000000 – 1 – 300(100 – 1) (i) The zero of x + 1 is -1. Question 9. (x + a) (x + b) = x2 + (a + b) x + ab We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz k = -2 – √2 = -(2 + √2), (iii) Here, p (x) = kx2 – √2 x + 1 (i) Given that p(y) = y2 – y + 1. = 3x(2x + 3) – 2(2x + 3) You have these advantages of browsing notes from our website. = (2a)3 – (b)3 – 3(2a)(b)(2a – b) They are in a list with arrows. Solution: (v) The degree of 3t is 1. ∴ p(-1) = 2(-1)3 + (-1)2 – 2(-1) – 1 (vi) p (x)= ax, a≠0 = 8a3 – 27b3 – 36a2b + 54ab2, Question 7. Thus, the required remainder is -π3 + 3π2 – 3π+1. Exercise 14.1 Solution. [Using a3 + b3 + 3 ab(a + b) = (a + b)3] ∴ p(3) = (3)3 – 4(3)2 + 3 + 6 Determine which of the following polynomials has (x +1) a factor. (ii) Area 35y2 + 13y – 12 10 Questions. ⇒ 3x = 0 ⇒ x = 0 (ii) x3 – 3x2 – 9x – 5 Ex 2.1 Class 9 Maths Question 2. Class 9 Maths Chapter 2 Polynomials This chapter guides you through algebraic expressions called polynomial and various terminologies related to it. Homework Help with Chapter-wise solutions and Video explanations. CBSE Worksheets for Class 9 Maths: One of the best teaching strategies employed in most classrooms today is Worksheets. Chapter wise assignments for class 9 Maths are given below updated for new academic session 2020-2021. = 3[-420] = -1260, (ii) We have, (28)3 + (-15)3 + (-13)3 Write the following cubes in expanded form Login to view more pages. = -2 + 1 + 2 -1 = 0 Note: Important questions have also been marked for your reference. = x3 + x2 – 4x2 – 4x – 5x – 5 , Extra questions for class 9 maths chapter 1 with solution. It is not a polynomial, because one of the exponents of y is -1, Click on exercise or topic link below to get started. = 0 + 0 + 0 + 1 = 1 Chapter -1 Sol. Chapter 13 Geometrical Constructions. (v) x10+ y3+t50 (ii) 4 – y2 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. Verify that (ii) x – x3 Chapter-9 Chapter-2 Sol. = (y – 1)(2y2 + 3y + 1) (ii) x4 + x3 + x2 + x + 1 Factorise So, (x + 1) is not a factor of x3 – x2 – (2 + √2) x + √2. (ii) 4y2-4y + 1 = 2 + 1 + 2 – 1 = 4 = (x + 1)[x(x – 5) + 1(x – 5)] (x+ a) (x+ b) = x2 + (a + b) x+ ab. (ii) Here, p (x) = 2x2 + kx + √2 Using the identity, = k – 3 + k (iv) We have, p(x) = (x + 1)(x – 2) = (- √2x)2 + (y)2 + (2 √2z)2y + 2(- √2x) (y)+ 2 (y) (2√2z) + 2 (2√2z) (- √2x) We hope the NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1, help you. Thus, 3x2 – x – 4 = (3x – 4)(x + 1), Question 5. Solution: Solution: ⇒ p (-1) ≠ 0 Find the remainder when x3 + 3x2 + 3x + 1 is divided by Since, p(x) = 0 = x2(x + 1) + 12x(x +1) + 20(x + 1) Area of a rectangle = (Length) x (Breadth) Thus, zero of 2x + 5 is \(\frac { -5 }{ 2 }\) . = (4a – 3b)3 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. 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