(i) x3 + y3 = (x + y)-(x2 – xy + y2) = x(x2 – 1) – 2(x2 -1) = (x2 – 1)(x – 2) = -1 + 3- 3 + 1 = 0 Get here MCQs on Class 9 Maths Chapter 2- Polynomials with answers. (iii) The zero of x is 0. ⇒ (x + y)[(x + y)2-3xy] = x3 + y3 So, (x+ 1) is a factor of x3 + x2 + x + 1. Thus, the possible length and breadth are (7y – 3) and (5y + 4). Class-IX CHAPTER – 1 Number System (Maths Assignment) 1. Evaluate the following products without multiplying directly = x3 + y3 + z3 – 3xyz = L.H.S. Contains solved exercises, review questions, MCQs, important board questions and chapter overview. So, it is a cubic polynomial. 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. We have, p(x) = 3x3+7x. (iii) p (x) = x2 – 1, x = x – 1 = 1000000000 – 8 – 6000(1000 – 2) We have, (3x + 4) (3x – 5) = (3x)2 + (4 – 5) x + (4) (- 5) Without actually calculating the cubes, find the value of each of the following (i) Volume 3x2 – 12x (i) x2+ x Ex 2.1 Class 9 Maths Question 4. = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) We have, (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z) ⇒ p (-1) ≠ 0 Since,( \(-\frac { 490 }{ 9 }\)) ≠ 0 Chapter 13 Geometrical Constructions. Question 13. = (x + 1)(x2 – 5x + x – 5) (i) ∵ (x + y)3 = x3 + y3 + 3xy(x + y) = (2x + 1)(x + 3) So, (x + 1) is not a factor of x3 – x2 – (2 + √2) x + √2. (ii) p(-1) = 5(-1) – 4(-1)2 + 3 Represent the following irrational numbers on number line. You have these advantages of browsing notes from our website. ∴ p(0) = (0 – 1)(0 + 1) = (-1)(1) = -1 These ncert book chapter wise questions and answers are very helpful for CBSE board exam. Download File. ∴ p (-1) = (-1)3 + (-1)2 + (-1) + 1 . ⇒ k = -2. So, the degree of the polynomial is 3. CBSE Worksheets for Class 9 Maths: One of the best teaching strategies employed in most classrooms today is Worksheets. = (3x + y)(3x + y), (ii) We have, 4y2 – 4y + 12 = 10000 + (10) x 100 + 21 Question 12. = (1000)3– (2)3 – 3(1000)(2)(1000 – 2) (ii) 2x2 + y2 + 8z2 – 2√2xy + 4√2yz – 8xz and zero of 7 + 3x is \(-\frac { 7 }{ 3 }\). (i) 10 (ii) 17 (iii)2+ 2 2. (v) (3 – 2x) (3 + 2x) Solution: (ii) x – x3 Question 2. (v) 5 + 2x Chapter-3 Chapter-11 Sol. (iii) (998)3 ⇒ 3x – 2 = 0 (i)We have, 103 x 107 = (100 + 3) (100 + 7) = 1000000 + 8 + 60000 + 1200 = 1061208, (iii) We have, 998 = 1000 – 2 Solution: [Using a3 + b3 + 3 ab(a + b) = (a + b)3] So, the degree of the polynomial is 2. = (3 – 5a)3 Check whether 7 + 3x is a factor of 3x3+7x. (viii) p (x) = 2x + 1, x = \(\frac { 1 }{ 2 }\) Factorise (i) 8a3 +b3 + 12a2b+6ab2 (i) We have, 9x2 + 6xy + y2 (x+ a) (x+ b) = x2 + (a + b) x+ ab. (x + a) (x + b) = x2 + (a + b) x + ab = (2 a + b)3 (x + a) (x + b) = x2 + (a + b) x + ab p(2) = (2)2 – 2 + 1 = 4 – 2 + 1 = 3 There is plenty to learn in this chapter about the definition and examples of polynomials, coefficient, degrees, and terms in a polynomial. (iii) Given that p(x) = x3 (vii) p (x) = cx + d, c ≠ 0 where c and d are real numbers. Chapter wise assignments for class 9 Maths are given below updated for new academic session 2020-2021. Extra questions along with questions of NCERT book complete the topic . = \(\frac { 1 }{ 2 }\) (x + y + z)[2(x2 + y2 + z2 – xy – yz – zx)] ∴ P(-1) = (-1)4 + (-1)3 + (-1)2 + (-1)+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. = (x + 1)(x2 – 4x – 5) Thus, 6x2 + 5x – 6 = (2x + 3)(3x – 2), (iv) We have, 3x2 – x – 4 = 3x2 – 4x + 3x – 4 Get NCERT solutions for Class 9 Maths free with videos of each and every exercise question and examples. Question 3. Question 10. Teachoo is free. = 1000000 – 1 – 300(100 – 1) = 3[-420] = -1260, (ii) We have, (28)3 + (-15)3 + (-13)3 (ii) (x+8) (x -10) = 4x2 + 25y2 + 9z2 – 20xy – 30yz + 12zx, Question 5. (ii) The given polynomial is 2 – x2 + x3. Since, p(0) = 0, so, x = 0 is a zero of x2. = (2a)3 + (b)3 + 3(2a)(b)(2a + b) (iii) We have, p(x) = 2x + 5. 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. So, the degree of the polynomial is 1. x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) Hindi Medium and English Medium both are available to free download. = -1 – 1 + 2 + √2 + √2 Chapter 4 Linear Equations in Two Variables. 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. Login to view more pages. (vii) P (x) = 3x2 – 1, x = – \(\frac { 1 }{ \sqrt { 3 } }\),\(\frac { 2 }{ \sqrt { 3 } }\) Since, p(x) = 0 Chapter-9 Chapter-2 Sol. Use the Factor Theorem to determine whether g (x) is a factor of p (x) in each of the following cases (iv) Given that p(x) = (x – 1)(x + 1) = 27 – 36 + 3 + 6 = 0 = (3y + 5z)(9y2 – 15yz + 25z2), (ii) We know that (ii) We have, 12ky2 + 8ky – 20k (i) We have, 99 = (100 -1) For (x – 1) to be a factor of p(x), p(1) should be equal to 0. = \(\frac { 1 }{ 2 }\)(x + y + z)[(x – y)2+(y – z)2+(z – x)2] Middle school is one of the most crucial periods a student undergoes in the course of their schooling years. These expert faculties solve and provide the NCERT Solution for class 9 so that it would help students to solve the problems comfortably. ∴ P(0) = (0)2 – 0 + 1 = 0 – 0 + 1 = 1 So, it is a quadratic polynomial. = 4k x (3y + 5) x (y – 1) We know that if x + y + z = 0, then, x3 + y3 + z3 = 3xyz We have, p(x) = x3 – ax2 + 6x – a and zero of x – a is a. We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz (i) We have, (-12)3 + (7)3 + (5)3 ⇒ x3 + y3 + z3 = 3xyz Chapter 2: Polynomials. Class 9 maths printable worksheets, online practice and online tests. x3 +y3 +z3 – 3xyz = \(\frac { 1 }{ 2 }\) (x + y+z)[(x-y)2 + (y – z)2 +(z – x)2] 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”. = (2x + 3)(3x – 2) power of the variable y is 2. = (y – 1)(2y2 + 2y + y + 1) Thus, 2y3 + y2 – 2y – 1 Question 16. CLASS 9 IX Sample Papers X Sample Papers CLASS 8 CLASS 07 Class 06 ... Chapter - 2: Polynomials. = (x + 1)(x + 2)(x + 10), (iv) We have, 2y3 + y2 – 2y – 1 NCERT Solutions for Class 9 Maths: The Class 9 Maths textbook solutions of exercises for all 15 chapters are included in this article. [Using a3 + b3 + 3 ab(a + b) = (a + b)3] (i) (x+2y+ 4z)2 [Using a2 – 2ab + b2 = (a- b)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. = 2y2(y – 1) + 3y(y – 1) + 1(y – 1) Learn about the degree of a polynomial and the zero of a polynomial with related Maths solutions. Since, p(-1) = 0, so, x = -1 is a zero of (x + 1)(x – 2). = (2x + 3y – 4z)2 = (2x + 3y + 4z) (2x + 3y – 4z), (ii) 2x2 + y2 + 8z2 – 2√2xy + 4√2yz – 8xz (i) The degree of x2 + x is 2. = 1000000000 – 8 – 6000000 +12000 (iii) We have 3 √t + t√2 = 3 √t1/2 + √2.t (vi) r2 Solution: [Using a3 – b3 – 3 ab(a – b) = (a – b)3] (iv) 3x2 – x – 4 ⇒ x + y = -z (x + y)3 = (-z)3 Factorise each of the following = (- √2x + y + 2 √2z) (- √2x + y + 2 √2z), Question 6. (v) (- 2x + 5y – 3z)2 ⇒ x3 + y3 + 3xy(x + y) = -z3 (i) We have, p (x)= 2x3 + x2 – 2x – 1 and g (x) = x + 1 Thus, 7 + 3x is not a factor of 3x3 + 7x. ∴ \(p(\frac { 1 }{ 2 } )\quad =\quad 2(\frac { 1 }{ 2 } )+1=\quad 1+1\quad =\quad 2\) Question 3. = -π3 + 3π2 + (-3π) + 1 ⇒ (x – y)3 + 3xy(x – y) = x3 – y3 ∴ The possible dimensions of the cuboid are 3, x and (x – 4). ⇒ x + 5 = 0 (iv) p (x) = (x-1) (x+1) The coefficient of x2 is -1. (iii) x3 + 13x2 + 32x + 20 = x2(x + 1) + 12x(x +1) + 20(x + 1) ∴ p(3) = (3)3 – 4(3)2 + 3 + 6 R.H.S (iii) p (x) = kx2 – √2 x + 1 = 3 x x x (x – 4) = 1 – 3 + 3 – 1 + 1 = 1 Using the identity, = 2(-1) + 1 + 2 – 1 (iii) We have, p(x) = x2 – 1 Chapter - 3 Pair of Linear Equations. (i) The given polynomial is 5x3 + 4x2 + 7x. All these questions are based on the important fundamental concepts given in NCERT Class 9 Maths. (iii) 104 x 96 ⇒ p (-1) ≠ 0 [Using (x + a)(x + b) = x2 + (a + b)x + ab] So, (x + 1) is not a factor of x4 + x3 + x2 + x+ 1. (iii) x = (4a – 3b)(4a – 3b)(4a – 3b). (i) 4x2 – 3x + 7 ⇒ x = \(\frac { 2 }{ 3 }\) Solution: Class 9 Maths Chapter 2 Polynomials This chapter guides you through algebraic expressions called polynomial and various terminologies related to it. The coefficient of x2 is 0. NCERT Solutions for Class 9 Maths Exercise 9.2 book solutions are available in PDF format for free download. (i) x3+x2+x +1 x3 + y3 + z3 = 3xyz, Question 14. (iv) The zero of x + π is -π. = 4x (3x – 1 ) -1 (3x – 1) = (3x)2 + 2(3x)(y) + (y)2 = (x – 1)(x + 1)(x – 2) (iv) Let p (x) = x3 – x2 – (2 + √2) x + √2 (ii) 35y2+ 13y -12 = 35y2 + 28y – 15y -12 = 27 – 4(9) + 3 + 6 NCERT Solutions Class 9 Maths Chapter 2 Polynomials. Since, p(1) = 2(1)2 + k(1) + √2 In this article, you will get the MCQs on Class 9 Maths Chapter 4: Linear Equations in Two Variables. Solution: Use suitable identities to find the following products Exercise 14.1 Solution. It is a complete package of solutions to problems of your really tough book. Solution: (i) p(x)=x+5 Write the following numbers in p/q form (i) 2.015 (ii) 0.235 Ans (399 235 ' 198 999) 4. (i) 2 + x2 + x (iv) x + π Find the remainder when x3 – ax2 + 6x – a is divided by x – a. 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. (v) We have, p(x) = x2 Solution: So, (x + 1) is not a factor of x4 + 3x3 + 3x2 + x+ 1. Since, p(x) = 0 (v) x10+ y3+t50 Factorise each of the following ⇒ 3x = 0 ⇒ x = 0 Thus, the required remainder is -π3 + 3π2 – 3π+1. NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12. Solution: (iv) 1 + x 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)] = 4k[3y2 – 3y + 5y – 5] = k – 3 + k (i) The zero of x + 1 is -1. 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(2x + 1) + 3(2x + 1) = a3 – a3 + 6a – a = 5a (i) 12x2 – 7x +1 = x2 + 14x+40, (ii) We have, (x+ 8) (x -10) So, the degree of the polynomial is 0. Teachoo provides the best content available! = (x + 1)(x2 + 2x + 10x + 20) = (3x + y)2 We have, 27y3 + 125z3 = (3y)3 + (5z)3 (iii) (- 2x + 3y + 2z)2 (i) We know that (ii) x4 + x3 + x2 + x + 1 ⇒ x3 + y3 + 3xy(-z) = -z3 [∵ x + y = -z] (ii) We have, p(x) = 5x – π For example, if you are weak in class 9 maths, you can’t make a great career in the field of engineering and mathematics. (iii) p(2) = 5(2) – 4(2)2 + 3 = 10 – 4(4) + 3 = (2a – b)3 Hence, verified. (iv) Since, 3 = 3x° [∵ x°=1] (ii) (2x – y + z)2 (i) (x + 4)(x + 10) (v) We have x10+  y3 + t50 Thus, the possible dimensions of the cuboid are 4k, (3y + 5) and (y -1). (ii) p (x) = 5x – π, x = \(\frac { 4 }{ 5 }\) (i) 5x3+4x2 + 7x (i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz [∵ (a2 – b2) = (a + b)(a-b)] (ii) x3 – y3 = (x – y) (x2 + xy + y2) The highest power of variable t is 1. Download NCERT Solutions For Class 9 Maths in PDF based on latest pattern of CBSE in 2020 - 2021. = (4a – 3b)3 = 2 + 1 + 2 – 1 = 4 = 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 = √2 -1, (iv) Here, p(x) = kx2 – 3x + k Thus, the value of 5x – 4x2 + 3 at x = -1 is -6. Hence, verified. = (3y + 5z)[(3y)2 – (3y)(5z) + (5z)2] (i) 9x2 + 6xy + y2 Thus, the value of 5x – 4x2 + 3 at x = 0 is 3. = -14 + 13 = 4 x k x (3y2 + 2y – 5) = (x + 1)(x – 5)(x + 1) (vii) The degree of 7x3 is 3. Since, x + y + z = 0 (v) p (x) = 3x = \(\frac { 1 }{ 2 }\) (x + y + 2)[(x2 + y2 – 2xy) + (y2 + z2 – 2yz) + (z2 + x2 – 2zx)] 27x3 + y3 + z3 – 9xyz = (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z) Thus, zero of x – 5 is 5. So, it is not a polynomial in one variable. = 3x(2x + 3) – 2(2x + 3) (iv) We have, p(x) = 3x – 2. [Using a2 + 2ab + b2 = (a + b)2] These assignments will be available in updated form along with new assignments and chapter wise tests with solutions. Verify Thus, the required remainder = \(\frac { 27 }{ 8 }\). 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. ⇒ x = \(\frac { -5 }{ 2 }\) (iii) y + y2+4 Question 4. (i) We have, x3 – 2x2 – x + 2 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(2) = 0, so, x = 2 is also a zero of (x + 1)(x – 2). = 2 + 0 + 0 – 0=2 = (3x -1) (4x -1) Thus, the required remainder = 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 ∴ p(-1) = (-1 +1) (-1 – 2) = (0)(- 3) = 0 = 8a3 – 27b3 – 36a2b + 54ab2, Question 7. = 1000000 -1 – 30000 + 300 So, it is a linear polynomial. = (2a + b)(2a + b)(2a + b), (ii) 8a3 – b3 – 12o2b + 6ab2 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, … (iv) We have y + \(y+\frac { 2 }{ y }\) = y + 2.y-1 = -1 + 1 – 1 + 1 = [(x)2 – (1)2](x – 2) (iv) p (x) = 3x – 2 Volume of a cuboid = (Length) x (Breadth) x (Height) (i) Here, p(x) = x2 + x + k p(1) = (1)2 – 1 + 1 = 1 – 1 + 1 = 1 (ii) Given that p(t) = 2 + t + 2t2 – t3 ⇒ k = \(\frac { 3 }{ 4 }\), Question 4. (ii) 64m3 – 343n3 (vi) We have, p(x) = ax, a ≠ 0. Ex 2.1 Class 9 Maths Question 3. = 1 – 1 + 1 – 1 + 1 Solution: (iii) \(\frac { \pi }{ 2 }\) x2 + x Classify the following as linear, quadratic and cubic polynomials. Hence, if x + y + z = 0, then CBSE Class 9 Maths Worksheet for students has been used by teachers & students to develop logical, lingual, … = x2 – 2x – 80, (iii) We have, (3x + 4) (3x – 5) (iii) 6x2 + 5x – 6 (viii) We have, p(x) = 2x + 1 = (100)3 – 13 – 3(100)(1)(100 -1) (i) p(y) = y2 – y +1 = (2y -1)2 If x + y + z = 0, show that x3 + y3 + z3 = 3 xyz. (i) Area 25a2 – 35a + 12 ∴p(0) = 2 + 0 + 2(0)2 – (0)3 ⇒ p(1) = k + 2 = 0 = (3 – 5a) (3 – 5a) (3 – 5a), (iv) 64a3 -27b3 -144a2b + 108ab2 because each exponent of x is a whole number. (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx, (i) (x + 2y + 4z)2 = x2 (x + 1) – 4x(x + 1) – 5(x + 1) = (x + 1)(x2 + 12x + 20) (ii) Let p (x) = x4 + x3 + x2 + x + 1 (ii) y2 + √2 Exercise 13.1 Solution. (i) 25a2 – 35a + 12 = 25a2 – 20a – 15a + 12 = 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3) Which of the following expressions are polynomials in one variable and which are not? Chapter-10 Chapter-3 Sol. Factorise (ii) 4 – y2 (v) The zero of 5 + 2x is \(-\frac { 5 }{ 2 }\) . (ii) We have, p(x) x3 + 3x2 + 3x + 1 and g(x) = x + 2 ⇒ 2x = -5 (ii) A monomial of degree 100 can be √2y100. (i) p(x) = 3x + 1,x = –\(\frac { 1 }{ 3 }\) 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. (ii) The degree of x – x3 is 3. Mathematics Part - II Solutions Textbook Solutions for Class 9 Math. (iii) p (x) = x3 – 4x2 + x + 6, g (x) = x – 3 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. [Using (a – b)3 = a3 – b3 – 3ab (a – b)] Thus, zero of 3x – 2 is \(\frac { 2 }{ 3 }\). Using identity, Give one example each of a binomial of degree 35, and of a monomial of degree 100. = (3x + y + z)(9x2 + y2 + z2 – 3xy – yz – 3zx). (i) Abmomial of degree 35 can be 3x35 -4. This solution is strictly revised in accordance … (i) We have, 3x2 – 12x = 3(x2 – 4x) Click on exercise or topic link below to get started. the remainder is not 0. (i) p(0) = 5(0) – 4(0)2 + 3 = 0 – 0 + 3 = 3 Chapter-1 Chapter-9 Sol. NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1 are part of NCERT Solutions for Class 9 Maths. (iii) p (x) = 2x + 5 (ii) p(x)= x3 + 3x2 + 3x + 1, g (x) = x + 2 Using identity, [Using (x + a)(x + b) = x2 + (a + b)x + ab] (v) 3t [Using (a – b)3 = a3 – b3 – 3ab (a – b)] (ii) 95 x 96 So, it is a quadratic polynomial. (vi) p (x)= ax, a≠0 We have, Solution: (i) Given that p(y) = y2 – y + 1. Ex 2.1 Class 9 Maths Question 5. = (y – 1)(2y2 + 3y + 1) = 2 + 2 + 8 – 8 = 4 (ii) p (x) = x – 5 The zero of x + 1 is -1. So, it is a cubic polynomial. (iv) p (x) = (x + 1) (x – 2), x = – 1,2 [Using a3 + b3 + 3 ab(a + b) = (a + b)3] (iii) (3x + 4) (3x – 5) (ii) 2 – x2 + x3 Rational Numbers, irrational Numbers, rationalize irrational numbres, operation on real numbers, laws of … Since, p(x) = 0 => ax = 0 => x-0 = -8 + 12 – 6 + 1 = (4a)3 – (3b)3 – 3(4a)(3b)(4a – 3b) ⇒ p(3) = 0, so g(x) is a factor of p(x). = x(3x – 4) + 1(3x – 4) = (3x – 4)(x + 1) = 2 + k + √2 =0 FREE Downloadable! (iv) 3 Solution: = (x + 1)[x(x – 5) + 1(x – 5)] Solution: Solution: Since, p(1) = 0, so x = 1 is a zero of x2 -1. = (2y)2 + 2(2y)(1) + (1)2 (i) 27y3 + 125z3 (iv) 64a3 -27b3 -144a2b + 108ab2 = 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) Find the zero of the polynomial in each of the following cases = (x + 1)[x(x + 2) + 10(x + 2)] (iii) The given polynomial is \(\frac { \pi }{ 2 } { x }^{ 2 }\) + x. = 10000 + 1000 + 21=11021, (ii) We have, 95 x 96 = (100 – 5) (100 – 4) is given, we can find the other two trigonometric ratios (i.e. p(-π) = (-π)3 + 3(- π)22 + 3(- π) +1 = (4m – 7n)(16m2 + 28mn + 49n2), Question 11. (vi) The degree of r2 is 2. Chapter-2 Chapter-10 Sol. Thus, zero of cx + d is \(-\frac { d }{ c }\), Question 1. ∴ p (- 1) =(- 1)3- (-1)2 – (2 + √2)(-1) + √2 = 7y(5y + 4) – 3(5y + 4) = (5 y + 4)(7y – 3) (v) The degree of 3t is 1. [Using (a + b)3 = a3 + b3 + 3ab (a + b)] = (- √2x)2 + (y)2 + (2 √2z)2y + 2(- √2x) (y)+ 2 (y) (2√2z) + 2 (2√2z) (- √2x) All the chapterwise questions with solutions to help you to revise complete CBSE syllabus and score more marks in Your board examinations. We know that Since, p(x) = 0 Let x = -12, y = 7 and z = 5. ⇒ 2x + 5 =0 (ii) (102)3 ∴ p(-2) = (-2)3 + 3(-2)2+ 3(-2) + 1 Let x = 28, y = -15 and z = -13. Solutions to all NCERT Exercise Questions and Examples of Chapter 2 Class 9 Polynomials are provided free at Teachoo. (i) (99)3 = 4k[(3y + 5) x (y – 1)] = 2k – 3 = 0 Thus, x3 – 2x2 – x + 2 = (x – 1)(x + 1)(x – 2), (ii) We have, x3 – 3x2 – 9x – 5 The highest (ii) 4y2-4y + 1 Solution: ∴ p(o) = (0)2 = 0 x3 + y3 = (x + y)(x2 – xy + y2) x3 – y3 = (x – y)(x2 + xy + y2) Unit 1 - Matrices & Determinants. (iv) We have, p(x) = (x + 1)(x – 2) Solution: Question 9. Thus, x3 + 13x2 + 32x + 20 Verify whether the following are zeroes of the polynomial, indicated against them. Question 1. = 8x3 + 1 + 6x(2x + 1) P(1) = 2 + 1 + 2(1)2 – (1)3 (vi) p (x) = 1x + m, x = – \(\frac { m }{ 1 }\) (ii) (28)3 + (- 15)3 + (- 13)3 Along with recalling the knowledge of linear … = (3)3 – (5a)3 – 3(3)(5a)(3 – 5a) = 10000 + (-900) + 20 = 9120, (iii) We have 104 x 96 = (100 + 4) (100 – 4) (i) We have, Telanagana State Board Class 9 Mathematics Textbook Solution Chapter-wise Telanagana SCERT Class IX Math Solution. Determine which of the following polynomials has (x +1) a factor. Question 1. (ii) Area 35y2 + 13y – 12 ∴ p(0) = (0)3 = 0, p(1) = (1)3 = 1 10 Questions. (iv) The given polynomial is √2 x – 1. We have, 64m3 – 343n3 = (4m)3 – (7n)3 Write the degree of each of the following polynomials. = 9x2 – x – 20, Question 2. ⇒ cx + d = 0 ⇒ cx = -d ⇒ \( x =-\frac { d }{ c }\) Chapter -1 Sol. 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. (iv) 2y3 + y2 – 2y – 1 = 10000 + (-9) + 20 = 9120 To help students in making easy preparations, we are providing the MCQs for class 9 NCERT Maths. Variables and expression are called as indeterminate and coefficients. We hope the NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1, help you. ∴ (28)3 + (-15)3 + (-13)3 = 3(28)(-15)(-13) 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. Verify that (vii) 7x3 = 2 x \(\frac { 1 }{ 2 }\) x (x + y + z)(x2 + y2 + z2 – xy – yz – zx) NCERT solutions for session 2019-20 is now available to download in PDF form. (i) 4 x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz (iii) x2 – \(\frac { { y }^{ 2 } }{ 100 }\) (v) p (x) = x2, x = 0 = 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) Solution: = 2y3 – 2y2 + 3y2 – 3y + y – 1 ⇒ (x + y)3 – 3(x + y)(xy) = x3 + y3 Thus, 3x2 – x – 4 = (3x – 4)(x + 1), Question 5. ∴ p(-1) = (-1)3 + 3(-1)2 + 3(-1) +1 = (2a – b) (2a – b) (2a – b), (iii) 27 – 125a3 – 135a + 225a2 = ( 100)2 + (3 + 7) (100)+ (3 x 7) ⇒ P (-1) ≠ 1 sin θ and tan θ) without evaluating θ. ! Thus, zero of 3x is 0. Class 9 Mathematics Notes for FBISE. Solution: Since, p(1) = (1)2 +1 + k (ii) Here, p (x) = 2x2 + kx + √2 = 8x3 + 12x2 + 6x + 1, (ii) (2a – 3b)3 = (2a)3 – (3b)3 – 3(2a)(3b)(2a – 3b) [By (2)] CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. (ii) 8a3 -b3-12a2b+6ab2 Thus, x3 – 3x2 – 9x – 5 = (x + 1)(x – 5)(x +1), (iii) We have, x3 + 13x2 + 32x + 20 Since, p(x) = 0 ⇒ x – 5 = 0 ⇒ x = -5 It is a polynomial in one variable i.e., y = – π3 + 3π2 – 3π +1 = (100)3 + (2)3 + 3(100)(2)(100 + 2) We have, (x + y)3 = x3 + y3 + 3xy(x + y) …(1) Since, p(1) = k(1)2 – (1) + 1 Thus, zero of ax is 0. Exercise 13.2 Solution. = x3 + x2 + 12x2 + 12x + 20x + 20 Since, p(x) = 0 ⇒ (x + y)(x2 + y2 – xy) = x3 + y3 = 0 + 0 + 0 + 1 = 1 (i) p (x) = x2 + x + k Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given = 1000300 – 30001 = 970299, (ii) We have, 102 =100 + 2 NCERT Book NCERT Sol. It is not a polynomial, because one of the exponents of y is -1, (iv) p (x) = kx2 – 3x + k Thus, zero of 2x + 5 is \(\frac { -5 }{ 2 }\) . The coefficient of x2 is 1. The coefficient of x2 is \(\frac { \pi }{ 2 }\). = (y – 1)(y + 1)(2y +1), Question 1. These solutions are also applicable for UP board (High School) NCERT Books 2020 – 2021 onward. Class 9 Chapter 2 Polynomials Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks. Find the remainder when x3 + 3x2 + 3x + 1 is divided by (i) x = 0 Thus, the required remainder = 0, (ii) The zero of \(x-\frac { 1 }{ 2 }\) is \(\frac { 1 }{ 2 }\) p(1) = (1 – 1)(1 +1) = (0)(2) = 0 These NCERT solutions are created by the BYJU’S expert faculties to help students in the preparation of their board exams. Solution: ∴ (-12)3 + (7)3 + (5)3 = 3[(-12)(7)(5)] Solution: (v) We have, p(x) = 3x. (iii) Let p (x) = x4 + 3x3 + 3x2 + x + 1 . (i) 8a3 +b3 +12a2b+6ab2 Solution: (iii) We have, = x3 – 4x2 + x + 6 and g (x) = x – 3 Homework Help with Chapter-wise solutions and Video explanations. Solution: = x2 + (2y)2 + (4z)2 + 2 (x) (2y) + 2 (2y) (4z) + 2(4z) (x) Practise the solutions to understand how to use the remainder theorem to work out the remainder of a polynomial. = (3x + y + z)[(3x)3 + y3 + z3 – (3x × y) – (y × 2) – (z × 3x)] 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 … Find the value of k, if x – 1 is a factor of p (x) in each of the following cases Find p (0), p (1) and p (2) for each of the following polynomials. (i) The given polynomial is 2 + x2 + x. = – 5x – 4x2 + 3 = -9 + 3 = -6 They are in a list with arrows. Terms of Service. (ii) 2x2 + 7x + 3 Chapter 14 Probability. Let p(x) = x3 + 3x2 + 3x +1 Solution: (i) We have, p(x) = x + 5. On signing up you are confirming that you have read and agree to Thus, the possible length and breadth are (5a – 3) and (5a – 4). 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