= 1000000000 – 8 – 6000000 +12000 Note: Important questions have also been marked for your reference. So, it is a quadratic polynomial. 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) y2 + √2 Thus, zero of ax is 0. In this … p(2) = (2)2 – 2 + 1 = 4 – 2 + 1 = 3 = -1 + 1 – 1 + 1 CBSE Class 11 Maths Worksheet for students has been used by teachers & students to develop logical, lingual, analytical, and … (i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz (vii) We have, p(x) = cx + d. Since, p(x) = 0 (iii) x4 + 3x3 + 3x2 + x + 1 = (1000)3– (2)3 – 3(1000)(2)(1000 – 2) Exercise 14.1 Solution. = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) Solution: = (x + 1)(x2 – 5x + x – 5) x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) Since, p(x) = 0 Since, \(p(\frac { 1 }{ 2 } )\) ≠ 0, so, x = \(\frac { 1 }{ 2 }\) is not a zero of 2x + 1. So, it is a cubic polynomial. = 2k – 3 = 0 ⇒ x = \(\frac { -5 }{ 2 }\) (ii) Given that p(t) = 2 + t + 2t2 – t3 = ( 100)2 + [(- 5) + (- 4)] 100 + (- 5 x – 4) Middle school is one of the most crucial periods a student undergoes in the course of their schooling years. = 1000300 – 30001 = 970299, (ii) We have, 102 =100 + 2 = (y – 1)(y + 1)(2y + 1) Chapter 4 Linear Equations in Two Variables. (i) (x+2y+ 4z)2 = x2(x + 1) + 12x(x +1) + 20(x + 1) ∴ p(0) = (0 – 1)(0 + 1) = (-1)(1) = -1 ⇒ (x – y)3 + 3xy(x – y) = x3 – y3 Extra questions for class 9 maths chapter 1 with solution. [Using (x + a)(x + b) = x2 + (a + b)x + ab] = 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) = (3x + y)(3x + y), (ii) We have, 4y2 – 4y + 12 (iii) We have, p(x) = 2x + 5. (vii) p (x) = cx + d, c ≠ 0 where c and d are real numbers. ⇒ 3x = 2 [Using (x + a)(x + b) = x2 + (a + b)x + ab] = (x + 1)(x2 + 2x + 10x + 20) It is a polynomial in one variable i.e., y (viii) p (x) = 2x + 1, x = \(\frac { 1 }{ 2 }\) = x2 – 2x – 80, (iii) We have, (3x + 4) (3x – 5) = (2y -1)2 = (3)3 – (5a)3 – 3(3)(5a)(3 – 5a) Which of the following expressions are polynomials in one variable and which are not? R.H.S ⇒ x = -5. Download NCERT Solutions For Class 9 Maths in PDF based on latest pattern of CBSE in 2020 - 2021. = (- √2x)2 + (y)2 + (2 √2z)2y + 2(- √2x) (y)+ 2 (y) (2√2z) + 2 (2√2z) (- √2x) (iv) The degree of 1 + x is 1. (x + a) (x + b) = x2 + (a + b) x + ab (iii) The given polynomial is \(\frac { \pi }{ 2 } { x }^{ 2 }\) + x. the remainder is not 0. Hence, verified. ∴ p (-1)= (-1)4 + 3 (-1)3 + 3 (-1)2 + (- 1) + 1 sin θ and tan θ) without evaluating θ. (ii) p (x) = 2x2 + kx + √2 1et p(x) = 5x – 4x2 + 3 = 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) Represent geometrically 8.1 on number line. = (3 – 5a)3 ∴ p(o) = (0)2 = 0 Write the following numbers in p/q form (i) 2.015 (ii) 0.235 Ans (399 235 ' 198 999) 4. (i) p (x)= 2x3 + x2 – 2x – 1, g (x) = x + 1 Unit 1 - Matrices & Determinants. (ii) The given polynomial is 2 – x2 + x3. Chapter-10 Chapter-3 Sol. Thus, the value of 5x – 4x2 + 3 at x = 2 is – 3. Thus, the possible length and breadth are (7y – 3) and (5y + 4). Volume of a cuboid = (Length) x (Breadth) x (Height) Since p(-1) = 0, so, x = -1, is also a zero of x2 – 1. [Using (a + b)3 = a3 + b3 + 3ab (a + b)] Find the zero of the polynomial in each of the following cases ⇒ k = √2 -1, (iv) Here, p(x) = kx2 – 3x + k (ii) Let p (x) = x4 + x3 + x2 + x + 1 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 … which is not a whole number. = (- √2x + y + 2 √2z)2 = (x + 1)(x2 + 12x + 20) = 2 + 0 + 0 – 0=2 Find the remainder when x3 + 3x2 + 3x + 1 is divided by NCERT Solutions for Class 9 Maths Exercise 9.2 book solutions are available in PDF format for free download. = 10000 + (-900) + 20 = 9120, (iii) We have 104 x 96 = (100 + 4) (100 – 4) (ii) 2x2 + y2 + 8z2 – 2√2xy + 4√2yz – 8xz (iv) (3a -7b – c)z Question 2. (iii) x ∴ (28)3 + (-15)3 + (-13)3 = 3(28)(-15)(-13) We hope the NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1, help you. CLASS 9 IX Sample Papers X Sample Papers CLASS 8 CLASS 07 Class 06 ... Chapter - 2: Polynomials. (i) 27y3 + 125z3 ∴ p(-2) = (-2)3 + 3(-2)2+ 3(-2) + 1 Question 1. It is not a polynomial, because one of the exponents of t is \(\frac { 1 }{ 2 }\), Thus, x3 – 3x2 – 9x – 5 = (x + 1)(x – 5)(x +1), (iii) We have, x3 + 13x2 + 32x + 20 (ii) p (x) = x – 5 = 10000 + (10) x 100 + 21 It is not a polynomial, because one of the exponents of y is -1, (iii) 27-125a3 -135a+225a2 (i) The zero of x + 1 is -1. (iii) p (x) = 2x + 5 ⇒ (x + y)(x2 + y2 – xy) = x3 + y3 = (3 – 5a) (3 – 5a) (3 – 5a), (iv) 64a3 -27b3 -144a2b + 108ab2 (iv) Since, 3 = 3x° [∵ x°=1] because each exponent of x is a whole number. (i) The degree of x2 + x is 2. = x(x2 – 1) – 2(x2 -1) = (x2 – 1)(x – 2) (i)We have, 103 x 107 = (100 + 3) (100 + 7) (i) x2+ x = 7y(5y + 4) – 3(5y + 4) = (5 y + 4)(7y – 3) [∵ (a2 – b2) = (a + b)(a-b)] (v) (- 2x + 5y – 3z)2 Thus, 2y3 + y2 – 2y – 1 = 3 x x x (x – 4) 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. Prepare effectively for your Maths exam with our NCERT Solutions for CBSE Class 9 Mathematics Chapter 2 Polynomials. Thus, zero of x + 5 is -5. (ii) We have, p(x) x3 + 3x2 + 3x + 1 and g(x) = x + 2 Mathematics Part - II Solutions Textbook Solutions for Class 9 Math. Solution: We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz The coefficient of x2 is \(\frac { \pi }{ 2 }\). Since, p(1) = (1)2 +1 + k State reasons for your answer. Since, x + y + z = 0 Write the following cubes in expanded form Hence, verified. = 8a3 – 27b3 – 18ab(2a – 3b) = 3(5460) = 16380. ∴ 993 = (100 – 1)3 Solution: (x+ a) (x+ b) = x2 + (a + b) x+ ab. Solution: (ii) x – x3 Solution: = (y – 1)(y + 1)(2y +1), Question 1. The coefficient of x2 is -1. [Using a3 + b3 + 3 ab(a + b) = (a + b)3] ⇒ 2x + 5 =0 So, (x + 1) is not a factor of x4 + 3x3 + 3x2 + x+ 1. Find p (0), p (1) and p (2) for each of the following polynomials. ⇒ p (-1) ≠ 0 Teachoo is free. Solution: P(2) = (2 – 1)(2 + 1) = (1)(3) = 3. (iv) 1 + x (ii) x3 – 3x2 – 9x – 5 Thus, 2×2 + 7x + 3 = (2x + 1)(x + 3), (iii) We have, 6x2 + 5x – 6 = 6x2 + 9x – 4x – 6 (v) We have, p(x) = x2 ∴ p(0) = (0)3 = 0, p(1) = (1)3 = 1 [Hint See question 9] ! So, the degree of the polynomial is 1. ∴ 3x3 + 7x is not divisib1e by 7 + 3x. 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… i.e. You have these advantages of browsing notes from our website. Thus, zero of 3x – 2 is \(\frac { 2 }{ 3 }\). Then, x + y + z = 28 – 15 – 13 = 0 (i) The given polynomial is 5x3 + 4x2 + 7x. Thus, the required remainder is \(-\frac { 27 }{ 8 }\) . = 1000000 – 1 – 300(100 – 1) = 1000000 + 8 + 600(100 + 2) Exercise 13.2 Solution. (iii) Let p (x) = x4 + 3x3 + 3x2 + x + 1 . Write the degree of each of the following polynomials. (i) x3+x2+x +1 (i) (- 12)3 + (7)3 + (5)3 We have, (x + 8) (x – 10) = x2 + [8 + (-10)] x + (8) (- 10) (iv) Given that p(x) = (x – 1)(x + 1) ⇒ (x – y)(x2 + y2 + xy) = x3 – y3 Rational Numbers, irrational Numbers, rationalize irrational numbres, operation on real numbers, laws of … (iv) 3 = (2 a + b)3 p(-π) = (-π)3 + 3(- π)22 + 3(- π) +1 = -8 + 12 – 6 + 1 It is a complete package of solutions to problems of your really tough book. = (2x + 3)(3x – 2) (v) 3t (ii) (x+8) (x -10) Login to view more pages. So, it is not a polynomial in one variable. We have, (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z) = a3 – a3 + 6a – a = 5a 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)] Then, x + y + z = -12 + 7 + 5 = 0 (i) We know that ⇒ x3 + y3 + 3xy(x + y) = -z3 (i) 25a2 – 35a + 12 = 25a2 – 20a – 15a + 12 = 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3) = (x + 1)[x(x – 5) + 1(x – 5)] Chapter -1 Sol. ⇒ x3 + y3 – 3xyz = -z3 Solution: (ii) 2 – x2 + x3 = -2 + 1 + 2 -1 = 0 Thus, the required remainder = 0, (ii) The zero of \(x-\frac { 1 }{ 2 }\) is \(\frac { 1 }{ 2 }\) ∴p(0) = 2 + 0 + 2(0)2 – (0)3 = (x + 1)[x(x + 2) + 10(x + 2)] = -1 + 3- 3 + 1 = 0 Thus, the required remainder is -π3 + 3π2 – 3π+1. (vi) The degree of r2 is 2. (vii) 7x3 Verify that = k – √2 + 1 = 0 Thus, the possible dimensions of the cuboid are 4k, (3y + 5) and (y -1). Ex 2.1 Class 9 Maths Question 3. (i) 12x2 – 7x +1 [Using a2 – 2ab + b2 = (a- b)2] The coefficient of x2 is 1. 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. (ii) 4 – y2 = \(\frac { 1 }{ 2 }\) (x + y + z)[2(x2 + y2 + z2 – xy – yz – zx)] Since, p(-1) = 0, so, x = -1 is a zero of (x + 1)(x – 2). (i) p(y) = y2 – y +1 (iii) 5t – √7 (i) The given polynomial is 2 + x2 + x. Solution: (ii) We have, p(x) = 5x – π ∴ p(-1) = 2(-1)3 + (-1)2 – 2(-1) – 1 (iv) We have, p(x) = (x + 1)(x – 2) Practise the solutions to understand how to use the remainder theorem to work out the remainder of a polynomial. It is a polynomial in one variable i.e., x Important questions in Number systems with video lesson. Question 2. ⇒ x + 5 = 0 (iv) The zero of x + π is -π. = (x + 1)(x + 2)(x + 10), (iv) We have, 2y3 + y2 – 2y – 1 (iii) p(2) = 5(2) – 4(2)2 + 3 = 10 – 4(4) + 3 = (3y + 5z)[(3y)2 – (3y)(5z) + (5z)2] (vi) r2 (i) Abmomial of degree 35 can be 3x35 -4. (iv) 64a3 -27b3 -144a2b + 108ab2 Solution: Telanagana State Board Class 9 Mathematics Textbook Solution Chapter-wise Telanagana SCERT Class IX Math Solution. (ii) ∵ (x – y)3 = x3 – y3 – 3xy(x – y) = (3x)2 + 2(3x)(y) + (y)2 = 4k[(3y + 5) x (y – 1)] We have, (3x + 4) (3x – 5) = (3x)2 + (4 – 5) x + (4) (- 5) Variables and expression are called as indeterminate and coefficients. [Using a3 – b3 – 3 ab(a – b) = (a – b)3] Answers to each and every question is explained in an easy to understand way, with videos of all the questions. ⇒ k = \(\frac { 3 }{ 4 }\), Question 4. = (- √2x + y + 2 √2z) (- √2x + y + 2 √2z), Question 6. Find the value of k, if x – 1 is a factor of p (x) in each of the following cases k = -2 – √2 = -(2 + √2), (iii) Here, p (x) = kx2 – √2 x + 1 = 4 x k x (3y2 + 2y – 5) NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1 are part of NCERT Solutions for Class 9 Maths. (v) We have x10+ y3 + t50 Ex 2.1 Class 9 Maths Question 1. Verify whether the following are zeroes of the polynomial, indicated against them. After the completion of chapters from NCERT Books and Exemplar Books, students should go for Assignments. (ii) (102)3 What are the possible expressions for the dimensions of the cuboids whose volumes are given below? = x3 + x2 – 4x2 – 4x – 5x – 5 , There is plenty to learn in this chapter about the definition and examples of polynomials, coefficient, degrees, and terms in a polynomial. The highest power of the variable x is 3. Since, p(x) = 0 (vii) P (x) = 3x2 – 1, x = – \(\frac { 1 }{ \sqrt { 3 } }\),\(\frac { 2 }{ \sqrt { 3 } }\) So, the degree of the polynomial is 3. = -π3 + 3π2 + (-3π) + 1 Review questions, MCQs, important board questions and answers are very helpful for CBSE board exam with... Chapter overview theorem to work out the remainder theorem class 9 maths chapter 2 assignment work out remainder. 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