(c) xy2 Question 13: The value of 2492 – 2482 is Solution: Find the zeroes of the polynomial p(x)= (x – 2)2 – (x+ 2)2. = (4x – 2y + 3z)(4x -2y + 3z), Question 30. (c) 4√2 So, it may have degree 5. we get p(0) = 10(0) – 4(0)2 – 3 = 0 – 0 – 3 = -3 If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p ( ½). 935k watch mins. (d) Now, (25x2 -1) + (1 + 5x)2 (iii) xy + yz + zx (a) 4 (b) 5 (c) 3 (d) 7 (ii) lf p(a) = Q then p(x) is a multiple of g(x) and f(p(a) # Q then p(x) is not a multiple of g(x) where ‘a’ is a zero of g (x)). ⇒ 3a = 6 (i) 9x2 + 4y2+16z2+12xy-16yz-24xz (A) 0 The polynomial p{x) = x4 – 2x3 + 3x2 – ax + 3a – 7 when divided by x + 1 leaves the remainder 19. = x2(x + 1) – 4(x + 1) Since, remainder ≠ 0, then p(x) is not a multiple of g(x). (c) (2x + 2) (2x + 5) (d) (2x -1) (2x – 3) = x2(x – 1) – 5x (x – 1) + 6(x – 1) Let p(x) = 2x4 – 5x3 + 2x2 – x+ 2 firstly, factorise x2-3x+2. It is not a polynomial, because exponent of x is 1/2 which is not a whole number. = (x + y) (3xy) = (x – 2)(x + 3)(2x – 5), (ii)We have, x3 – 6x2 + 11x – 6 = (x + y) (3xy) (c) any real number (iv) Degree of polynomial y3(1 – y4) or y3 – y7 is 7, because the maximum exponent of y is 7. Solution: Question 1. Hence, p -1 is a factor of h(p). = 3 (b + c)[a(a + b) + c(a + b)] (a) 5 + x (b) 5 – x (c) 5x -1 (d) 10x Hence, one of the zeroes of the polynomial p(x) is ½. = (2x-1)(x+ 4) and p( 2) = 2(2)4 – 5(2)3 + 2(2)2 -2 + 2 26a = 26 a = -1. Zero of the zero polynomial is Factorise e.g. (ii) p(x) = 2x3 – 11x2 – 4x+ 5, g(x) = 2x + l. Thinking Process Question 14. This solution is strictly revised in accordance with the recently updated syllabus issued by CBSE. ∴ Coefficient of x² in 3x² – 7x + 4 is 3. 27a+41 = 15+a Let p(x) = 8x4 + 4x3 – 16x2 + 10x + m (iii) (-x + 2y-3z)2 (ii) 2x – 3 is a factor of x + 2x3 -9x2 +12 (a) 1 Hence, √2 is a polynomial of degree 0, because exponent of x is 0. (a) 0 If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. Download the NCERT Exemplar Problem Solutions for Class 9 Maths Chapter 2 - Polynomials solved by Mathematics Expert Teachers at Mathongo.com as per CBSE (NCERT) Book guidelines. If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. Question 1: (i) We have, (4a – b + 2c)2 = (4a)2 + (-b)2 + (2c)2 + 2(4a)(-b) + 2(-b)(2c) + 2(2c)(4a) 2y= 0 Question 2. Solution: Question 25: Hence, zero of the polynomial p(x) is -5/2. dceta.ncert@nic.in 011 2696 2580 NCERT, Sri Aurobindo Marg, New Delhi-110016 011 2696 2580 NCERT, Sri Aurobindo Marg, New Delhi-110016 We know that, = 2x(2x+ 3) + 1 (2x+ 3) p(1) = (1 + 2)(1-2) Hence, the zeroes of y² + y – 6 are 2 and – 3. = (2x + 3) (2x + 1). Therefore, we can not exactly determine the highest power of variable, hence cannot define the degree of zero polynomial. [using identity, a3+b3 + c3 – 3 abc = (a + b + c)(a2 + b2 + c2 – ab – be -ca)] = 0 + 3abc [∴ a + b + c = 0, given] Important questions in Polynomials with video lesson. Polynomials in one variable, zeroes of polynomial, Remainder Theorem, Factorization, and Algebraic Identities. Which of the following is a factor of (x+ y)3 – (x3 + y3)? (a) 3 We know that [using identity, (a – b)3 = a3 -b3 – 3ab and (a + b)3 =a3 +b3 + 3ab ] = (2x)3 – (5y)3 – 30xy(2x – 5y) – (2x)3 – (5y)3 – 30xy (2x + 5y) Solution: (C) It is not […] NCERT Exemplar Class 9 Maths Chapter 2 Polynomials are part of NCERT Exemplar Class 9 Maths. = x3 – x2 – 5x2 + 5x + 6x – 6 = -2[r(r + 7) -6(r + 7)] Solution: (iv) Zero of a polynomial is always 0. (D) Not defined Hence, one of the factor of given polynomial is 3xy. Solution: Question 14: (a) -3 = (100)2 + (1 + 2)100 + (1)(2) NCERT Solutions Class 9 Maths Chapter 2 Polynomials are provided here. (iv) Polynomial 4- 5y2 is a quadratic polynomial, because maximum exponent of y is 2. (a) 0 (b) 1 Solution: Question 15: (iv) Given polynomial h(y) = 2 y Here, zero of g(x) is 3. = (x+ y)(x2+ y2+ 2xy- x2+ xy- y2) (ii) 6x2 + 7x – 3 and p(-2) =10 (-2) -4 (-2)2 – 3 NCERT Exemplar class 9 maths is designed to give teachers and students more problems that are of higher aptitude and have a greater focus on the application of concepts learned in class. Solution: Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. Solution: Write the coefficient of x2 in each of the following = 50x2 + 10x = 10x (5x+ 1) then (5)2 = a2 + b2+ c2 + 2(10) = 1000000 + 27 + 900(103) (i) Given, polynomial is (ii) y3 – 5y 5 Questions. (a) Degree of 4x4 + Ox3 + Ox5 + 5x + 7 is equal to the highest power of variable x. (iv) 0 and 2 are the zeroes of t2 – 2t Thinking Process Solution: Question 32: (i) False (ii) binomial of degree 20. = -2(r2 + 7r – 6r – 42) NCERT Class 9 New Books for Maths Chapter 2 Polynomials includes all the questions given in CBSE syllabus. Expand the following Fi nd (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz). NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12. (D): In zero polynomial, the coefficient of any power of variable is zero i.e., 0x², 0x5 etc. (iv) h(y) = 2y ⇒ a2 + b2 + c2 = 81 – 52 = 29, Question 31. (a) 0 (a) 2 (b)½ (c)-1 (d)-2 Question 8: (c) 2/5 (vi) 2 + x (iii) x3 + x2-4x-4 Solution: NCERT Exemplar ProblemsNCERT Exemplar MathsNCERT Exemplar Science, Kerala Syllabus 9th Standard Physics Solutions Guide, Kerala Syllabus 9th Standard Biology Solutions Guide, NCERT Solutions for class 9 Maths in Hindi, NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions InText Questions, NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area InText Questions, NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers InText Questions, NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities InText Questions, NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles InText Questions, NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties InText Questions, NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles InText Questions, NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations InText Questions, NCERT Solutions for Class 7 Maths Chapter 3 Data Handling InText Questions, NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals InText Questions, NCERT Solutions for Class 7 Maths Chapter 1 Integers InText Questions, NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4, NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3, NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2, NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5. = 2x2 + 8x – x – 4 [by splitting middle term] = 4a2 + 4a – 3 SOLUTION: (i) The given polynomial is 5x³ + 4x² + 7x The highest power of the variable x is 3. Factorise NCERT Exemplar Class 9 Maths. [Hint: Factorise x2 – 3x + 2] Find the value of the polynomial 3x3 – 4x2 + 7x – 5, when x = 3 and also when x = -3. Factorise Question 19. ⇒ (a + b + c)2 = (9)2 [Squaring on both sides] Solution: Question 21. The type of questions that will be asked from NCERT Class 9 Maths Unit 2 are displayed in the below provided NCERT Exemplar Class 9 Maths Unit 2. Hence, zero of x – 3 is 3. [using identity, a3 + b3 + c3 – 3 abc = (a + b + c)(a2 + b2 + c2 – ab – be – ca)] (c) 18 (i) p(x) = 10x – 4x2 – 3 (ii) p(y) = (y + 2)(y – 2) Solution: Question 9. (i) Polynomial 2 – x² + x³ is a cubic polynomial, because its degree is 3. [∴ a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2] NCERT Books for Class 9 Maths Chapter 2 Polynomials can be of extreme use for students to understand the concepts in a simple way.Class 9th Maths NCERT Books PDF Provided will help you during your preparation for both school … On putting x = 0,1 and – 2, respectively in Eq. Since, p(x) is divisible by (x+2), then remainder = 0 Solution: Solution: And to make that base strong students are advised to solve NCERT Exemplar class 9 Maths. (iv) p(x) = x3 – 6x2 + 2x – 4, g(x) = 1 – (3/2) x (ii) Substitute these factors in place of x in given polynomial and find the minimum factor which satisfies given polynomial and write it in the form of a linear polynomial. (a) 5 + x [using identity, (a + b)3 = a3 + b3 + 3ab (a + b)] ∴ (0.2)³ + (-0.3)³ + (0.1)3 = 3(0.2) (-0.3) (0.1) (ii) Polynomial 3x3 is a cublic polynomial, because maximum exponent of x is 3. and p(-2) = (-2 + 2)(-2 -2) Solution: When we divide p(x) by x+1, we get the remainder p(-1) (ii) True Solution: (ii) 9x2 – 12x + 4 ⇒ (-2)3 – 2m(-2)2 + 16 = 0 Write whether the following statements are true or false. We have, p(x) = x4 – 2x3 + 3x2 – ox + 3a – 7 -[(2x)3 + (5y)3 + 3(2x)(5y)(2x+5y)] Classify the following polynomials as polynomials in one variable, two variables etc. Justify your answer. (i) 9x2 -12x+ 3 (ii) 9x2 -12xy + 4 Hence, p(x) is divisible by x2 – 3x + 2. When we divide p(x) by x+1, we get the remainder p(-1) At x= 3, p(3) = 3(3)3 – 4(3)2 + 7(3) – 5 = (x-2)(x- 1) Write the coefficient of x² in each of the following Question 7. (c) 49 (-1)3 + (-1)2 + (-1) + 1 = 0 => -1+1-1 + 1 = 0 => 0 = 0 Hence, our assumption is true. (ii) Further, put the factors equals to zero, then determine the values of x. (ii) p(x) = 2x3 – 11x2 – 4x + 5, g(x) = 2x + 1 Hence, zero of polynomial is 4. 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Without actually calculating the cubes, find the value of 36xy-36xy = 0 Hence, zero of polynomial is m = 1 (d) Given p(x) = x + 3, put x = -x in the given equation, we get p(-x) = -x + 3 (i) 2x – 1 Therefore, (x-2y)3 + (2y-3z)3 + (3z-x)3 = 3(x-2y)(2y-3z)(3z-x). (d) 1/2 = (x – 1)(3x2 + 2x – 1) Polynomials | Maths | NCERT Exemplar Solutions | Class 9. (iv) Polynomial NCERT Books for Class 10 Maths Chapter 2 Polynomials can be of extreme use for students to understand the concepts in a simple way.NCERT Textbooks for Class 10 Maths are highly recommended as they help cover the entire … Provided absolutely free … Extra questions for CBSE Class 9 Maths Solutions are created by the BYJU S... Notes, CBSE chemistry notes for free in school or equal to one 3x°! -1, which is ( i ) the coefficient of x2 in 3x ncert exemplar class 9 maths polynomials! Solutions are being updated for new academic year 2020-21 True because a polynomial can exactly. 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