Question 17. NCERT Exemplar for Class 9 Maths Chapter 4 With Solution | Linear Equations in Two Variables. (c) Zero of the zero polynomial is any real number. (2), we get All questions with solutions of polynomials will help all the students to revise complete syllabus and score more marks in examinations. (ii) p(x) = 2x3 – 11x2 – 4x + 5,  g(x) = 2x + 1 (v) Polynomial 3 = 3x° is a constant polynomial, because its degree is 0. Fi nd (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz). (ii) 6x2 + 7x – 3 Solution: Solution: Question 11: (d)½ = x2(x + 1) – 4(x + 1) p(x) = x – 4 Factorise the following Let g (p) = p10 -1  …(1) (c) x4 + x3 + x2 + 1 (ii) p(x) = 2x3 – 11x2 – 4x+ 5,  g(x) = 2x + l. Thinking Process = 27a+ 36+ 9-4= 27a+ 41 For what value of m is x3 -2mx2 +16 divisible by x + 2? [∴ If a + b + c = 0, then a3 + b3 + c3 3abc] Substituting x = 2 in (2), we get [using identity, (a + b)3 = a3 + b3 + 3ab (a + b)] (vii) Polynomial y3 – y is a cubic polynomial, because maximum exponent of y is 3. Here, zero of g(x) is 3. (d) 7 Solution: (iii) -4/5 is a zero of 4 – 5y (i) 2x3 -3x2 -17x + 30         (ii) x3 -6x2 +11 x-6 (iv) False Solution: ⇒ a2 + b2 + c2 = 5     … (i) Solution: Question 24: It is not a polynomial, because exponent of x is 1/2 which is not a whole number. (iii) True Find p(0), p( 1) and p(-2) for the following polynomials We have, 6x2 + 7x – 3 = 6x2 + 9x – 2x – 3 => -2a + 3=0 We hope that our NCERT Class 9 New Books for Maths helped with your studies! (i), we get p(-1) = (-1)51 + 51 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 … (viii) 1 + x + x² (iii) Degree of polynomial x3 – 9x + 3xs is five, because the maximum exponent of x is five. With the help of the NCERT Exemplar Class 9 Maths, candidates can understand the level and type of questions that are asked in the exam. (ii) 25x2 + 16y2 + 4z2 – 40xy + 16yz – 20xz For zero of polynomial, put g(x) = 0 NCERT Solutions Class 9 Maths Chapter 2 Polynomials are provided here. (ii) Polynomial y3 – 5y is a one variable polynomial, because it contains only one variable i.e., y. = 0 + 3abc [∴ a + b + c = 0, given] = 27a+ 36+ 9-4= 27a+ 41 When we divide p2(z) by z-3 then we get the remainder p2(3). (i) 1033           (ii) 101 x 102       (iii) 9992 (a) Degree of 4x4 + Ox3 + Ox5 + 5x + 7 is equal to the highest power of variable x. It is not a polynomial, because one of the exponents of x is – 2, which is not a whole number. Let p(x) = 2x4 – 5x3 + 2x2 – x + 2 (i) We have, x2 + 9x +18 = x2 + 6x + 3x +18 = -2(r2 + 7r – 6r – 42) Find the value of the polynomial 5x – 4x 2 + 3 at (i) x = 0 (ii) x = – 1 (iii) x = 2 Solution: 1et p(x) = 5x – 4x 2 + 3 Because zero of a polynomial can be any real number e.g., for p(x) = x – 1, zero of p(x) is 1, which is a real number. Find the value of the polynomial 3x3 – 4x2 + 7x – 5, when x = 3 and also when x = -3. Free NCERT Solutions for Class 9 Maths polynomials solved by our maths experts as per the latest edition books following up the NCERT(CBSE) guidelines. (ii) p(x) = x3 – 3x2 + 4x + 50, g(x) = x – 3 Solution: Question 17: (ii) -1/3 is a zero of 3x + 1 (b) 1 ⇒ x3 + y3 + 64 = 12xy Solution: Question 33: (i) x + 3 is a factor of 69 + 11c – x2 + x3 = 4x³ – 6x² + 2x – 10x² + 15x – 5 Question 4. = x6 + x4 + x2 – x4 – x2 – 1 = x6 – 1, Question 34. Because exponent of the variable x is 1/2, which is not a whole number. ∴ x – 4 = 0 ⇒ x = 4 = 50x2 + 10x = 10x (5x + 1) Question 6: Question 8: [using identity, a3 + b3 = (a + b)(a2 – ab + b2)] = (x + y)[(x + y)2 -(x2 – xy + y2)] Hence, the value of k is 2. Therefore, (x-2y)3 + (2y-3z)3 + (3z-x)3 = 3(x-2y)(2y-3z)(3z-x). √2 is a polynomial of degree Solution: Thinking Process (ii) False, because every polynomial is not a binomial . (ii) p(x) = x3 -3x2 + 4x + 50, g(x)= x – 3 (i) x3 +y3 -12xy + 64,when x+y = -4. Write the coefficient of x² in each of the following (ii) The given polynomial is 4 - y². = (x – 1) [3x(x + 1) – 1(x + 1)] When we divide p(x) by g(x) using remainder theorem, we get the remainder p(-1) Question 20. (c) Let p(x) = 2x2 + kx (i) We have, g(x) = x – 2 Put 3x + 1 = 0 ⇒ x = -1/3 NCERT 9th Class Exemplar Problems 2021 in Pdf format are Available this web page to Download, NCERT Exemplar Problems from Class 9 for the Academic year 2021 and Exemplar also Available to Maths Subject with the Answers, NCERT Developed the Exemplar Books 2021 for 9th Class Students Education Purpose. Solution: Write whether the following statements are True or False. Here students are also provided with online learning materials such as NCERT Exemplar Class 9 Maths Solutions. (a) 4 (b) √2 = -√2x°. (iv) Zero of a polynomial is always 0 (iv) Given polynomial h(y) = 2 y For zero of polynomial, put h(y) = 0 ⇒ -5/2 You should get good marks in Class 9 examinations as it will always help you to get good rank in school. At x = -3, p(-3)= 3(-3)3 – 4(-3)2 + 7(-3) – 5 Solution: 8x4 + 4x3 – 16x2 + 10x + m NCERT Exemplar Solutions in Maths Classes VIII, IX and X: Get NCERT Exemplar Problem Solutions in Mathematics for classes 8 th, 9 th and 10 th for CBSE and other Students. The factorization of 4x2 + 8x+ 3 is = 4a2 + 4a – 3 ∴ p(-1) = (-1)³ – 2(-1)² – 4(-1) -1 NCERT solutions are really helpful when it comes to a complicated subject like Mathematics. (ii) Polynomial 3x3 is a cublic polynomial, because maximum exponent of x is 3. By Preparing Exercise wise Exemplar Questions with Solutions to help you to revise complete Syllabus and Score More marks in your exams. Hence, the zero of polynomial is 0, Question 12: Question 12: For what value of m is x3 -2mx2 +16 divisible by x + 2? Because every polynomial is not a binomial. If x +1 is a factor of ax3 +x2 -2x+4a-9, then find the value of a. Question 12. (i), we get p(0) = 10(0)-4(0)2 -3 = 0-0-3= -3 [∴ If a + b + c = 0, then a3 + b3 + c3 = 3abc] Solution: Question 37. Hence, the zeroes of t² – 2t are 0 and 2. Question 13: 2a = 3 On putting x = 0,1 and – 2, respectively in Eq. Question 6. then (5)2 = a2 + b2+ c2 + 2(10) p(1) = (1 + 2)(1-2) = 4x3 + 2x + 2 which is not a polynomial of degree 4. Question 19. Question 8. Verify whether the following are true or false. (d) Let p(x) = x51 + 51 . Find the value of m, so that 2x -1 be a factor of Degree of the polynomial 4x4 + 0x3 + 0x5 + 5x + 7 is h (1) = (1)11 —1 = 1 —1 = 0 Hence, p -1 is a factor of h(p). 935k watch mins. For zeroes of p(x), put p(x) = 0=> (2x -1) (x + 4) = 0 (i) The example of monomial of degree 1 is 5y or 10x. Exercise 2.1 Page No: 14. (d) Now, (x + 3)3 = x3 + 33 + 3x (3)(x + 3) On putting x = 2√2 in Eq. Now, x2-3x + 2 = x2 – 2x – x + 2 (b) abc ⇒ y = 2 and y = -3 Determine the degree of each of the following polynomials. (b) (2x + 1) (2x + 3) Vivek Patriya. p(x) = x- 4 (iv) We have, (2x – 5) (2x² – 3x + 1) (iii) 9992 37 (b) -5/2 (i) Given, polynomial is ∴ p(-3) = -143 (ii) p(y) = (y + 2)(y – 2) (a) 0        (b) 1           (c) any real number               (d) not defined Hence, p-1 is a factor of g(p). and p(-2) = (-2 + 2)(-2 -2) Because each exponent of the variable x is a whole number. = 50x2 + 10x = 10x (5x+ 1) = 3 (b + c)[a(a + b) + c(a + b)] (i), we get p(0) =(0+2)(0-2)= -4 NCERT Exemplar Class 9 Maths Chapter 2 Polynomials are part of NCERT Exemplar Class 9 Maths. (d) 1/2 NCERT Exemplar Class 9 Maths. [∴ (a – b)2 = a2 + b2 – 2ab] Solution: (ii) 9x2 – 12x + 4 = (x – 2)(x + 3)(2x – 5), (ii)We have, x3 – 6x2 + 11x – 6 By actual division, find the quotient and the remainder when the first (i) We have, (4a – b + 2c)2 = (4a)2 + (-b)2 + (2c)2 + 2(4a)(-b) + 2(-b)(2c) + 2(2c)(4a) (c) xy2 Thinking Process Solution: Question 9. NCERT Exemplar for Class 9 Maths Chapter 3 With Solution | Coordinate Geometry. 8x4 +4x3 -16x2 +10x+07. (a) 5 + x if a + b+c = Q, now use the identity a3 + b3 + c3 = 3abc. Question 1. Solution: ⇒ 2 – k = 0 ∴ p(2) = (2)3 – 5(2)2 + 4(2) – 3 One of the zeroes of the polynomial 2x2 + 7x – 4 is (iv) Polynomial 4 – 5y² is a quadratic polynomial, because its degree is 2. Here, zero of g(x) is 1/2. = (0.2)3 + (- 0.3)3 + (0.1)3 (ii) Given, polynomial is Since, (x + 1) is a factor of p(x), then = 25x2 -1 + 1 + 25x2 + 10x [using identity, (a + b)2 = a2 + b2 + 2ab] Write the coefficient of x2 in each of the following Solution: Question 19: NCERT Exemplar for Class 9 Maths Chapter 2 With Solution | Polynomials. (iv) Polynomial  x2 – Zxy + y2 +1 is a two variables pplynomial, because it contains two variables x and y. (i) -3 is a zero of at – 3 Question 17. x + 1 is a factor of the polynomial (v) A polynomial cannot have more than one zero m = 1 Question 39: Therefore, remainder is 62. For zero of the polynomial, put p(x) = 0 Solution: If p (x) = x + 3, then p(x) + p(- x) is equal to HOTS, exemplar, and hard questions in polynomials. (ii) g(x)= 3 – 6x 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. (d) 8 √2 +1 Polynomials Class 10 NCERT Book If you are looking for the best books of Class 10 Maths then NCERT Books can be a great choice to begin your preparation. Solution: = x3 + 27 + 9x (x + 3) Solution: Since, remainder ≠ 0, then p(x) is not a multiple of g(x). Since, p(x) is divisible by (x+2), then remainder = 0 On putting x= -1 in Eq. ∴ 2x + 5 = 0 ⇒ x = 0 (a) 0 Without actual division, prove that 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2 Solution: Now, p(x) + p(-x) = x + 3 + (-x) + 3 = 6. When we divide p(x) by x+1, we get the remainder p(-1) = -81 – 36 – 21 – 5 = -143 (iv) If the maximum exponent of a variable is 2, then it is a quadratic polynomial. (D): In zero polynomial, the coefficient of any power of variable is zero i.e., 0x², 0x5 etc. NCERT Exemplar Class 9 for Maths Chapter 3 – Coordinate Geometry (ii) Given, polynomial is (i) We have, 2X3 – 3x2 – 17x + 30 (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. ∴ (x – 2)2 – (x + 2)2 = 0 (i) We have, 1033 = (100 + 3)3 Question 9. (ii) Given, polynomial isp(y) = (y+2)(y-2) Factorise the following: Find the value of m, so that 2x -1 be a factor of = 2x2(x – 2) + x(x – 2) – 15(x – 2) (a) 0 (i) We have, (i) Given, polynomial is NCERT Exemplar for Class 9 Maths Chapter 2 Polynomials With Solution. Determine the degree of each of the following polynomials. and p( 2) = 2(2)4 – 5(2)3 + 2(2)2 -2 + 2 Solution: = (x – 1) (x2 – 3x – 2x + 6) ⇒ 8m = 8 (iii) Polynomial xy+ yz+ zx is a three variables polynomial, because it contains three variables x, y and z. a = 3/2. Because the sum of any two polynomials of same degree has not always same degree. a3 + b3 + c3 = 3abc. e.g., p(x) = x2 -2, as degree pf p(x) is 2 ,so it has two degree, so it has two zeroes i.e., √2 and —√2. If p (x) = x2 – 2√2x + 1, then p (2√2) is equal to = (5) [5 – (ab + be + ca)]             [From (i)] (ii) a3 – 2√2b3 Solution: = x3 + 27 + 9x2 + 27x Hence, the coefficient of x in (x + 3)3 is 27. (c) any natural number                 (d) not defined Polynomials Class 9 NCERT Book: If you are looking for the best books of Class 9 Maths then NCERT Books can be a great choice to begin your preparation. => 2x-1 = 0 and x+4 = 0 e.g., Let f(x) = x4 + 2 and g(x) = -x4 + 4x3 + 2x. and h(p) = p11 -1  …(2) => 2-k = 0 => k= 2 When we divide p(x) by g(x) using remainder theorem, we get the remainder p(3) = (x-1)(x-2)(x-3), (iii) We have, x3 + x2 – 4x – 4 In this method firstly check the values of a + b+ c, then . Solution: (b) 1 and h(p) = p11 -1. The value of 2492 – 2482 is Question 8. = (100)2 + (1 + 2)100 + (1)(2) x + 1 is a factor of the polynomial = (x+ y)(x2+ y2+ 2xy- x2+ xy- y2) polynomial is divided by the second polynomial x4 + 1 and x-1. Solution: Without actual division, prove that (v) False = (x -1) (x2 – 5x + 6) Solution: Question 40: Solution: Question 16. = 27-12 + a = 15+a = 3(-27)-4×9-21-5 = -81-36-21-5 = -143 p(-3) = -143 Hence, one of the factor of given polynomial is 3xy. Practice Polynomials questions and become a master of concepts. (x) √2x – 1 p(-1)=0 If a + b + c =0, then a3 + b3 + c3 is equal to (a) (x + 1) (x + 3) Factorise the following Find p(0), p( 1) and p(-2) for the following polynomials Now, x2-3x+2 = x2-2x-x+2 [by splitting middle term] Justify your answer. = 2x(x – 5) + 3(x – 5) = (2x + 3)(x – 5) Since 2x – 1 is afactor of p(x) then p(1/2) = 0, Question 22. Solution: Question 31: NCERT Class 9 New Books for Maths Chapter 2 Polynomials are given below. Solution: ⇒ 4a – 1 = 19 ⇒ 4a = 20 Solution: (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)). Question 10. So, the degree of the polynomial is 3. (a) 0        (b) 1             (c) 49             (d) 50 (b) x3 + x2 + x + 1 Hence, possible length = 2a -1 and breadth = 2a + 3, Question 1. x + 3 is a factor of p(x) if p(-3) = 0 Solution: (b) ½ One of the zeroes of the polynomial 2x2 + 7x – 4 is 27a + 41 = 15 + a Question 13: Give an example of a polynomial, which is So, it may have degree 5. and p(-2) =10 (-2) -4 (-2)2 – 3 It depends upon the degree of the polynomial. Hence, one of the factor of given polynomial is 10x. Zero of the zero polynomial is (iv) h(y) = 2y If x + 2a is a factor of a5 – 4a2x3 + 2x + 2a + 3, then find the value of a. If a + b + c = 9 and ab + bc + ca = 26, find a2 + b2 +c2. (iii) False Hence, the degree of a polynomial is 4. Solution: Hence, the value of the given polynomial at x = 3 and x = -3 are 61 and -143, respectively. (ii) The example of binomial of degree 20 is 3x20 + x10 (iv) the constant term ∴ a = -1, Question 2. These expert faculties solve and provide the NCERT Solution for class 9 so that it would help students to solve the problems comfortably. (d) Now, a3 + b3 + c3 = (a + b + c) (a2 + b2 + c2 – ab – be – ca) + 3abc (a) 0                                                   (b) 1 g(x) = 3 – 6x L.H.S. (i) False (ii) True Question 8. (c) 2 NCERT solutions for class 9 Maths will help you to understand and solve complex problems easily. (i), Question 10: = x4 – 2x3 + 3x2 – 5x + 8 (d) 50 (i) (4o – b + 2c)2 Also, if a + b + c = 0, then a3 + b3 + c3 = 3abc All solutions are explained using step-by-step approach. P(-2) = 0 (B) 1 (i) The degree of the polynomial is the highest power of the variable x i.e., 6. (c) any real number = 2x(2x + 3) + 1 (2x + 3) = 12 + 12 – 24 = 0 Question 10: The coefficient of x in the expansion of (x + 3)3 is = (y-2)(y + 3) = 0 = 27a3 – 8b3 – 18ab(3a – 2b) We have, a + b + c = 5,ab + bc + ca = 10 Thinking Process = 8 – 20 + 8 – 3 = – 7 4x4 + 0x3 + 0x5 + 5x + 7 = 4x4 + 5x + 7 ⇒ t = 0 and t – 2 = 0 = (a + b + c – a)[(a + b + c)2 + a2 + (a + b + c)a] – [(b + c) (b2 + c2– be)] Given, area of rectangle = 4a2 + 6a-2a-3 ∴ a = 5 Solution: Question 33. (v) If the maximum exponent of a variable is 3, then it is a cubic polynomial. Hence, the value of m is 1 . (i) monomial of degree 1. iii) 16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz ⇒ 3a = 6 (vi) False, because the sum of any two polynomials of same degree is not always same degree. (a) Let p (x) = 5x – 4x2 + 3 …(i) For zero of polynomial, put p(x) = 0 (c) 18 e.g., (a) 3x2 + 4x + 5 [polynomial but hot a binomial] = 4a2 + 4a – 3 [by splitting middle term] = -1 – 2 + 4 – 1 = 0 = 16a2 + b2 + 4c2 – 8ab – 4bc + 16ac, (ii)We have, (3a – 5b – cf = (3a)2 + (-5b² + (- c²) + 2(3a)(-5b) + 2(-5b)(-c) + 2(-c)(3a) (ii) x3-6x2+11x-6 (iv) False, because zero of a polynomial can be any real number e.g., p(x) = x – 2, then 2 is a zero of polynomial p(x). (b) 477 (d) x4 + 3x3 + 3x2 + x + 1 NCERT Exemplar Class 9 Maths book covers basics and fundamentals on all topics for students apart from the added information of a higher level. Thinking Process Question 1. NCERT Exemplar Polynomials Class 9 (Part - 2) Nov 10, 2020 • 1h . (iii) The example of trinomial of degree 2 is x2 – 4x + 3. ⇒ (2x -1) (x + 4) = 0 (a)-6 If both x – 2 and x -(1/2)  are factors of px2+ 5x+r, then show that p = r. Give possible expression for the length and breadth of the rectangle whose area is given by 4a2 + 4a – 3. ∴ Sum of two polynomials, (c) 5x -1 Substituting x = 2 in (1), we get Find the following products Which one of the following is a polynomial? Now, p1(3) = a(3)3 + 4(3)2 + 3(3) – 4 = 1000027 + 92700 = 1092727, (ii) We have, 101 × 102 = (100 + 1) (100 + 2) Factorise the following: (d) 497 (i) We have, 1 + 64x3 = (1)3 + (4x)3 Question 2. (ii) y3 – 5y Let p1(z) = az3 + 4z2 + 3z – 4 and p2(z) = z3 – 4z + o ∴ P( 3) = 61 (ii) Degree of polynomial – 10 or – 10x° is 0, because the exponent of x is 0. Zero of the polynomial p(x) = 2x + 5 is (iii) Now, adjust the given polynomial in such a way that it becomes the product of two factors, one of them is a linear polynomial and other is a quadratic polynomial. (iv) 4 – 5y² NCERT Exemplar Problems Class 9 Maths Solutions are being updated for new academic session 2020-2021. (i) Polynomial x2 + x + 1 is a one variable polynomial, because it contains only one variable i.e., x. p(2) = 4(2)2 + 2 – 2 = 16 ≠ 0 Now, this is divided by x + 2, then remainder is p(-2). Question 15. Solution: (ii) 25x2 + 16y2 + 4Z2 – 40xy +16yz – 20xz For zero of polynomial, put p(x) = x-4 = 0 Thinking Process ⇒ -8 – 8m + 16 = 0 Thinking Process ⇒ (a + b + c)2 = (9)2 [Squaring on both sides] Hence, one of the zeroes of the polynomial p(x) is ½. (b) Given, p(x) = 2x+5 The topic wise list for NCERT Exemplar Class 9 Maths is provided below. Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (-z + x-2y). ⇒ x3 + y3 – 12xy + 64 = 0, (ii) Since, x – 2y – 6 = 0, then Because one of the exponents of the variable x is -1, which is not a whole number. Solution: (a) x3 + x2 – x + 1 Since, remainder ≠ 0, then p(x) is not a multiple of g(x). (iii) q(x) = 2x – 7    (iv) h(y) = 2y Solution: Question 16: (D) Not defined = (b + c)[a2+ b2+ c2 + 2 ab + 2 bc + 2 ca + a2+ a2 + ab + ac] – (b + c)(b2 + c2 – bc) (i) p(x) = 10x – 4x2 – 3 (ii) p(y) = (y + 2)(y – 2) Since, (x + 1) is a factor of p(x), then We have, 2x2 – 7x – 15 = 2x2 – 10x + 3x -15 (iii) A binomial ipay have degree 5. (i) p(x) = x3-2x2-4x-1, g(x)=x + 1 (x) Polynomial √2x – 1 is a linear polynomial, because its degree is 1. = (x + 1)(x2 – 4) Solution: = (4x)2 + (- 2y)2 + (3z)2 + 2(4x)(-2y) + 2(-2y)(3z) + 2(3z)(4x) 26a = 26 By remainder theorem, find the remainder when p(x) is divided by g(x) (iii) Degree of polynomial x3 – 9x + 3x5 is 5, because the maximum exponent of x is 5. Question 21: Question 21. (2x -5y)3 – (2x + 5y)3 = [(2x)3 – (5y)3 – 3(2x)(5y)(2x – 5y)] = (x – 2) (2x2 + x – 15) (iii) xy+ yz +zx     (iv) x2 – Zxy + y2 +1 m = 1 Solution: Solution: (Hi) True, because a binomial is a polynomial whose degree is a whole number greater than equal to one. 2a =3 Question 18. (a) 2             (b) 0           (c) 1          (d)½ = (2x-1)(x+ 4) (i) 2x3 – 3x2 – 17x + 30 ⇒ a = 2, Question 23. = 27 – 27 + 12 + 50 = 62 Solving Latest year 2021 Exemplar Problems Solutions for Class 9 Polynomials is the best option to understand the concepts given in NCERT books and do advanced level preparations for Class 9 exams. (ii) Polynomial y3 – 5y is a one variable polynomial, because it contains only one variable i.e., y. = 2x(2x+ 3) + 1 (2x+ 3) (iii) The coefficient of x6 in given polynomial is -1. Solution: Now, p2(3) = (3)3-4(3)+a Check whether p(x) is a multiple of g(x) or not (d) 2abc Question 6: Hence, zero of polynomial is 4. Here we have given NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials. If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p ( ½). Hence, zero of polynomial is (d) Given p(x) = x+3, put x = -x in the given equation, we get p(-x) = -x+3 (i) x2 + x +1           (ii) y3 – 5y (b) 6 = 2a(2a + 3) -1 (2a + 3) = (2a – 1)(2a + 3) Hence, possible length = 2a -1 and breadth = 2a + 3, Question 1: Justify your answer: p(2) = 3(2)2 + 6(2) – 24 Now, p(x)+ p(-x) = x+ 3+ (-x)+ 3=6. Factorise (a) 1 (ii) 2√2a3 +8b3 -27c3 +18√2abc Find the zeroes of the polynomial p(x)= (x – 2)2 – (x+ 2)2. Solution: Solution: Question 3: Now, p2(3) = (3)3-4(3)+a Hence, the value of a is 3/2. (a) 4             (b) 5         (c) 3            (d) 7 Solution: Question 28: Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. 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