ncert exemplar class 9 maths polynomials

(ii) The given polynomial is 4 - y². Hence, possible length = 2a -1 and breadth = 2a + 3, Question 1. Find the values of a. Number Systems | NCERT Exemplar Solutions | Class 9 Exercise 1.1 Page No: 2 Write the correct answer in each of the following: 1. (c) 2 Question 4: Practise the solutions to understand how to use the remainder theorem to work out the remainder of a polynomial. = 2a(2a + 3) -1 (2a + 3) = (2a – 1)(2a + 3) Hence, possible length = 2a -1 and breadth = 2a + 3, Question 1: Solution: (a) 2 Therefore, remainder is 0. [∴ If a + b + c = 0, then a3 + b3 + c3 3abc] ⇒ t = 0 and t – 2 = 0 Solution: (i), we get Hence, zero of polynomial q(x) is 7/2. (ii) Further, put the factors equals to zero, then determine the values of x. (i) -3 is a zero of at – 3 a3 + b3 + c3 = 3abc. (iii) In both the case if remainder is zero, then biquadratic polynomial is divisible by = 9a2 + 25b2 + c2 – 30ab + 10bc – 6ac, (iii) We have, (- x + 2y – 3z)2 = (- x)2 + (2y)2 + (-3z)2 + 2(-x)(2y) + 2(2y)(- 3z) + 2(- 3z)(- x) Given, polynomial is p(x) = (x – 2)2 – (x + 2)2 We have, 84 – 2r – 2r2 = – 2 (r2 + r – 42) If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is Without actual division, prove that 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2 = (x – 1) [3x(x + 1) – 1(x + 1)] ⇒ x = ½ and x = -4 (d) 1/2 (i) We have, p(x) = x³ – 2x² – 4x – 1 and g(x) = x + 1 We know that, We have, (v) 3 Solution: Solution: (a) x2 + y2 + 2 xy (b) x2 + y2 – xy (c) xy2 (d) 3xy Show that p-1 is a factor of p10 -1 and also of p11 -1. 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Question 9. = (2x-1)(x+ 4) So, x = -1 is zero of x3 + x2 + x+1 = 4x³ – 6x² + 2x – 10x² + 15x – 5 x + 3 is a factor of p(x) if p(-3) = 0 Since, x + 2a is a factor of p(x), then put p(-2a) = 0 (-2a)5 – 4a2 (-2a)3 + 2(-2a) + 2a + 3 = 0 => -32a5 + 32a5 -4a + 2a+ 3 = 0 x³ + (-2y)3 + (-6)3 = 3x(-2y)(-6) and p(-2) =10 (-2) -4 (-2)2 – 3 ⇒ p(x) is divisible by x2 – 3x + 2 i.e., divisible by x – 1 and x – 2, if p(1) = 0 and p(2) = 0 x3 – y3 = (x – y)(x2 + y2 + xy) and = (100)2 + (1 + 2)100 + (1)(2) (ii) y3 – 5y (iii) trinomial of degree 2. P(2√2) = (2√2)2– (2√2)(2√2) + 1 =8-8+1=1, Question 6: On putting x = 2√2 in Eq. Using suitable identity, evaluate the following Thinking Process p(x) = 10x – 4x2 – 3 Because exponent of the variable x is 1/2, which is not a whole number. Write the coefficient of x² in each of the following = (4x – 2y + 3z)2 [∴ a3 + b3 = (a + b)(a2 – ab + b2)] (ii) 6x2 + 7x – 3 (i) The example of monomial of degree 1 is 5y or 10x. [∴ a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2] NCERT Exemplar Class 9 Maths. Solution: (a) 0 (b) 1 (c) 4√2 (d) 8 √2 +1 Since p(x) is divided by x + 1, then remainder is p(-1). Solution: (d) 7 and h(p) = p11 -1. Hence, zero of the polynomial p(x) is -5/2. (v) Polynomial 3 = 3x° is a constant polynomial, because its degree is 0. = 4a2 + 6a – 2a – 3 Solution: Also, if a + b + c = 0, then a3 + b3 + c3 = 3abc Solution: = (x – 2) (2x2 + 6x – 5x – 15) Justify your answer. = -20-4×4-3 =-20-16-3=-39 Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (-z + x-2y). Question 20: Solution: (c) 49 Question 36. Solution: Question 15: (v) -3 is a zero of y2 + y – 6 and h(p) = p11 -1 …(2) (d) (2x – 1) (2x – 3) (iii) If the maximum exponent of a variable is 1, then it is a linear polynomial. Question 17. = 3×27-4×9 + 21-5 = 81-36+21-5 P( 3) =61 26a = 26 Let g (p) = p10 -1 …(1) Using suitable identity, evaluate the following: Simplify (2x- 5y)3 – (2x+ 5y)3. (b) -1 e.g. (i), we get p(0) = 10(0)-4(0)2 -3 = 0-0-3= -3 (a) 1 we see that (x – 2y) +(2y – 3z)+ (3z – x) = 0 (ii) Polynomial (i) A binomial can have atmost two terms (b) √2 = -√2x°. (c) 4√2 (iv) y3(1 – y4) ⇒ a2 + b2 + c2 + 2 (ab + bc + ca) = 81 NCERT Exemplar Problems Class 9 Maths Solutions are being updated for new academic session 2020-2021. = 1000000 + 1 – 2000 = 998001, Question 27. (i), we get and p(-2) = (-2 + 2)(-2 -2) = x3 + 27 + 9x (x + 3) NCERT Exemplar for Class 9 Maths Chapter 3 With Solution | Coordinate Geometry. NCERT solutions are really helpful when it comes to a complicated subject like Mathematics. Question 8. (2), we get (i) False, because a binomial has exactly two terms. (d) Now, a3 + b3 + c3 = (a + b + c) (a2 + b2 + c2 – ab – be – ca) + 3abc (i) Firstly, find the zero of g(x) and then put the value of in p(x) and simplify it. [∴ (a + b)3 = a3 + b3 + 3ab(a + b)] (b) Let p (x) = 2x2 + 7x-4 Prove that (a +b +c)3 -a3 -b3 – c3 =3(a +b)(b +c)(c +a). = 2(6 – r)(r + 7) or 2(6 – r) (7 + r), Question 24. If a + b + c =0, then a3+b3 + c3 is equal to = 2a(2a + 3) -1 (2a + 3) = (2a – 1)(2a + 3) Solution: (c) Let p(x) = 2x2 + kx (c) any real number Hence, the zeroes of y² + y – 6 are 2 and – 3. Solution: (i) a3 -8b3 -64c3 -2Aabc (d) Now, 2492 – 2482 = (249 + 248) (249 – 248) [using identity, a2 – b2 = (a – b)(a + b)] We hope the NCERT Exemplar Class 9 Maths Chapter 2 Polynomials will help you. We have prepared chapter wise solutions for all characters are given below. Expand the following: p(1) = 10 (1) — 4 (1 )2 -3 = x(x + 6) + 3(x + 6) = (x + 3) (x + 6) p(-1) = 5(-1) -4(-1)2 + 3= -5-4+3 = -6, Question 7: The topic wise list for NCERT Exemplar Class 9 Maths is provided below. (d) 27 Question 18. Here we have given NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials. ⇒ 2 – k = 0 Solution: NCERT Exemplar Class 9 Maths is a very important resource for students preparing for 9th Class Examination. Question 3. In this method firstly check the values of a + b+ c, then . Find the zeroes of the polynomial in each of the following, Check whether p(x) is a multiple of g(x) or not Solution: Question 21. Since, remainder ≠ 0, then p(x) is not a multiple of g(x). Question 6. = 16 + 16 + 12 + 10 + 8 = 62. NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials Exercise 2.1 Question 1. On putting x = 0, 1 and – 2, respectively in Eq. Solution: (i) Polynomial x2+ x+ 1 is a one variable polynomial, because it contains only one variable i.e., x. (iv) x2 – Zxy + y2 + 1 x3 + y3 = (x + y)(x2 + y2 – xy) ∴ Sum of two polynomials, f(x) + g(x) = x5 + 2 + (-x5 + 2x2) = 2x2 + 2, which is not a polynomial of degree 5. (iv) Degree of polynomial y3(1 – y4) or y3 – y7 is 7, because the maximum exponent of y is 7. if a + b+c = Q, now use the identity a3 + b3 + c3 = 3abc. Justify your answer. (c) 5x -1 Degree of the polynomial 4x4 + 0x3 + 0x5 + 5x + 7 is Show that p-1 is a factor of p10 -1 and also of p11 -1. (Hi) True, because a binomial is a polynomial whose degree is a whole number greater than equal to one. 37 = (3a)3 – (2b)3 – 3(3a)(2b)(3a – 2b) (a) -3 (b) 4 (c) 2 (d)-2 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. Here, we see that (x-2y) +(2y-3z)+ (3z-x) = 0 and p(-2) =10 (-2)- 4 (-2)2 – 3 e.g., Let f(x) = x4 + 2 and g(x) = -x4 + 4x3 + 2x. Question 13: Expand the following = 27a+ 36+ 9-4= 27a+ 41 When we divide p2(z) by z-3 then we get the remainder p2(3). (a)-6 (b) 6 (c) 2 (d) -2 Question 16: p(2) = 4(2)2 + 2 – 2 = 16 ≠ 0 Question 18: Since, (x + 1) is a factor of p(x), then (c) 487 Question 7: Factorise the following (v) True Let p(x) = 2x4 – 5x3 + 2x2 – x + 2 At x= 3, p(3) = 3(3)3 – 4(3)2 + 7(3) – 5 Substituting x = 2 in (1), we get = 10000 + 300 + 2 = 10302, (iii) We have, (999)2 = (1000 -1)2 (i) p(x) = x3 – 5x2 + 4x – 3, g(x) = x – 2. Practice Polynomials questions and become a master of concepts. (d) -2 (iv) 0 and 2 are the zeroes of t2 – 2t p(x) = x – 4 (i) Firstly check the maximum exponent of the variable.. If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is (d) 3xy Solution: ⇒ t = 0 and t = 2 Now, p1(3) = a(3)3 + 4(3)2 + 3(3) – 4 Hence, √2 is a polynomial of degree 0, because exponent of x is 0. e.g., (a) 3x2 + 4x + 5 [polynomial but hot a binomial] Visit FlexiPrep for more files and information on Subject-Wise-NCERT-Books-PDF: Mathematics Hence, √2 is a polynomial of degree 0, because exponent of x is 0. (ii) If the maximum exponent of a variable is 0, then it is a constant polynomial. = (y-2)(y + 3) = 0 Question 5: Solution: Question 30: Hence, one of the factor of given polynomial is 3xy. Here are all questions are solved with a full explanation and available for free to download. Solution: (c) 0 = -x2z + x3 – 2 x2y – 4y2z + 4xy2 – 81y3 – z3 + xz2 – 2yz2– 2xyz + 2x2y – 4xy2 – xz2 + x2z – 2xyz + 2yz2 – 2xyz + 4y2z ⇒ x3 + y3 – 12xy + 64 = 0, (ii) Since, x – 2y – 6 = 0, then (i) (4o – b + 2c)2 = 2x(x – 5) + 3(x – 5) = (2x + 3)(x – 5) (iv) p(x) = x3 – 6x2 + 2x – 4, g(x) = 1 – (3/2) x (i), we get Question 17: ⇒ a = 2, Question 23. Solution: Question 40: (i) Degree of polynomial 2x-1 is one, Decause the maximum exponent of x is one. (i) 2x3 – 3x2 – 17x + 30 h (1) = (1)11 —1 = 1 —1 = 0 Hence, p -1 is a factor of h(p). Solution: Question 35: 8x4 + 4x3 – 16x2 + 10x + m Zero of the polynomial p(x) = 2x + 5 is Let p(x) =a5 -4a2x3 +2x + 2a +3 (iii) q(x) = 2x – 7 (iv) h(y) = 2y Hence, one of the factor of given polynomial is 10x. [∴ a3 – b3 = (a – b)(a2 + ab + b2)] (b) Given, p(x) = 2x+5 Solution: (i) Polynomial 2 – x2 + x3 is a cubic polynomial, because maximum exponent of x is 3. Question 4. Question 8: = 3 x (-1) = -3 (x – 2)2 – (x+ 2)2=0 Question 4. Question 3: (d) Given p(x) = x + 3, put x = -x in the given equation, we get p(-x) = -x + 3 (iii) q(x) = 2x – 7 (ii) Every polynomial is a binomial (iv) Polynomial x2 – Zxy + y2 +1 is a two variables pplynomial, because it contains two variables x and y. = x6 + x4 + x2 – x4 – x2 – 1 = x6 – 1, Question 34. Here we have given NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials. a = 3/2. So, it may have degree 5. These NCERT solutions are created by the BYJU’S expert faculties to help students in the preparation of their board exams. Verify whether the following are true or false. Solution: Classify the following polynomials as polynomials in one variable, two variables etc. (b) 0 (d) Now, (x + 3)3 = x3 + 33 + 3x (3)(x + 3) 2y= 0 Factorise the following: Polynomials in one variable, zeroes of polynomial, Remainder Theorem, Factorization, and Algebraic Identities. (i) Since, x + y + 4 = 0, then (iii) The example of trinomial of degree 2 is x2 – 4x + 3. One of the factors of (25x2 – 1) + (1 + 5x)2 is (b) Given, p(x) = x2 – 2√2x + 1 …(i) (d) Now, (x + 3)3 = x3 + 33 + 3x (3)(x + 3) = x3 + 27 + 9x (x + 3) (iii) 16x2 + 4)^ + 9z2-^ 6xy – 12yz + 24xz (c) x4 + x3 + x2 +1 (d) x4 + 3x3 + 3x2 + x +1 = 8x³ + 2xy2 + 18xz2 + 4x2y + 6xyz – 12x2z – 4x2y – y3 – 9yz2 – 2xy2 – 3y2z + 6xyz + 12x2z + 3y2z + 27z3 + 6xyz + 9yz2 – 18xz2 (iii) 5t – √7 = (1000)2 + (1)2 – 2(1000)(1) = 27a3 – 54a2b + 36ab2 – 8b3. We hope that our NCERT Class 9 New Books for Maths helped with your studies! (b) Let assume (x + 1) is a factor of x3 + x2 + x+1. NCERT Exemplar Class 9 for Maths Chapter 3 – Coordinate Geometry On putting x = -1 in Eq. (c) 3abc (d) 2abc NCERT Exemplar for Class 9 Maths Chapter 2 With Solution | Polynomials. For zeroes of p(x), put p(x) = 0=> (2x -1) (x + 4) = 0 Hence, we cannot exactly determine the degree of variable. Solution: = (x + 1) (x – 2) (x + 2)[∴ a2– b2 = (a – b) (a + b)], (iv) We have, 3x3 – x2 – 3x + 1 = 3x3 – 3x2 + 2x2– 2x – x + 1 = 32 – 40 + 8 = 40 – 40 = 0 All questions with solutions of polynomials will help all the students to revise complete syllabus and score more marks in examinations. On putting x = -1 in Eq. = (x – 2)(x + 3)(2x – 5), (ii)We have, x3 – 6x2 + 11x – 6 (c) any natural number (d) not defined = -81 – 36 – 21 – 5 = -143 = (b + c)[a2+ b2+ c2 + 2 ab + 2 bc + 2 ca + a2+ a2 + ab + ac] – (b + c)(b2 + c2 – bc) If p (x) = x2 – 2√2x + 1, then p (2√2) is equal to (c) 2/5 NCERT 9 Mathematics Exemplar Problem Text book Solutions. Therefore zero of the polynomial is p(x) is 0. Again, putting p = 1 in Eq. If x + 1 is a factor of ax3 + x2 – 2x + 4o – 9, find the value of a. Verify whether the following are true or false. NCERT 9th class Mathematics exemplar book solutions for chapter 2 Polynomials are available in PDF format for free download. (iv) Degree of polynomial y3(1-y4) or y3 – y7 is seven, because the maximum exponent of y is seven. Question 21. (c) 2 m = 1 e.g., Let us consider zero polynomial be 0(x-k), where k is a real number For determining the zero, putx-k = 0=>x = k Hence, zero of the zero polynomial be any real number. (b) x2 + y2 – xy (a) Degree of 4x4 + Ox3 + Ox5 + 5x + 7 is equal to the highest power of variable x. Because 8 = 8x°, then exponent of the variable x is 0, which is a whole number. ⇒ 2x – 1 = 0 and x + 4 = 0 Question 1: => 2-k = 0 => k= 2 (-1)3 + (-1)2 + (-1) + 1 = 0 Therefore, (x + 3) is a factor of p(x). ∴ 3 – 6x = 0 ⇒ x = 3/6 = 1/2. (b) 5 (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). ⇒ 3a = 6 If x + 2a is a factor of a5 – 4a2x3 + 2x + 2a + 3, then find the value of a. (ii) p(x) = 2x3 – 11x2 – 4x + 5, g(x) = 2x + 1 Solution: Question 9: Solution: (ii) Polynomial 3x3 is a cublic polynomial, because maximum exponent of x is 3. Which of the following is a factor of (x+ y)3 – (x3 + y3)? p1(3) = p2(3) Find the value of m, so that 2x -1 be a factor of When we divide p(x) by x+1, we get the remainder p(-1) ⇒ x = 0 Solution: (d) Now, (25x2 -1) + (1 + 5x)2 p(-1)=0 Question 15. = 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) If you have any query regarding NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials, drop a comment below and we will get back to you at the earliest, RD Sharma Class 11 Solutions Free PDF Download, NCERT Solutions for Class 12 Computer Science (Python), NCERT Solutions for Class 12 Computer Science (C++), NCERT Solutions for Class 12 Business Studies, NCERT Solutions for Class 12 Micro Economics, NCERT Solutions for Class 12 Macro Economics, NCERT Solutions for Class 12 Entrepreneurship, NCERT Solutions for Class 12 Political Science, NCERT Solutions for Class 11 Computer Science (Python), NCERT Solutions for Class 11 Business Studies, NCERT Solutions for Class 11 Entrepreneurship, NCERT Solutions for Class 11 Political Science, NCERT Solutions for Class 11 Indian Economic Development, NCERT Solutions for Class 10 Social Science, NCERT Solutions For Class 10 Hindi Sanchayan, NCERT Solutions For Class 10 Hindi Sparsh, NCERT Solutions For Class 10 Hindi Kshitiz, NCERT Solutions For Class 10 Hindi Kritika, NCERT Solutions for Class 10 Foundation of Information Technology, NCERT Solutions for Class 9 Social Science, NCERT Solutions for Class 9 Foundation of IT, PS Verma and VK Agarwal Biology Class 9 Solutions, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, Periodic Classification of Elements Class 10, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, CBSE Previous Year Question Papers Class 12, CBSE Previous Year Question Papers Class 10. = -2[r(r + 7) -6(r + 7)] (a) 2 Solution: (ii) We have, (0.2)3 – (0.3)3 + (0.1)3 NCERT Class 9 Maths Unit 2 is for Polynomials. Question 11. (a) 12 (b) 477 (c) 487 (d) 497 (iv) Polynomial Solution: Question 27: g(1)=110-1= 1-1=0 Hence, p-1 is a factor of g(p). Here, zero of g(x) is 2. Exercise 2.1: Multiple Choice Questions (MCQs). Therefore, (x – 2y)³ + (2y – 3z)³ + (3z – x)³ = 3(x – 2y)(2y – 3z)(3z – x). Hence, the degree of a polynomial is 4. = (x+ y)(x2+ y2+ 2xy – x2+ xy – y2) 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. Using long division method Question 2: Coefficient of x² in 4x³ – 16x² + 17x – 5 is -16. = (x + 1)(x2 – 4) Put 4 – 5y = 0 ⇒ y = 4/5 [∴ a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2] (iii) xy+ yz +zx (iv) x2 – Zxy + y2 +1 For zeroes of p(x), put p(x) = 0 (iii) We have, p(x) = 4x³ – 12x² + 14x – 3 and g(x) = 2x -1 Classify the following as constant, linear, quadratic and cubic polynomials: (x2 + 4y2 + z2 + 2xy + xz – 2yz)(- z + x – 2y) L.H.S. If a + b + c = 5 and ab + bc + ca = 10, then prove that a3 + b3 + c3 – 3abc = -25. ⇒ a2 + b2 + c2 + 2ab + 2bc + 2ca = 81 Question 14: NCERT Exemplar Solutions For Class 9 Maths. Because every polynomial is not a binomial. Solution: Question 2. = -20- 4 × 4 – 3 = -20 – 16 – 3 = -39 Question 14. lf a+b+c=9 and ab + bc + ca = 26, find a2 + b2 + c2. = -1 + 51 = 50 Hence, the value of m is 1 . ∴ a = 5 (i) p(x) = 10x – 4x2 – 3 (ii) p(y) = (y + 2)(y – 2) One of the zeroes of the polynomial 2x2 + 7x – 4 is All solutions are explained using step-by-step approach. ∴ P( 3) = 61 [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: (i) The given polynomial is 5x³ + 4x² + 7x The highest power of the variable x is 3. (ii) Polynomial 3x³ is a cubic polynomial, because its degree is 3. Factorise: Solution: = (1 + 4x)[(1)2 – (1)(4x) + (4x)2] (iv) 84 – 2r – 2r2 p(x) = x4 – 2x3 + 3x2 – 5x + 3(5) – 7 (v) Polynomial 3 is a constant polynomial, because the exponent of variable is 0. We have, 2x2 – 7x – 15 = 2x2 – 10x + 3x -15 Question 13. You should get good marks in Class 9 examinations as it will always help you to get good rank in school. [∴ (x + a)(x + b) = x2 + (a + b)x + ab] ⇒ y – 2 = 0 and y + 3 = 0 So, the degree of the polynomial is 3. For zero of polynomial, put g(x) = 0 (d) 10x (a) x3 + x2 – x + 1 (i) 2x – 1 = 25x2 -1 + 1 + 25x2 + 10x [using identity, (a + b)2 = a2 + b2 + 2ab] (iv) False The class will be conducted in Hindi and the notes will be provided in English. It depends upon the degree of the polynomial a3+b3 + c3 = 3abc, Exercise 2.2: Short Answer Type Questions. (a) 1 Prove that (a +b +c)3 -a3 -b3 – c3 =3(a +b)(b +c)(c +a). NCERT Exemplar Class 9 Maths Unit 2 Polynomials. Solution: HOTS, exemplar, and hard questions in polynomials. Solution: Solution: then (5)2 = a2 + b2+ c2 + 2(10) Hence, zero of polynomial is 4. = (x – 1) (x + 1)(3x -1), Question 25. Question 14. = 497 x 1 = 497. (vi) False (i) We have, 2X3 – 3x2 – 17x + 30 So, x = -1 is zero of x3 + x2 + x + 1 These expert faculties solve and provide the NCERT Solution for class 9 so that it would help students to solve the problems comfortably. Question 1: and p(2) = 2(2)4 – 5(2)3 + 2(2)2 – 2 + 2 = 2x16-5x8+2x4+ 0 = 32 – 40 + 8 = 40 – 40 =0 Show that, (i), we get p(-1) = (-1)51 + 51 (ii) Degree of polynomial -10 or -10x° is zero, because the exponent of x is zero. ∴ p(-1) = (-1)³ – 2(-1)² – 4(-1) -1 (i) We have, g(x) = x – 2 = 2x(x+ 4)-1(x+ 4) Hence, the value of k is 2. If p (x) = x2 – 2√2x + 1, then p (2√2) is equal to Justify your answer: Hence, one of the factor of given polynomial is 10x. Hence, the value of the given polynomial at x = 3 and x = -3 are 61 and -143, respectively. (ii) The two different values of zeroes put in biquadratic polynomial. Without actually calculating the cubes, find the value = 27a+ 36+ 9-4= 27a+ 41 This solution is strictly revised in accordance with the recently updated syllabus issued by CBSE. (a) 2 (b)½ (c)-1 (d)-2 Solution: Question 9. On putting x= -1 in Eq. Thinking Process (d) Let p(x) = x51 + 51 . When we divide p1(z) by z – 3, then we get the remainder p,(3). Given, area of rectangle = (Length) × (Breadth) (a) -2/5 [∴ (a – b)2 = a2 + b2 – 2ab] NCERT Class 9 Maths Solutions develop logical thinking skills so that students cable to solve all the sums once the concept is clear. Which of the following expressions are polynomials? (ii) Given, polynomial is Solution: Find the value of Solution: The coefficient of x in the expansion of (x + 3)3 is Solution: Question 33: 2a =3 (ii) Given, polynomial isp(y) = (y+2)(y-2) 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. Fi nd (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz). Solution: Question 26: (ii) 9x2 – 12x + 4 Solution: (i) 9x2 +4y2 + 16z2 +12xy-16yz -24xz At x = -3, p(-3)= 3(—3)3 – 4(-3)2 + 7(-3)- 5 (ii) -1/3 is a zero of 3x + 1 If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p ( ½). Expand the following = (a + b + c)(a2 + b2 + c2 – ab – be – ca) For zero of polynomial, put p(x) = 0 (vii) y³ – y (i) Given, polynomial is (i) We have, (3a – 2b)3 p(1) = (1 + 2)(1-2) Put 3x + 1 = 0 ⇒ x = -1/3 (i), we get If x51 + 51 is divided by x + 1, then the remainder is (iii) Polynomial 5t – √7 is a linear polynomial, because its degree is 1. [using identity, a3 + b3 = (a + b)(a2 -ab+ b2)] = (x+ y)[(x+ y)2 -(x2 -xy+ y2)] (a) Let p (x) = 5x – 4x2 + 3 …(i) Hence, x – 2 is not a factor of p(x). Let p1(z) = az3 +4z2 + 3z-4 and p2(z) = z3-4z + o (a) -3 Put t² – 2t = 0 ⇒ t(t – 2) = 0 If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. Factorise: Find the value of m, so that 2x -1 be a factor of (i) Given, polynomial is (d) Now, (x+ y)3 – (x3 + y3) = (x + y) – (x + y)(x2– xy + y2) Since, p(x) is divisible by (x+2), then remainder = 0 ⇒ a2 + b2 + c2 = 5 … (i) Factorise Question 16. Because a binomial is a polynomial whose degree is a whole number which is greater than or equal to one. Question 1. (i) x2 + 9x + 18 (a) -2/5 (b) -5/2 (c)2/5 (d)5/2 Here students are also provided with online learning materials such as NCERT Exemplar Class 9 Maths Solutions. = 2x(2x+ 3) + 1 (2x+ 3) Thus, required polynomial, (ii) 2√2a3 +8b3 -27c3 +18√2abc = 27-12 + a = 15+a According to’ the question, both the remainders are same. = 2 – 5 + 2 – 1 + 2 = 6 – 6 = 0 [using identity, a3+b3 + c3 – 3 abc = (a + b + c)(a2 + b2 + c2 – ab – be -ca)] = 0 + 3abc [∴ a + b + c = 0, given] (iii) A binomial may have degree 5 CBSE Maths notes, CBSE physics notes, CBSE chemistry notes. (i) False Solution: Polynomials | Maths | NCERT Exemplar Solutions | Class 9. = 2x (2x² – 3x + 1) – 5(2x² – 3x + 1) (i) We have, (4a – b + 2c)2 = (4a)2 + (-b)2 + (2c)2 + 2(4a)(-b) + 2(-b)(2c) + 2(2c)(4a) Factorise the following Solution: Question 8: e.g., Let f(x) = x5 + 2 and g(x) = -x5 + 2x2 For zeroes of polynomial, put p(x) = 0 (a) 0 (b) 1 (c) 49 (d) 50 (i) monomial of degree 1. for p(x) = x² – 4, degree is 2, so it has two zeroes i.e., 2 and -2. (d) Now, a3+b3 + c3= (a+ b + c) (a2 + b2 + c2 – ab – be – ca) + 3abc (a) 0 The value of the polynomial 5x – 4x2 + 3, when x = – 1 is All the solutions in … (d) 497 (iii) p(x) = x3 – 12x2 + 14x – 3, g(x) = 2x – 1 – 1 ⇒ a2 + b2 + c2 = 81 – 52 = 29, Question 31. (b) (2x + 1) (2x + 3) Since 2x – 1 is afactor of p(x) then p(1/2) = 0, Question 22. (d) x4 + 3x3 + 3x2 + x + 1 NCERT solutions for class 9 Maths is available to download for free from the links below. Solution: (ii) -1/3 is a zero of 3x+1 ⇒ -5/2 ⇒ y(y + 3) – 2(y + 3) = 0 All the chapterwise questions with solutions to help you to revise complete CBSE syllabus and score more marks in Your board examinations. 27a+41 = 15+a [∴ a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2] = (- 5x + 4y + 2z)(- 5x + 4y + 2z), (iii) We have, 16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz On putting y =0,1 and -2, respectively in Eq. Published by Administrator on October 27, 2020 October 27, 2020. (ii) x3 – 8y3 – 36xy-216,when x = 2y + 6. = (x – 2) [2x(x + 3) – 5(x + 3)] Question 12: = (4x)2 + (- 2y)2 + (3z)2 + 2(4x)(-2y) + 2(-2y)(3z) + 2(3z)(4x) = (2x + 3) (2x + 1). (c) Zero of the zero polynomial is any real number. Question 5: = x3 – 8y3 – z3 – 6xyz. Solution: Question 2: Solution: Question 9: Question 10. quadratic polynomial. Solution: (ix) Polynomial t² is a quadratic polynomial, because its degree is 2. 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