Hence, zero of polynomial q(x) is 7/2. (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. (ii) Every polynomial is a Binomial. (D) Not defined = 2x(x – 5) + 3(x – 5) = (2x + 3)(x – 5) (b) Given, p(x) = 2x+5 (ii) Given, polynomial isp(y) = (y+2)(y-2) =(-5x)2 + (4y)2 + (2z)2 + 2(-5x)(4y) + 2(4y)(2z) + 2(2z)(-5x) ∴ x – 4 = 0 ⇒ x = 4 Thinking Process Question 19: Solution: Question 26: (ii) The given polynomial is 4 - y². Because the sum of any two polynomials of same degree has not always same degree. (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx (iv) Given polynomial h(y) = 2 y Zero of the zero polynomial is Since, remainder ≠ 0, then p(x) is not a multiple of g(x). Solution: then (5)2 = a2 + b2+ c2 + 2(10) Hence, the values of p(0), p(1) and p(-2) are respectively, -3,3 and – 39. In this class, Vivek Patriya will discuss NCERT Exemplar Polynomials Class 9 (Part - 2) .The class will be helpful for the aspirants of CBSE 9th. (iv) We have, (2x – 5) (2x² – 3x + 1) Solution: Question 23: Give possible expression for the length and breadth of the rectangle whose area is given by 4a2 + 4a – 3. Write the coefficient of x2 in each of the following (viii) Polynomial 1 + x + x² is a quadratic polynomial, because its degree is 2. Given, polynomial is p(x) = (x – 2)2 – (x + 2)2 (iii) -4/5 is a zero of 4 – 5y (b) 5 Hence, the remainder is 50. (ii) Polynomial 3x3 is a cublic polynomial, because maximum exponent of x is 3. We know that (iii) Degree of polynomial x3 – 9x + 3x5 is 5, because the maximum exponent of x is 5. Solution: (c) 2 ⇒ y(y + 3) – 2(y + 3) = 0 (a) 4 = x(x-2)-1 (x-2)= (x-1)(x-2) Using suitable identity, evaluate the following: We have, p(x) = x4 – 2x3 + 3x2 – ox + 3a – 7 = (3x)2 + (2y)2 + (-4z)2 + 2(3x)(2y) + 2(2y)(-4z) + 2(-4z)(3x) 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. Question 4. So, x = -1 is zero of x3 + x2 + x+1 (d) 5/2 p(2) = 4(2)2 + 2 – 2 = 16 ≠ 0 ∴ p(-1) = (-1)³ – 2(-1)² – 4(-1) -1 = x2 (x4 + x2 + 1) – 1(x4 + x2 + 1) (ii) x3-6x2+11x-6 a = 3/2. Question 19. Question 3. At x = -3, p(-3)= 3(—3)3 – 4(-3)2 + 7(-3)- 5 Show that, (a) x3 + x2 – x +1 (b) x3 + x2 + x+1 = 10-4-3= 10-7= 3 (iii) 9992 Solution: = x6 + x4 + x2 – x4 – x2 – 1 = x6 – 1, Question 34. Since p(x) is divided by x + 1, then remainder is p(-1). [using identity, a3+b3 + c3 – 3 abc = (a + b + c)(a2 + b2 + c2 – ab – be -ca)] = 0 + 3abc [∴ a + b + c = 0, given] Hence, we cannot exactly determine the degree of variable. Simplify (2x- 5y)3 – (2x+ 5y)3. = -2[r(r + 7) -6(r + 7)] Question 17. (-1)3 + (-1)2 + (-1) + 1 = 0 => -1+1-1 + 1 = 0 => 0 = 0 Hence, our assumption is true. Now, this is divided by x + 2, then remainder is p(-2). Question 6: Put 4 – 5y = 0 ⇒ y = 4/5 Thinking Process With the help of it, candidates can prepare well for the examination. Question 8: Solution: (a) x2 + y2 + 2 xy Question 4: p(x) = x – 4 If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p ( ½). Solution: Question 15: NCERT Exemplar Problems textbook is very helpful to understand the basic concepts of Mathematics. => 2x-1 = 0 and x+4 = 0 When we divide p2(z) by z-3 then we get the remainder p2(3). Hence, zero of polynomial is 4. iii) 16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz = (1000)2 + (1)2 – 2(1000)(1) Hence, 0 of x2-3x+2 are land 2. NCERT Class 9 Maths Unit 2 is for Polynomials. Factorise: Solution: Hence, zero of polynomial is => x = ½ and x = -4 Factorise the following (vi) Not polynomial ⇒ a2 + b2 + c2 = 25 – 20 (ii) y3 – 5y We know that, Solution: (b) 2x ⇒ -8 – 8m + 16 = 0 (ii) the coefficient of x3 (a) Degree of 4x4 + Ox3 + Ox5 + 5x + 7 is equal to the highest power of variable x. (c) 1 Question 12: Find p(0), p( 1) and p(-2) for the following polynomials a3+b3 + c3 = 3abc, Exercise 2.2: Short Answer Type Questions. p(x) = 10x – 4x2 – 3 Question 2: Let p(x) = 8x4 + 4x3 – 16x2 + 10x + m p1(3) = p2(3) Solution: Question 2: = (- 5x + 4y + 2z)2 (i) Polynomial (a)-6 NCERT Exemplar Problems Class 9 Maths Solutions are being updated for new academic session 2020-2021. HOTS, exemplar, and hard questions in polynomials. Hence, the value of the given polynomial at x = 3 and x = -3 are 61 and -143, respectively. Solution: For zeroes of p(x), put p(x) = 0=> (2x -1) (x + 4) = 0 = (x – 2) (2x2 + x – 15) Write the coefficient of x² in each of the following = (100)2 + (1 + 2)100 + (1)(2) Question 10: BeTrained.in has solved each questions of NCERT Exemplar very thoroughly to help the students in solving any question from the book with a team of well experianced subject matter experts. = 3(3x2 – 3x – x + 1) (v) True For zero of polynomial, put p(x) = 0 Solution: = 8x³ + 2xy2 + 18xz2 + 4x2y + 6xyz – 12x2z – 4x2y – y3 – 9yz2 – 2xy2 – 3y2z + 6xyz + 12x2z + 3y2z + 27z3 + 6xyz + 9yz2 – 18xz2 (i) Given, polynomial is p(-1)=0 Hence, one of the zeroes of the polynomial p(x) is ½. NCERT Class 9 New Books for Maths Chapter 2 Polynomials are given below. = (a + b + c – a)[(a + b + c)2 + a2 + (a + b + c)a] – [(b + c) (b2 + c2– be)] Solution: ⇒ x = 0 L.H.S. Let p(x) = x3 -2mx2 +16 (i) Firstly, adjust the given number either in the farm 0f a3 + b3 or a3 -b3 = 2 – 5 + 2 – 1 + 2 = 6 – 6 = 0 (ii) 6x2 + 7x – 3 lf a+b+c=9 and ab + bc + ca = 26, find a2 + b2 + c2. ⇒ a2 + b2 + c2 = 5 … (i) = x3 +27 + 9x2+27x Hence, the coefficient of x in (x + 3)3 is 27. [using identity, (a + b)2 = a2 + b2 + 2 ab)] NCERT Solutions for Class 9 Maths Chapter 2 Polynomials (बहुपद) (Hindi Medium) Ex 2.1. (d) Now, (25x2 -1) + (1 + 5x)2 ⇒ (2x)(-4) = 0 Solution: = x(x + 6) + 3(x + 6) = (x + 3) (x + 6) (i) Since, x + y + 4 = 0, then Question 13: (i), we get (ii) 2√2a3 +8b3 -27c3 +18√2abc 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 … Solution: (b) 9 (b) 1 Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. (b) 3x2 + 5 [polynomial and also a binomial] Thinking Process (a) -3 (b) 4 (c) 2 (d)-2 Solution: (i) p(x) = x3 – 2x2 – 4x – 1, g(x) = x + 1 = 2a(2a + 3) -1 (2a + 3) = (2a – 1)(2a + 3) Hence, possible length = 2a -1 and breadth = 2a + 3, Question 1: (vii) y³ – y (i), we get p(0) = 10(0)-4(0)2 -3 = 0-0-3= -3 [∴ a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2] Question 5: The highest power of … Let p1(z) = az3 +4z2 + 3z-4 and p2(z) = z3-4z + o (iii) The example of trinomial of degree 2 is x2 – 4x + 3. (a) Let p (x) = 5x – 4x2 + 3 …(i) = (x – 1) [3x(x + 1) – 1(x + 1)] Solution: we get p(0) = 10(0) – 4(0)2 – 3 = 0 – 0 – 3 = -3 (b) Now, 4x2 + 8x + 3= 4x2 + 6x + 2x + 3 [by splitting middle term] (d) The degree of zero polynomial is not defined, because in zero polynomial, the coefficient of any variable is zero i.e., Ox2 or Ox5,etc. (i) We have, p(x) = x³ – 2x² – 4x – 1 and g(x) = x + 1 (ii) The coefficient of x3 in given polynomial is 1/5. Question 8. ⇒ (x – 2 + x + 2)(x – 2 – x – 2) = 0 Here are all questions are solved with a full explanation and available for free to download. = [(a + b + c)3 – a3] – (b3+ c3) Give an example of a polynomial, which is = -2 (5y)3 – 30xy(2x – 5y + 2x + 5y) Hence, the value of m is 1. 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. For zeroes of polynomial, put p(x) = 0 = 4x³ – 6x² + 2x – 10x² + 15x – 5 Because one of the exponents of the variable x is -1, which is not a whole number. NCERT Class 9 Maths Solutions develop logical thinking skills so that students cable to solve all the sums once the concept is clear. Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (-z + x-2y). Question 9: (v) 3 ⇒ t = 0 and t = 2 (iii) Given, polynomial is q(x) = 2x -7 For zero of polynomial, put q(x) = 2x-7 = 0 ⇒ (-2)3 – 2m(-2)2 + 16 = 0 Which one of the following is a polynomial? ∴ Coefficient of x² in 3x² – 7x + 4 is 3. Solution: (iii) (-x + 2y-3z)2 Therefore, (x-2y)3 + (2y-3z)3 + (3z-x)3 = 3(x-2y)(2y-3z)(3z-x). e.g., Let us consider zero polynomial be 0(x – k), where k is a real number For determining the zero, put x – k = 0 ⇒ x = k Therefore, we can not exactly determine the highest power of variable, hence cannot define the degree of zero polynomial. ∴ p(-3) = -143 Solution: Question 28: Question 6. We have prepared chapter wise solutions for all characters are given below. Question 6: Find the value of If the polynomials az3 +4z2 + 3z-4 and z3-4z + o leave the same remainder when divided by z – 3, find the value of a. 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. Then it is a polynomial can have any number of zeroes Maths book covers basics fundamentals... Practice Polynomials questions and become a master of concepts Class 6, 7, 8 9... 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