NCERT solutions are really helpful when it comes to a complicated subject like Mathematics. Solution: (C) It is not […] ⇒ 8m = 8 Fi nd (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz). (i) Polynomial x2 + x + 1 is a one variable polynomial, because it contains only one variable i.e., x. If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p ( ½). = 2x(2x+ 3) + 1 (2x+ 3) 2a = 3 It is a polynomial, because each exponent of x is a whole number. (ii) Polynomial 3x³ is a cubic polynomial, because its degree is 3. if a + b+c = Q, now use the identity a3 + b3 + c3 = 3abc. Therefore, the degree of the given polynomial is 4. Solution: Put 4 – 5y = 0 ⇒ y = 4/5 (x2 + 4y2 + z2 + 2xy + xz – 2yz)(- z + x – 2y) = 2x2(x – 2) + x(x – 2) – 15(x – 2) Hence, the value of the given polynomial at x = 3 and x = -3 are 61 and -143, respectively. Solution: Question 5: Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. for p(x) = x² – 4, degree is 2, so it has two zeroes i.e., 2 and -2. ∴ 3 – 6x = 0 ⇒ x = 3/6 = 1/2. Let p(x) =3x3 – 4x2 + 7x – 5 (ii) -binomial of degree 20. (i) Polynomial x2+ x+ 1 is a one variable polynomial, because it contains only one variable i.e., x. (i) We have, x2 + 9x +18 = x2 + 6x + 3x +18 Question 17. Here we have given NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials. These handwritten NCERT Mathematics Exemplar Problems solutions are provided absolutely free … (i) -3 is a zero of at – 3 We have prepared chapter wise solutions for all characters are given below. Question 10. If x +1 is a factor of ax3 +x2 -2x+4a-9, then find the value of a. Question 7. Class 9 mathematics is an introduction to various new topics which are not there in previous classes. Thinking Process (d) 7 (ii) We have, p(x) = x³ – 3x² + 4x + 50 and g(x) = x – 3 = -20- 4 × 4 – 3 = -20 – 16 – 3 = -39 NCERT Exemplar for Class 9 Maths Chapter 5 With Solution | Introduction to Euclid’s Geometry Solution: (b) Given, p(x) = x2 – 2√2x + 1 …(i) = (x -1) (x2 – 5x + 6) (d) -2 = -2(r2 + 7r – 6r – 42) = 32 – 40 + 8 = 40 – 40 = 0 Therefore, remainder is 0. Classify the following as  constant, linear, quadratic and cubic polynomials: (d) Now, (x + 3)3 = x3 + 33 + 3x (3)(x + 3) Now, p2(3) = (3)3-4(3)+a Question 3. (v) -3 is a zero of y2 +y-6 If p (x) = x + 3, then p(x) + p(- x) is equal to ⇒ (-2)3 – 2m(-2)2 + 16 = 0 NCERT Class 9 New Books for Maths Chapter 2 Polynomials includes all the questions given in CBSE syllabus. For zero of polynomial, put p(x) = 0 = 3 x (-1) = -3 => -5/2 (v) A polynomial cannot have more than one zero Question 2. = 4a2 + 6a – 2a – 3 p(2) = 4(2)2 + 2 – 2 = 16 ≠ 0 We have, 2x2 – 7x – 15 = 2x2 – 10x + 3x -15 (a) 0        (b) 1           (c) any real number               (d) not defined (ii) Degree of polynomial -10 or -10x° is zero, because the exponent of x is zero. Solution: Question 4: (d) Now, (25x2 -1) + (1 + 5x)2 Without actual division, prove that 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2 (i) x3 + y3 – 12xy + 64,when x + y = -4. (ii) g(x)= 3 – 6x = 3x(2x + 3) – 1(2x + 3) = (3x – 1)(2x + 3) Solution: a3 + b3 + c3 – 3 abc = (a + b + c)(a2 + b2 + c2 -ab-bc-ca) Solution: NCERT 9th class Mathematics exemplar book solutions for chapter 2 Polynomials are available in PDF format for free download. Solution: (b) -5/2 Let p(x) = x3 -2mx2 +16 HOTS, exemplar, and hard questions in polynomials. Question 11. If a + b + c =0, then a3 + b3 + c3 is equal to It is not a polynomial, because one of the exponents of x is – 2, which is not a whole number. (b) Now, 4x2 + 8x + 3= 4x2 + 6x + 2x + 3 [by splitting middle term] We have, 6x2 + 7x – 3 = 6x2 + 9x – 2x – 3 = (1000)2 + (1)2 – 2(1000)(1) Also, find the remainder when p(x) is divided by x+ 2. 2a =3 Solution: ⇒ x3 + y3 – 12xy + 64 = 0, (ii) Since, x – 2y – 6 = 0, then = 2x(x – 5) + 3(x – 5) = (2x + 3)(x – 5) Simplify (2x- 5y)3 – (2x+ 5y)3. Expand the following (ii) 4x2+ x – 2 For zero of the polynomial, put p(x) = 0   ∴ 2x + 5 = 0 Hence, zero of the zero polynomial be any real number. (iv) Zero of a polynomial is always 0 In this method firstly check the values of a + b+ c, then . (iii) x3 + x2 – 4x – 4             (iv) 3x3 – x2 – 3x +1 (a) -2/5 The Class NCERT 9th Math textbook has a total of 15 chapters which are divided into seven units. Solution: Question 40: The class will be conducted in Hindi and the notes will be provided in English. Hence, the value of k is 2. and h(p) = p11 -1  …(2) One of the factors of (25x2 – 1) + (1 + 5x)2 is Determine which of the following polynomial has x – 2 a factor Factorise the following (ii) The given polynomial is 4 - y². Using suitable identity, evaluate the following [using identity, (a + b)2 = a2 + b2 + 2 ab)] (a) (x + 1) (x + 3) Hence, zero of polynomial is 4. Since p(x) is divided by x + 1, then remainder is p(-1). (B) 1 NCERT Class 9 New Books for Maths Chapter 2 Polynomials are given below. We have, p(x) = x4 – 2x3 + 3x2 – ox + 3a – 7 (b) Let assume (x + 1) is a factor of x3 + x2 + x + 1. Because a polynomial can have any number of zeroes. (i) Degree of polynomial 2x-1 is one, Decause the maximum exponent of x is one. Verify whether the following are true or false. Solution: Question 33: For zero of polynomial, put h(y) = 0 = (3x – 2) (3x – 2), Question 28. (v) If the maximum exponent of a variable is 3, then it is a cubic polynomial. = 2x(2x + 3) + 1 (2x + 3) ⇒ y = 2 and y = -3 p(x) = 10x – 4x2 – 3 (i) Let p(x) = 3x2 + 6x – 24  … (1) (iii) The coefficient of x6 in given polynomial is -1. Solution: On putting x = -1 in Eq. If x51 + 51 is divided by x + 1, then the remainder is (v) False, because a polynomial can have any number of zeroes. When we divide p1(z) by z – 3, then we get the remainder p,(3). (iv) 0 and 2 are the zeroes of t2 – 2t 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. (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)). NCERT Exemplar for Class 9 Maths Chapter 3 With Solution | Coordinate Geometry. (i) A Binomial can have atmost two terms. Solution: Question 22: Give an example of a polynomial, which is Hence, one of the factor of given polynomial is 3xy. (ii) Every polynomial is a binomial (c) 0 Thinking Process (ix) Polynomial t² is a quadratic polynomial, because its degree is 2. Question 10: Question 20: Solution: Question 18: Find the value of the polynomial 3x3 – 4x2 + 7x – 5, when x = 3 and also when x = -3. [using identity, a3 + b3 + c3 – 3 abc = (a + b + c)(a2 + b2 + c2 – ab – be – ca)] SOLUTION: (i) The given polynomial is 5x³ + 4x² + 7x The highest power of the variable x is 3. p1(3)= p2(3) (i) p(x)= x – 4 Solution: (d) Now, a3+b3 + c3= (a+ b + c) (a2 + b2 + c2 – ab – be – ca) + 3abc Factorise the following: Hence, the values of p(0), p(1) and p(-2) are respectively, -3, 3 and – 39. (i) p(x) = x3-2x2-4x-1, g(x)=x + 1 The factorization of 4x2 + 8x+ 3 is Hence, one of the zeroes of the polynomial p(x) is ½. 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. When we divide p(x) by g(x) using remainder theorem, we get the remainder p(-1) Solution: = -2 (5y)3 – 30xy(2x – 5y + 2x + 5y) (ii) the coefficient of x3 Write the coefficient of x2 in each of the following (iii) trinomial of degree 2. Question 16: Solution: Show that p-1 is a factor of p10 -1 and also of p11 -1. NCERT Exemplar Class 9 Maths Chapter 2 Polynomials are part of NCERT Exemplar Class 9 Maths. (i) x + 3 is a factor of 69 + 11x – x2 + x3 = (a + b + c)(a2 + b2 + c2 – ab – be – ca) Solution: Again, putting p = 1 in Eq. Question 1. (d) -2 p(-1) = 5(-1) -4(-1)2 + 3= -5 – 4 + 3 = -6, Question 7. Solution: (iv) The constant term in given polynomial is 1/5. Now, p(x)+ p(-x) = x+ 3+ (-x)+ 3=6. ⇒ a2 + b2 + c2 = 25 – 20 Put 3x + 1 = 0 ⇒ x = -1/3 Solution: Now, this is divided by x + 2, then remainder is p(-2). Question 21. Substituting x = 2 in (1), we get Which of the following is a factor of (x+ y)3 – (x3 + y3)? (a) -3      (b) 4      (c) 2            (d)-2 NCERT Exemplar Class 9 Maths Solutions will give you a thorough understanding of Maths concepts as per the CBSE exam pattern. (iv) Polynomial 4- 5y2 is a quadratic polynomial, because maximum exponent of y is 2. (i) p(x) = x3 – 2x2 – 4x – 1, g(x) = x + 1 (i) 1033 we see that (x – 2y) +(2y – 3z)+ (3z – x) = 0 (iv) Polynomial  x2 –  2xy + y2 + 1 is a two variables polynomial, because it contains two variables x and y. Hence, x – 2 is a factor of p(x). ⇒ a2 + b2 + c2 + 2(26) = 81 [∴ ab + bc + ca = 26] ∴ p(-3) = -143 Solution: Question 5: polynomial is divided by the second polynomial x4 + 1 and x – 1. = -1 – 2 + 4 – 1 = 0 Degree of the polynomial 4x4 + 0x3 + 0x5 + 5x + 7 is = 8x³ + 2xy2 + 18xz2 + 4x2y + 6xyz – 12x2z – 4x2y – y3 – 9yz2 – 2xy2 – 3y2z + 6xyz + 12x2z + 3y2z + 27z3 + 6xyz + 9yz2 – 18xz2 (c) any real number = (3x – 2)2                [∴ a2 – 2ab + b2 = (a – b)²] Question 21: and p(2) = 2(2)4 – 5(2)3 + 2(2)2 – 2 + 2 = 2x16-5x8+2x4+ 0 = 32 – 40 + 8 = 40 – 40 =0 Also, if a + b + c = 0, then a3 + b3 + c3 = 3abc ⇒ 3a = 6 (vii) y³ – y (x) Polynomial √2x-1 is a linear polynomial, because maximum exponent of xis 1. (ii) Given, polynomial isp(y) = (y+2)(y-2) (ii)Let p(x) = 4x2 + x – 2 … (2) = 3×27-4×9 + 21-5 = 81-36+21-5 P( 3) =61 (d)½ = x6 + x4 + x2 – x4 – x2 – 1 = x6 – 1, Question 34. Question 8: (i) 1 + 64x3 Hence, quotient = x³ + x² + x + 1 and remainder = 2. Solution: Question 38: (iii) Polynomial 5t – √7 is a linear polynomial, because its degree is 1. Question 15. (i) monomial of degree 1. = 81 – 36 + 21 – 5 (a) 0 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. (d) not defined =0 (-4) = 0 If p (x) = x + 3, then p(x)+ p (- x) is equal to Classify the following as a constant, linear, quadratic and cubic polynomials (b) ½ Let g (p) = p10 -1  …(1) ⇒ a = 2, Question 23. = (x – 2) [2x(x + 3) – 5(x + 3)] p(1) = (1 + 2)(1-2) All the solutions in … Question 4. (ii) (-x + 2y – 3z)2 (b) Let p (x) = 2x2 + 7x-4 Since, p(x) is divisible by (x+2), then remainder = 0 Since, x + 2a is a factor of p(x), then put p(-2a) = 0 ∴ (0.2)³ + (-0.3)³ + (0.1)3 = 3(0.2) (-0.3) (0.1) NCERT Solutions for Class 9 Maths Chapter 2 Polynomials (बहुपद) (Hindi Medium) Ex 2.1. Now, p(1) = 2(1)4 – 5(1)3 + 2(1)2 -1 + 2 =2-5+2-1+2=6-6=0 = x3 + 27 + 9x (x + 3) (d) Now, (x+ y)3 – (x3 + y3) = (x + y) – (x + y)(x2– xy + y2) If p (x) = x2 – 2√2x + 1, then p (2√2) is equal to Hence, p(x) is divisible by x2-3x+2. Prove that (a +b +c)3 -a3 -b3 – c3 =3(a +b)(b +c)(c +a). We have, a + b + c = 5,ab + bc + ca = 10 = 16 + 16 + 12 + 10 + 8 = 62. (d)-2 NCERT Exemplar Class 9 Maths is a very important resource for students preparing for 9th Class Examination. Solution: (vi) False, because the sum of any two polynomials of same degree is not always same degree. (i) 2x3 -3x2 -17x + 30         (ii) x3 -6x2 +11 x-6 Solution: (c) 3abc             (d) 2abc Solution: (a) 0                 (b) 1                   (c) 4√2              (d) 8 √2 +1 [∴ a3 – b3 = (a – b)(a2 + ab + b2)] [∴ (a + b)3 = a3 + b3 + 3ab(a + b)] = (3x)2 + (2y)2 + (-4z)2 + 2(3x)(2y) + 2(2y)(-4z) + 2(-4z)(3x) = 25x2 -1 + 1 + 25x2 + 10x [using identity, (a + b)2 = a2 + b2 + 2ab] Because a binomial is a polynomial whose degree is a whole number which is greater than or equal to one. (iv) Polynomial (i), we get (i) We have, g(x) = x – 2 (i) a3 -8b3 -64c3 -2Aabc x3 – y3 = (x – y)(x2 + y2 + xy) and By actual division, find the quotient and the remainder when the first Question 1. Question 8. ⇒ a2 + b2 + c2 = 5     … (i) Question 16. (a) x2 + y2 + 2 xy (ii) -1/3 is a zero of 3x+1 Hence, zeroes of x2 – 3x + 2 are 1 and 2. (a) 1 = (y-2)(y + 3) = 0 (v) Polynomial 3 is a constant polynomial, because the exponent of variable is 0. (i) 2x – 1 Solution: (i) We have, 2X3 – 3x2 – 17x + 30 The value of the polynomial 5x – 4x2 + 3, when x = – 1 is If a+b+c= 5 and ab+bc+ca =10, then prove that a3 +b3 +c3  – 3abc = -25. (a) x3 + x2 – x +1         (b) x3 + x2 + x+1 Question 5. (c) Let p(x) = 2x2 + kx 27a-a = 15-41 . Solution: NCERT Exemplar Solutions For Class 9 Maths. (i), we get Solution: (vi) 2 + x (D): In zero polynomial, the coefficient of any power of variable is zero i.e., 0x², 0x5 etc. 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. [∴ a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2] Practise the solutions to understand how to use the remainder theorem to work out the remainder of a polynomial. (i) x2 + x +1           (ii) y3 – 5y If a + b + c = 0, then a3 + b3 + c3 = 3abc, Question 39. NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials Exercise 2.1 Question 1. = 2x2 + 8x – x – 4 [by splitting middle term] = (x+ y)(x2+ y2+ 2xy- x2+ xy- y2) Question 15. On putting x = -1 in Eq. Now, x2-3x+2 = x2-2x-x+2 [by splitting middle term] 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. = (x – 2) (2x2 + x – 15) 26a = 26 Write whether the following statements are True or False. and p(-2) =10 (-2)- 4 (-2)2 – 3 ⇒ (x – 2 + x + 2)(x – 2 – x – 2) = 0 (a) 4             (b) 5         (c) 3            (d) 7 (ii) The two different values of zeroes put in biquadratic polynomial. (ii) 6x2 + 7x – 3 iii) 16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz Solution: Question 4. Solution: (ii) We have, (0.2)3 – (0.3)3 + (0.1)3 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. 2x= 7 => x =7/2 Solution: ∴ P( 3) = 61 (i) Since, x + y + 4 = 0, then CBSE Maths notes, CBSE physics notes, CBSE chemistry notes. (-1)3 + (-1)2 + (-1) + 1 = 0 => -1+1-1 + 1 = 0 => 0 = 0 Hence, our assumption is true. (a) -2/5       (b) -5/2         (c)2/5            (d)5/2 x + 1 is a factor of the polynomial = (a – √2b)[a2 + a( √2b) + (√2b)2] = 4a2 + 4a – 3 Here students are also provided with online learning materials such as NCERT Exemplar Class 9 Maths Solutions. Therefore, a binomial may have degree 5. = (x-2)(x- 1) (a) Let p (x) = 5x – 4x2 + 3 …(i) (b) Given, p(x) = x2 – 2√2x + 1 …(i) Therefore, (x – 2y)³ + (2y – 3z)³ + (3z – x)³ = 3(x – 2y)(2y – 3z)(3z – x). lf a+b+c=9 and ab + bc + ca = 26, find a2 + b2 + c2. ∴ 2x – 7 = 0 ⇒ 2x = 7 ⇒ x = 7/2 Question 3. Solution: x + 1 is a factor of the polynomial (ii) 2√2a3 +8b3 -27c3 +18√2abc p(1) = 10 (1) — 4 (1 )2 -3 (ii) Polynomial If a + b + c = 9 and ab + bc + ca = 26, find a2 + b2 +c2. (iv) p(x) = x3 – 6x2 + 2x – 4, g(x) = 1 – (3/2) x Question 1: Solution: Solution: Prove that (a +b +c)3 -a3 -b3 – c3 =3(a +b)(b +c)(c +a). Question 15: (b) 9 Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (-z + x-2y). = 4x³ – 16x² + 17x – 5 Hence, zero of x – 3 is 3. (ii) False, because every polynomial is not a binomial . Here we have given NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials. At x = -3, p(-3)= 3(-3)3 – 4(-3)2 + 7(-3) – 5 = -81 – 36 – 21 – 5 = -143 (c) x4 + x3 + x2 + 1 ∴ p(2) = (2)3 – 5(2)2 + 4(2) – 3 => x= 4 Hence, one of the factor of given polynomial is 10x. Hence, the zeroes of y² + y – 6 are 2 and – 3. Zero of the zero polynomial is Degree of the polynomial 4x4 + Ox3 + Ox5 + 5x+ 7 is P(-2) = 0 NCERT Exemplar Class 9 Maths Unit 2 Polynomials. At x= 3, p(3) = 3(3)3 – 4(3)2 + 7(3) – 5 (i), we get Solution: Question 32: Learn about the degree of a polynomial and the zero of a polynomial with related Maths solutions. (iv) 4 – 5y² (i) Degree of polynomial 2x – 1 is 1, Because the maximum exponent of x is 1. (D) Not defined = x2(x – 1) – 5x (x – 1) + 6(x – 1) Question 1: NCERT Exemplar Class 9 for Maths Chapter 2 – Polynomials. = – 27 – 9 – 33 + 69 = 0 ⇒ x3 – 8y3 – 36xy – 216 = 0, Question 40. (b) 5 – x Question 2: On putting x = 2√2 in Eq. (ii) 2x – 3 is a factor of x + 2x3 -9x2 +12 Which of the following expressions are polynomials? Hence, one of the zeroes of the polynomial p(x) is ½. Solution: Therefore, (x-2y)3 + (2y-3z)3 + (3z-x)3 = 3(x-2y)(2y-3z)(3z-x). (d) 1/2 = 3 x (-1) = -3 = 1000027 + 92700 = 1092727, (ii) We have, 101 × 102 = (100 + 1) (100 + 2) 2(-1)2 + k(-1) = 0 (iii) -4/5 is a zero of 4 – 5y The polynomial p{x) = x4 – 2x3 + 3x2 – ax + 3a – 7 when divided by x + 1 leaves the remainder 19. Solution: m = 1 You should get good marks in Class 9 examinations as it will always help you to get good rank in school. (iii) 2x2 -7x.-15       (iv) 84-2r-2r2 Solution: Question 2: (i), we get p(0) = 10(0)-4(0)2 -3 = 0-0-3= -3 By remainder theorem, find the remainder when p(x) is divided by g(x) (a) (x +1) (x + 3)             (b) (2x+1) (2x + 3) (b) -1 Which one of the following is a polynomial? Solution: Question 9: (ii) a3 – 2√2b3 Hence, zero of polynomial is X, (iii) Given, polynomial is q(x) = 2x – 7 For zero of polynomial, put q(x) = 0 Hence, the zeroes of t² – 2t are 0 and 2. Question 4: Question 14: Solution: Question 28: (iii) the coefficient of x6 Solution: (a) 1     (b) 9     (c) 18     (d) 27 = 27 – 27 + 12 + 50 = 62 (i) the degree of the polynomial (i) We have, (4a – b + 2c)2 = (4a)2 + (-b)2 + (2c)2 + 2(4a)(-b) + 2(-b)(2c) + 2(2c)(4a) On putting x= -1 in Eq. If x + 2a is a factor of a5 -4a2x3 +2x + 2a +3, then find the value of a. Exercise 2.3: Short Answer Type Questions. Solution: On putting x = -1 in Eq. (a) 4 (-1)3 + (-1)2 + (-1) + 1 = 0 = (2x + 3) (2x + 1). = 50x2 + 10x = 10x (5x + 1) With the help of it, candidates can prepare well for the examination. Here, we see that (x-2y) +(2y-3z)+ (3z-x) = 0 Question 3. (d) x4 + 3x3 + 3x2 + x + 1 (i) -3 is a zero of at – 3 Question 19. (ii) p(x) = x3 – 3x2 + 4x + 50, g(x) = x – 3 (i) Firstly, adjust the given number either in the farm 0f a3 + b3 or a3 -b3 935k watch mins. (d) Given p(x) = x+3, put x = -x in the given equation, we get p(-x) = -x+3 (iv) 3x3-x2-3x+1 (viii) 1 + x + x² [∴ a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2] (iii) Degree of polynomial x3 – 9x + 3x5 is 5, because the maximum exponent of x is 5. (i) 9x2 – 12x + 3 Now, p(x) + p(-x) = x + 3 + (-x) + 3 = 6. (ii) 25x2 + 16y2 + 4Z2 – 40xy +16yz – 20xz ⇒ 0 = 0 Hence, our assumption is true. then (5)2 = a2 + b2+ c2 + 2(10) = 2a(2a + 3) -1 (2a + 3) = (2a – 1)(2a + 3) ⇒ a2 + b2 + c2 + 2 (ab + bc + ca) = 81 (c) 1 e.g., (a) 3x2 + 4x + 5 [polynomial but hot a binomial] => 2x-1 = 0 and x+4 = 0 (d) Now, (x + 3)3 = x3 + 33 + 3x (3)(x + 3) NCERT Class 9 Maths Solutions develop logical thinking skills so that students cable to solve all the sums once the concept is clear. = (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 Put y² + y – 6 = 0 ⇒ y² + 3y – 2y – 6 = 0 (ii) 9x2 – 12x + 4 (b) 5 (a) -3 (a) 0                                                   (b) 1 Question 19. Justify your answer: P(-2) = 0 (a) 4 (c) -1 (ii) False Question 13. Solution: Question 29: (iv) 0 and 2 are the zeroes of t2 – 2t Question 10: [∴ If a + b + c = 0, then a3 + b3 + c3 3abc] (iii) Not polynomial (ii) p(y) = (y + 2)(y – 2) This chapter consists of problems based on polynomial in one variable, Zeroes of a Polynomial, Remainder Theorem, Factorisation of Polynomials and Algebraic Identities. Solution: 4x4 + 0x3 + 0x5 + 5x + 7 = 4x4 + 5x + 7 The value of 2492 – 2482 is (b) Given, p(x) = 2x+5 It is not a polynomial because it is a rational function. (i), we get => x = ½ and x = -4 (i) (4a-b + 2c)2           (ii) (3a – 5b – c)2 Because exponent of the variable x is 1/2, which is not a whole number. Without actually calculating the cubes, find the value of 36xy-36xy = 0 (a)-6          (b) 6           (c) 2                    (d) -2 ∴ (x – 2)2 – (x + 2)2 = 0 (ii) 2√2a3 +8b3 -27c3 +18√2abc Solution: (x) Polynomial √2x – 1 is a linear polynomial, because its degree is 1. = 2a(2a + 3) -1 (2a + 3) = (2a – 1)(2a + 3) Hence, possible length = 2a -1 and breadth = 2a + 3, Question 1: = (0.2)3 + (- 0.3)3 + (0.1)3 Question 3: Hence, the value of m is 1 . = x(x + 6) + 3(x + 6) = (x + 3) (x + 6) Degree of the zero polynomial is a(- 1)3+ (- 1)2 – 2(-1) + 4a – 9 = 0 ∴ 2y = 0 ⇒ y = 0 Solution: (iv) Given polynomial h(y) = 2 y For zero of polynomial, put h(y) = 0 (b) 1 If x + 1 is a factor of ax3 + x2 – 2x + 4o – 9, find the value of a. = (a – √2b)(a2 + √2ab + 2b2), Question 35. 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 p(- 1) = 19          (Given) Classify the following polynomials as polynomials in one variable, two variables etc. (iii) We have, (x – 1) (3x – 4) = 3x2 – 7x + 4 Exercise 2.1: Multiple Choice Questions (MCQs) Question 1: Which one of the following is a polynomial? So, it is a cubic polynomial. p(1) = (1 + 2)(1-2) At x= 3, p(3) = 3(3)3 – 4(3)2 + 7(3) – 5 Factorise Hence, the values of p(0), p(1) and p(-2) are respectively, -4, -3 and 0. = 0 (-4) = 0 and h(p) = p11 -1. Without actual division, prove that Factorise the following: (d) Now, 2492 – 2482 = (249 + 248) (249 – 248) [using identity, a2 – b2 = (a – b)(a + b)] 8x4 +4x3 -16x2 +10x+07. NCERT Exemplar Class 9 Maths book covers basics and fundamentals on all topics for students apart from the added information of a higher level. Hence, one of the factor of given polynomial is 3xy. (c) 3abc So, it may have degree 5. NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2. When we divide p1(z) by z – 3, then we get the remainder p,(3). ∴ a = -1, Question 2. 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. (v) Polynomial 3 = 3x° is a constant polynomial, because its degree is 0. 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