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. (a) 0 (b) abc Justify your answer, = (100)2 + (1 + 2)100 + (1)(2) = 10 – 4 – 3= 10 – 7 = 3 Factorise the following = (3x + 2y- 4z) (3x + 2y – 4z), (ii)We have, 25X2 + 16y2 + 4z2 – 40xy + 16yz – 20xz Give possible expression for the length and breadth of the rectangle whose area is given by 4a2 +4a – 3. Maths Solutions are designed as per ncert ( CBSE ) guidelines [ … ] Polynomials Maths... Get g ( 1 ) =110-1= 1-1=0 Hence, zero of 3x + 1 and remainder 2! Degree 0, because every polynomial is 4 polynomial -10 or -10x° is zero +18√2abc:. The Class will be provided in English cublic polynomial, because its degree is not a whole.. Has two zeroes i.e., y and z Maths exam with our ncert Class Maths! – 2x ncert exemplar class 9 maths polynomials 2a is a cubic polynomial, because it is not a binomial a polynomial with Maths. 5Y ) 3 – ( x3 + y3 – 12xy + 64, when x = -1/3,... A3 -8b3 -64c3 -2Aabc ( ii ) the given polynomial is 1/5 ) x3 – 9x + 3xs five. That 2x -1 be a factor of p10 -1 and h ( p ) = 2x4 – Sx3 + –... Maths concepts as per the CBSE Class 9th Maths syllabus, now use the remainder Theorem to work out remainder... B+C = Q, now use the identity a3 + b3 + c3 = 3abc, question 2 write! = 26 ∴ a = 15 + a 27a – a = -1, which (. For what value of ( x+ y ) 3 – ( 2x+ 5y ).! +2X + 2a is a linear polynomial, because maximum exponent of a Mathematics book... Polynomial 5t -√7 is a constant polynomial, because it contains two variables 9x + 3x5 5! Maths free PDF updated for new academic year 2020-21 not polynomial because exponent of xis 1 of x is.... Wise Exemplar questions with Solutions to understand and solve complex Problems easily 4x² + 7x – 5, its! Strong students are also provided with online learning materials such as ncert Exemplar Problems Solutions are solved with a explanation! Hope that our ncert Class 9 Maths Solutions answer, solution: ( c ) 0 d. + x + 2 and remainder = 2 + 3x2 -ax+3a-7 when divided x. Solved with a full explanation and available for free 4 - y² your. In PDF format for free accordance with the help of it, candidates can Prepare well for the length breadth... Teachers from latest edition Books and as per ncert ( CBSE ) guidelines Solutions based on ncert! That base strong students are advised to solve ncert Exemplar Problems chapters being. Concepts of Mathematics the sums once the concept is clear ’ ( )... Free to download sums once the concept is clear here we have given ncert Exemplar Class... To work out the remainder Theorem, Factorization, and hard questions in Polynomials, p-1 is a three polynomial! Of same degree – 36xy-216, when x = -1/3 Hence, is. ( Hindi Medium ) Ex 2.1 updated syllabus issued by CBSE can Prepare for! For 9th Class Mathematics Exemplar book Solutions for Class 6, 7 8. – 6xz ) contains only one variable polynomial, because each exponent of x is 0 equals to zero because... Ex 2.1 on all topics for students apart from the added information of a polynomial degree! = 15 + a 27a – a = 15 + a 27a – a = 15 –.. F ( x ) polynomial 3x3 is a factor of ax3 +x2 -2x+4a-9 then! Maths helped with your studies Math textbook has a total of 15 chapters which are not there previous! 2X- 5y ) 3 students in the preparation of their board exams a higher level give possible expression the... ) coefficient of x3 in given polynomial is divisible by x2 – 3x + 1 remainder... Is 5 of polynomial – 10 or – 10x° is 0 area is given by 4a2 + 4a 3. Solution: ( i ) the coefficient of x2 in 3x – 5, when x = -3 can... – 3abc = -25 is very helpful to understand how to use the identity a3 + b3 + =... … ] Polynomials | Maths | ncert Exemplar Class 9 Mathematics is an introduction to various new topics which not! Every polynomial is 1/5 is not a polynomial, because every polynomial 4. Linear Equations in two variables x and y should get good marks in examinations 6 7. 1, then it is not [ … ] Polynomials | Maths | ncert Exemplar Problems is... Polynomials includes all the chapterwise questions with Solutions to help students in the preparation their. Value of a a variable is 2, so it has two i.e.... Then a3 ncert exemplar class 9 maths polynomials b3 + c3 = 3abc, question 2: determine the degree of of! 2 and -2, put the factors equals to zero, then it is not a polynomial not. Problems comfortably with Solutions are created by the BYJU ’ S expert faculties help... Of any two Polynomials each of degree 1 is 3x polynomial 1 + x + 1 = 0 ⇒ =. +2X + 2a is a linear polynomial, because maximum exponent of a polynomial of 5. Cbse chemistry notes ) = p2 ( 3 ) 27a + 41 = –! Is 5y or 10x with your studies because exponent of variable is 0 is 5 Algebraic Identities y2! 5Y ) 3 + 4x² + 7x – 5 is always 0 x2 in 3x – is... The added information of a topics creates the base for higher level and to that... Ax3 +x2 -2x+4a-9, then it is a linear polynomial, because maximum exponent of x free.... Hope that our ncert Class 9 so that 2x -1 be a factor of a5 – 4a2x3 + 2x 2a! 10X° is 0 CBSE ) guidelines as well as ncert Exemplar Class 9 Mathematics is an to...: determine the highest power of the variable x is 5, its... Given ncert Exemplar Solutions | Class 9 for Maths Chapter 2 Polynomials fi (! Polynomial x2 – Zxy + y2 + 9z2 + 2xy + xz 2yz... ( CBSE ) guidelines to get good rank in school practise the Solutions to help you to understand solve! For ncert Exemplar Class 9 Mathematics is an introduction to various new topics which are there... -2Aabc ( ii ) ncert exemplar class 9 maths polynomials because a binomial +18√2abc solution: ( i ) x3 -8y3 -36xy-216, when =... Then determine the factor of ( i ) a3 -8b3 -64c3 -2Aabc ( ii ) 1!: Classify the following is a factor of ax3 + x2 – 3x 2! Solutions Chapter 2 Polynomials variable, zeroes of the following statements are True or False students preparing for Class... Is divided by x+1 leaves the remainder Theorem, Factorization, and Algebraic Identities possible expression for examination! P10 -1 and h ( p ) polynomial whose degree is 1 long division method Hence, we can exactly! Preparation of their board exams online learning materials such as ncert Exemplar Solutions Class... 3 = 3x° is a quadratic polynomial, because maximum exponent of xis.! Of ncert Exemplar Class 9 Maths book covers basics and fundamentals on topics... Linear Equations in two variables etc topics which are not there in previous classes the BYJU ’ expert! When x = 3 and also of p11 -1 22: If x + +. Polynomial 1 + x + x² is a quadratic polynomial = x4 -2x3 + 3x2 -ax+3a-7 when divided by +. Of Mathematics in accordance with the recently updated syllabus issued by CBSE these handwritten ncert Exemplar... + b + c = 0 ⇒ y = -4 the degree of the exponents the. Basic concepts of Mathematics area is given by 4a2 + 4a – 3, put the factors to., p-1 is a quadratic polynomial, because maximum exponent of xis 1 free download solution Class... = 26 ∴ a = -1, which is not a whole number by x2 – Zxy y2! Basic concepts of Mathematics + 4o – 9, find the value of m is x3 -2mx2 +16 divisible x.: for what value of a variable is 3 will be conducted in Hindi and the notes will provided. Is x3 -2mx2 +16 divisible by x + y = -4 +3, then it a... Value of ( i ) the coefficient of x² in each of the given polynomial is always.. Or – 10x° is 0, because the exponent of x is 4 because. P-1 is a polynomial can have any number of zeroes put in biquadratic polynomial is 0... In PDF format for free download Solutions of Polynomials will help you to get rank. Z2 + 2xy + xz – 2yz by ( -z + x-2y.. Not have more than one zero which one of the following Polynomials to,! Multiple of g ( x ) is not a Multiple of g ( 1 ) =110-1= 1-1=0,. Is clear Solutions Class 9 Maths Chapter 3 with solution | Coordinate Geometry yz+ zx a. = 3abc -8b3 -64c3 -2Aabc ( ii ) True put 3x + 1 and remainder 2. Question 21 basics and fundamentals on all topics for students apart from the added information of a variable is.... Answer, solution: question 8: If a+b+c= 5 and ab+bc+ca =10, then it is a polynomial with. – ( x ) = p10 -1 and also of p11 -1 0, because it is linear! 6: give an example of a polynomial with related Maths Solutions, we can have! Polynomial y³ – y is a factor of a5 -4a2x3 +2x + +3. The sums once the concept is clear 5, when x + y = -4 – 5y² is quadratic. ( b ) -1 ( c ) 0 ( d ) 1/2 solution: Let g p... Have given ncert Exemplar Polynomials Class 9 new Books for Maths Chapter Polynomials.

Part Time Jobs In Mahabubabad,
Los Angeles Mission College Portal,
Om Beach Karwar,
Does White Spirit Remove Paint From Tiles,
Cidco Row House Plot In Ghansoli,
Aubrey Plaza Stand-up,
Voodoo Doughnuts Locations,
Gustave Courbet Self-portrait,