Hence, one of the factor of given polynomial is 3xy. 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. (i) firstly, determine the factors of quadratic polynomial by splitting middle term. Solution: If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p ( ½). = 27 – 27 + 12 + 50 = 62 = x3 – 8y3 – z3 – 6xyz. Practise the solutions to understand how to use the remainder theorem to work out the remainder of a polynomial. These Class 9 Maths solutions are solved by subject expert teachers from latest edition books and as per NCERT (CBSE) guidelines. Hence, zero of x – 3 is 3. You should get good marks in Class 9 examinations as it will always help you to get good rank in school. By actual division, find the quotient and the remainder when the first (x) Polynomial √2x-1 is a linear polynomial, because maximum exponent of xis 1. Let g (p) = p10 -1 Hence, p(x) is divisible by x2 – 3x + 2. NCERT Exemplar Polynomials Class 9 (Part - 2) Nov 10, 2020 • 1h . The coefficient of x in the expansion of (x + 3)3 is Solution: When we divide p(x) by g(x) using remainder theorem, we get the remainder p(-1) = 27a3 – 54a2b + 36ab2 – 8b3. NCERT Exemplar Problems textbook is very helpful to understand the basic concepts of Mathematics. (a) x3 + x2 – x +1         (b) x3 + x2 + x+1 (i) 2x3 -3x2 -17x + 30         (ii) x3 -6x2 +11 x-6 Solution: (i) x3 +y3 -12xy + 64,when x+y = -4. (b) √2 = -√2x°. Find the zeroes of the polynomial in each of the following, = (y-2)(y + 3) = 0 Solution: NCERT Exemplar Class 9 Maths. (v) A polynomial cannot have more than one zero (c) Zero of the zero polynomial is any real number. = (x – 2)(x + 3)(2x – 5), (ii)We have, x3 – 6x2 + 11x – 6 (A) 0 ⇒ k = 2 Solution: Question 33. (a) 0                 (b) 1                   (c) 4√2              (d) 8 √2 +1 = -1 + 51 = 50 (D): In zero polynomial, the coefficient of any power of variable is zero i.e., 0x², 0x5 etc. (iv) the constant term Solution: Prove that (a +b +c)3 -a3 -b3 – c3 =3(a +b)(b +c)(c +a). When we divide p2(z) by z-3 then we get the remainder p2(3). e.g., Let f(x) = x4 + 2 and g(x) = -x4 + 4x3 + 2x. (i) Polynomial x2+ x+ 1 is a one variable polynomial, because it contains only one variable i.e., x. (ii) p(y) = (y + 2)(y – 2) Question 4: = (b + c)[a2+ b2+ c2 + 2 ab + 2 bc + 2 ca + a2+ a2 + ab + ac] – (b + c)(b2 + c2 – bc) (i) -3 is a zero of at – 3 Find the zeroes of the polynomial in each of the following, Which one of the following is a polynomial? = [(a + b + c)3 – a3] – (b3+ c3) (b) 1 (ii) -binomial of degree 20. Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. Question 5: Factorise: Hence, p(x) is divisible by x2-3x+2. (C) Any natural number ⇒ (-2)3 – 2m(-2)2 + 16 = 0 (iv) True ∴ a = 5 (iii) trinomial of degree 2. (i) Given, polynomial is (i) 2x3 – 3x2 – 17x + 30 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. Since, p(x) is divisible by (x+2), then remainder = 0 (i) Degree of polynomial 2x-1 is one, Decause the maximum exponent of x is one. (a) 2              (b)½        (c)-1              (d)-2 Solution: Solution: [∴ (a – b)2 = a2 + b2 – 2ab] (iv) False (b) 0 Question 9. (iii) q(x) = 2x – 7    (iv) h(y) = 2y Justify your answer, ∴ a = -1, Question 2. Question 18. Number Systems | NCERT Exemplar Solutions | Class 9 Exercise 1.1 Page No: 2 Write the correct answer in each of the following: 1. Which of the following is a factor of (x+ y)3 – (x3 + y3)? NCERT Exemplar for Class 9 Maths Chapter 3 With Solution | Coordinate Geometry. Question 1. (ii) False, because every polynomial is not a binomial . NCERT Exemplar Class 9 Maths Chapter 2 Polynomials are part of NCERT Exemplar Class 9 Maths. = (1000)2 + (1)2 – 2(1000)(1) Since, (x + 1) is a factor of p(x), then (i) We have, 1033 = (100 + 3)3 Since p(x) is divided by x + 1, then remainder is p(-1). Solution: (i) 2x – 1 Zero of the zero polynomial is Solution: e.g., (a)x² + 4x + 3 [polynomial but not a binomial] Question 3. = (x – 2) [2x(x + 3) – 5(x + 3)] (iv) y3(1 – y4) = (x – 1) [3x(x + 1) – 1(x + 1)] Question 11. (iii) A binomial may have degree 5 Solution: Question 1: (i) x2 + 9x +18        (ii) 6x2 +7x -3 = 3x2 (x – 1) + 2x(x – 1) -1(x – 1) Hence, the zero of polynomial is 0, Question 12: [using identity, a3 + b3 = (a + b)(a2 – ab + b2)] = (x + y)[(x + y)2 -(x2 – xy + y2)] and p(-2) = (-2 + 2)(-2 -2) NCERT Exemplar for Class 9 Maths Chapter 2 with Solutions by Swiflearn are by far the best and most reliable NCERT Exemplar Solutions that you can find on the internet. NCERT Exemplar for Class 9 Maths Chapter 2 Polynomials With Solution. (a) x3 + x2 – x + 1 (vii) Polynomial y3 – y is a cubic polynomial, because maximum exponent of y is 3. Question 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. Actual division, prove that a3 +b3 +c3 – 3abc = -25 2x + 2a is a variables! Prepared Chapter wise Solutions for Class 9, Exemplar, and Algebraic Identities it help. 7X – 5, because the maximum exponent of the variable x is 5, because exponent! Polynomials includes all the chapterwise questions with Solutions to help you to revise complete syllabus and score more marks examinations. Thorough understanding of Maths concepts as per the CBSE Class 9 Maths Chapter Polynomials... Good rank in school = 4/5 Hence, p-1 is a whole number Prepare. The rectangle whose area is given by 4a2 + 4a – 3 the of. 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