(i) The example of monomial of degree 1 is 5y or 10x. Since, remainder ≠ 0, then p(x) is not a multiple of g(x). = 3 x (-1) = -3 = 1000000 + 1 – 2000 = 998001, Question 27. Let p(x) = x3 -2mx2 +16 (iv) the constant term (ii) Coefficient of x2 in 3x – 5 is 0. Question 10: Since 2x – 1 is afactor of p(x) then p(1/2) = 0, Question 22. Solution: Hence, the value of the given polynomial at x = 3 and x = -3 are 61 and -143, respectively. = 2x(x+ 4)-1(x+ 4) = (- 5x + 4y + 2z)2 = 3(a + b)(b + c)(c + a) = R.H.S. and p(-2) =10 (-2) -4 (-2)2 – 3 Therefore, (x + 3) is a factor of p(x). = (x – 1)(3x2 + 2x – 1) 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. [∴ a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2] On putting x = -1 in Eq. Here students are also provided with online learning materials such as NCERT Exemplar Class 9 Maths Solutions. = (x + y) (3xy) Check whether p(x) is a multiple of g(x) or not If a + b + c = 0, then a3 + b3 + c3 = 3abc, Question 39. Therefore, the degree of the given polynomial is 4. (c) 2 (a) 0 p(-1)=0 If a+b+c= 5 and ab+bc+ca =10, then prove that a3 +b3 +c3  – 3abc = -25. = 8 – 20 + 8 – 3 = – 7 We hope the NCERT Exemplar Class 9 Maths Chapter 2 Polynomials will help you. p(x) = x- 4 (ii) Given, polynomial is (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). Without actually calculating the cubes, find the value (iv) False Solution: Question 2: √2 is a polynomial of degree (a) 2 (b) 0 (c) 1 (d)½ Solution: (b) √2 = -√2x°. Because each exponent of the variable a is a whole number. All the chapterwise questions with solutions to help you to revise complete CBSE syllabus and score more marks in Your board examinations. Given, polynomial is p(x) = (x – 2)2 – (x + 2)2 = (x+ y)(x2+ y2+ 2xy- x2+ xy- y2) If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p ( ½). Question 5. and h(p) = p11 -1  …(2) By remainder theorem, find the remainder when p(x) is divided by g(x) = 2a(2a + 3) -1 (2a + 3) = (2a – 1)(2a + 3) Also, if a + b + c = 0, then a3 + b3 + c3 = 3abc (ii) Given, polynomial is We have, a + b + c = 9 (i) a3 -8b3 -64c3 -2Aabc and p(2) = 2(2)4 – 5(2)3 + 2(2)2 – 2 + 2 = 2x16-5x8+2x4+ 0 = 32 – 40 + 8 = 40 – 40 =0 Here are all questions are solved with a full explanation and available for free to download. Classify the following as a constant, linear, quadratic and cubic polynomials (i), we get (ii) x3 – 8y3 – 36xy-216,when x = 2y + 6. Question 17. (d)-2 Question 12. Justify your answer, (iv) 84 – 2r – 2r2 (ii) x3 -8y3 -36xy-216,when x = 2y + 6. (iii) p(x) = x3 – 12x2 + 14x -3, g(x)= 2x – 1 – 1 Therefore, (x-2y)3 + (2y-3z)3 + (3z-x)3 = 3(x-2y)(2y-3z)(3z-x). Question 2: (iii) 5t – √7 Solution: Solution: => 8=8 m => x = ½ and x = -4 Question 21. ∴ 2y = 0 ⇒ y = 0 = (x – 2) (2x2 + 6x – 5x – 15) x + 3 is a factor of p(x) if p(-3) = 0 ⇒ -5/2 NCERT solutions are really helpful when it comes to a complicated subject like Mathematics. (ii) The example of binomial of degree 20 is 6x20 + x11 or x20 +1 For zeroes of polynomial, put p(x) = 0 Let p(x) =a5 – 4a2x3 + 2x + 2a + 3 You can also Download NCERT Solutions for class 9 Maths in Hindi to help you to revise complete Syllabus and score more marks in your examinations. One of the zeroes of the polynomial 2x2 + 7x – 4 is The highest power of … Without actually calculating the cubes, find the value of 36xy-36xy = 0 = (x – 1) (3x2 + 3x – x – 1) (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)). Question 3. If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p ( ½). Solution: = x2(x + 1) – 4(x + 1) Since, remainder ≠ 0, then p(x) is not a multiple of g(x). (b) -5/2 (a) 12 (iii) trinomial of degree 2. Practice Polynomials questions and become a master of concepts. = (x – 2) [2x(x + 3) – 5(x + 3)] Zero of the polynomial p(x) = 2x + 5 is 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 Question 12. Solution: p(-1) = 5(-1) -4(-1)2 + 3= -5 – 4 + 3 = -6, Question 7. (ii) False, because every polynomial is not a binomial . ⇒ x3 + y3 + 64 = 12xy (ix) Polynomial t² is a quadratic polynomial, because its degree is 2. = 27 – 27 + 12 + 50 = 62 Because a binomial has exactly two terms. Polynomials | Maths | NCERT Exemplar Solutions | Class 9. (b) 6 = x2 + 4y2 + 9z2 – 4xy – 12yz + 6xz, Question 29. Now, p2(3) = (3)3-4(3)+a Find the following products: The polynomial p{x) = x4 – 2x3 + 3x2 – ax + 3a – 7 when divided by x + 1 leaves the remainder 19. (i) p(x) = 10x – 4x2 – 3 = x3 – 8y3 – z3 – 6xyz. (d) 6 Solution: Solution: Question 30: Question 15. Visit FlexiPrep for more files and information on Subject-Wise-NCERT-Books-PDF: Mathematics (i) 2x – 1 Solution: Hence, the remainder is 50. (b) x² + 5 [polynomial and also a binomial]. Degree of the zero polynomial is if a + b+c = Q, now use the identity a3 + b3 + c3 = 3abc. (b) x2 + y2 – xy then (5)2 = a2 + b2+ c2 + 2(10) Exercise 2.3: Short Answer Type Questions. (x) Polynomial √2x – 1 is a linear polynomial, because its degree is 1. g(x) = 3 – 6x Hence, one of the factor of given polynomial is 3xy. (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx (a)-6          (b) 6           (c) 2                    (d) -2 (b) 4 (viii) Polynomial 1 + x+ x2 is a quadratic polynomial, because maximum exponent of xis 2. = (x – 1) [x(x – 3) – 2(x – 3)] Simplify (2x- 5y)3 – (2x+ 5y)3. (viii) 1 + x + x² (iv) The constant term in given polynomial is 1/5. Find the value of the polynomial 3x3 – 4x2 + 7x – 5, when x = 3 and also when x = -3. Solution: Question 9. (iv) x2 – Zxy + y2 + 1 = -1 + 51 = 50 (i) x2 + x +1           (ii) y3 – 5y Download NCERT Solutions for Class 9 Maths Free PDF updated for 2020 - 21. [using identity, (a + b)2 = a2 + b2 + 2 ab)] ∴ P( 3) = 61 (-1)3 + (-1)2 + (-1) + 1 = 0 => -1+1-1 + 1 = 0 => 0 = 0 Hence, our assumption is true. Solution: Question 32: = x2 (- z + x – 2y) + 4y2(- z + x – 2y) + z2(- z + x – 2y) + 2xy(- z + x – 2y) + xz(- z + x – 2y) – 2yz (- z + x – 2y) Hence, one of the factor of given polynomial is 10x. (i) -3 is a zero of at – 3 Put 3x + 1 = 0 ⇒ x = -1/3 NCERT Class 9 New Books for Maths Chapter 2 Polynomials includes all the questions given in CBSE syllabus. X+ x2 is a polynomial, because every polynomial is 4 x² + x + 2 y ) –! Exam pattern 9, find the value of m is x3 -2mx2 +16 divisible by x + 2 nd 2x. 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