POLYNOMIALS (A) Main Concepts and Results Meaning of a Polynomial Degree of a polynomial Coefficients Monomials, Binomials etc. Constant, Linear, Quadratic Polynomials etc. Value of a polynomial for a given value of the variable Zeroes of a polynomial Remainder theorem Factor theorem Factorisation of a quadratic polynomial by splitting the middle term Factorisation of algebraic expressions by using the Factor theorem Algebraic identities – (x + y)2 = x 2 + 2xy + y 2 (x – y)2 = x2 – 2xy + y2 x 2 – y 2 = (x + y) (x – y) (x + a) (x + b) = x 2 + (a + b) x + ab (x + y + z)2 = x 2 + y 2 + z 2 + 2xy + 2yz + 2zx 2 233(x + y)3 = x 3 + 3xy + 3xy + y = x + y 3 + 3xy (x + y) (x – y)3 = x3 – 3x2 y + 3xy2 – y3 = x3 – y3 – 3xy (x – y) x 3 + y 3 = (x + y) (x 2 – xy + y 2) x 3 – y 3 = (x – y) (x 2 + xy + y 2) x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx) (B) Multiple Choice Questions Sample Question 1 : If x 2 + kx + 6 = (x + 2) (x + 3) for all x, then the value of k is (A)1 (B)–1 (C)5 (D)3 Solution : Answer (C) EXERCISE 2.1 Write the correct answer in each of the following : 1. Which one of the following is a polynomial? x 22 (A) – (B)21 2 x 3 2 x − 1 (C) x 2 + 3x (D) x + 1 x 2. 2 is a polynomial of degree 1 (A)2 (B)0 (C)1 (D) 2 3. Degree of the polynomial 4x 4 + 0x 3 + 0x 5 + 5x + 7 is (A)4 (B)5 (C)3 (D)7 4. Degree of the zero polynomial is (A) 0 (B) 1 (C) Any natural number (D) Not defined 5. If p ( x)= x 2–2 2x + 1, then p(22) is equal to (A)0 (B)1 (C) 42 (D)82 + 1 6. The value of the polynomial 5x – 4x2 + 3, when x = –1 is (A)– 6 (B)6 (C)2 (D)–2 7. If p(x) = x + 3, then p(x) + p(–x) is equal to (A)3 (B)2x 8. Zero of the zero polynomial is (A) 0 (C) Any real number 9. Zero of the polynomial p(x) = 2x + 5 is 25 ––(A) (B)52 10. One of the zeroes of the polynomial 2x21 (A) 2 (B) 2 (C) 0 (D) 6 (B) 1 (D) Not defined 2 (C) 5 + 7x –4 is 1 (C) – 2 11. If x 51 + 51 is divided by x + 1, the remainder is (A)0 (B)1 (C)49 5 (D) (D) –2 (D) 50 2 12. If x + 1 is a factor of the polynomial 2x 2 + kx, then the value of k is (A)–3 (B)4 (C)2 (D)–2 13. x + 1 is a factor of the polynomial (A) x 3 + x 2 – x + 1 (C) x 4 + x 3 + x 2 + 1 14. One of the factors of (25x (A) 5 + x (B) (B) x 3 + x 2 + x + 1 (D) x 4 + 3x 3 + 3x 2 + x + 1 2 – 1) + (1 + 5x)2 is 5 – x (C) 5x – 1 (D) 10x 15. The value of 2492 – 2482 is (A) 12 (B) 477 (C) 487 (D) 497 16. The factorisation of 4x 2 + 8x + 3 is (A) (x + 1) (x + 3) (B) (2x + 1) (2x + 3) (C) (2x + 2) (2x + 5) (D) (2x –1) (2x –3) 17. Which of the following is a factor of (x + y)3 – (x3 + y3)? (A) x 2 + y 2 + 2xy (B) x 2 + y 2 – xy (C) xy 2 (D) 3xy 18. The coefficient of x in the expansion of (x + 3)3 is (A) 1 (B) 9 (C) 18 (D) 27 19. If x y + –1y x = ( ), 0x y ≠ , the value of x3 – y3 is 1 (A)1 (B)–1 (C)0 (D) 2 ⎛ 1 ⎞⎛ 1 ⎞ 20. If 49x2 – b = ⎜ 7x + ⎟⎜ 7x – ⎟ , then the value of b is ⎝ 2 ⎠⎝ 2 ⎠ 11 1 (A) 0 (B) (C) (D)24 2 21. If a + b + c = 0, then a 3 + b3 + c 3 is equal to (A) 0 (B) abc (C) 3abc (D) 2abc (C) Short Answer Questions with Reasoning Sample Question 1 : Write whether the following statements are True or False. Justify your answer. 3 1 1 6 +x2 (i) x 2 +1 is a polynomial (ii) is a polynomial, x ≠ 0 5 x Solution : (i) False, because the exponent of the variable is not a whole number. 3 6 + x2 (ii) True, because = 6 + x , which is a polynomial. x EXERCISE 2.2 1. Which of the following expressions are polynomials? Justify your answer: 2(i)8 (ii) 3x –2 x (iii) 1– 5x 1 ( x –2 )( x –4 ) 1 (iv) –2 + 5x +7 (v) (vi)5x xx + 1 1322 1 a – a +4a – 7 (viii)73 2x (vii) 2. Write whether the following statements are True or False. Justify your answer. (i) A binomial can have atmost two terms (ii) Every polynomial is a binomial (iii) A binomial may have degree 5 (iv) Zero of a polynomial is always 0 (v) A polynomial cannot have more than one zero (vi) The degree of the sum of two polynomials each of degree 5 is always 5. (D) Short Answer Questions Sample Question 1 : (i) Check whether p(x) is a multiple of g(x) or not, where p(x) = x 3 – x + 1, g(x) = 2 – 3x (ii) Check whether g(x) is a factor of p(x) or not, where x 1 p(x) = 8x 3 – 6x 2 – 4x + 3, g(x) = − 3 4 Solution : (i) p(x) will be a multiple of g(x) if g(x) divides p(x). 2 Now, g(x) = 2 – 3x = 0 gives x = 3 ⎛ 2 ⎞⎛ 2 ⎞3 ⎛ 2 ⎞ Remainder = p ⎜⎟ = ⎜⎟ − ⎜⎟ + 1 ⎝ 3 ⎠⎝ 3 ⎠⎝ 3 ⎠ 82 17 −+1= = 27 3 27 Since remainder ≠ 0, so, p(x) is not a multiple of g(x). x 13 (ii) g(x) = −=0 gives x = 34 4 ⎛ 3 ⎞ g(x) will be a factor of p(x) if p ⎜⎟ = 0 (Factor theorem)⎝ 4 ⎠ ⎛ 3 ⎞⎛ 3 ⎞3 ⎛ 3 ⎞2 ⎛ 3 ⎞ Now, p ⎜⎟ = 8⎜⎟ − 6 ⎜⎟ − 4⎜⎟ + 3 ⎝ 4 ⎠⎝ 4 ⎠⎝ 4 ⎠⎝ 4 ⎠ 27 9 = 8 ×− 6 ×− 3 +3 = 0 64 16 ⎛ 3 ⎞ Since, p ⎜⎟ = 0, so, g(x) is a factor of p(x).⎝ 4 ⎠ Sample Question 2 : Find the value of a, if x – a is a factor of x 3 – ax 2 + 2x + a – 1. Solution : Let p(x) = x3 – ax2 + 2x + a – 1 Since x – a is a factor of p(x), so p(a) = 0. i.e., a 3 – a(a)2 + 2a + a – 1 = 0 a 3 – a 3 + 2a + a – 1 = 0 3a = 1 1 Therefore, a = 3 Sample Question 3 : (i)Without actually calculating the cubes, find the value of 483 – 303 – 183. (ii)Without finding the cubes, factorise (x – y)3 + (y – z)3 + (z – x)3. Solution : We know that x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx). If x + y + z = 0, then x 3 + y 3 + z 3 – 3xyz = 0 or x 3 + y 3 + z 3 = 3xyz. (i) We have to find the value of 483 – 303 – 183 = 483 + (–30)3 + (–18)3. Here, 48 + (–30) + (–18) = 0 So, 483 + (–30)3 + (–18)3 = 3 × 48 × (–30) × (–18) = 77760 (ii) Here, (x – y) + (y – z) + (z – x) = 0 Therefore, (x – y)3 + (y – z)3 + (z – x)3 = 3(x – y) (y – z) (z – x). EXERCISE 2.3 1. Classify the following polynomials as polynomials in one variable, two variables etc. (i) x 2 + x + 1 (ii) y 3 – 5y (iii) xy + yz + zx (iv) x2 – 2xy + y2 + 1 2. Determine the degree of each of the following polynomials : (i) 2x – 1 (ii) –10 (iii) x 3 – 9x + 3x 5 (iv) y 3 (1 – y 4) 3. For the polynomial x 3 + 2 x + 17 2 6 – x – x , write 52 (i) the degree of the polynomial (ii) the coefficient of x3 (iii) the coefficient of x6 (iv) the constant term 4. Write the coefficient of x 2 in each of the following : π 2 (i) x + x –1 (ii) 3x – 5 6 (iii) (x –1) (3x –4) (iv) (2x –5) (2x2 – 3x + 1) 5. Classify the following as a constant, linear, quadratic and cubic polynomials : (i) 2 – x 2 + x 3 (ii) 3x 3 (iii) 5t – 7 (iv) 4 – 5y 2 (v) 3 (vi) 2 + x (vii) y3 – y (viii) 1 + x + x2 (ix) t2 (x) 2x –1 6. Give an example of a polynomial, which is : (i) monomial of degree 1 (ii) binomial of degree 20 (iii) trinomial of degree 2 7. Find the value of the polynomial 3x3 – 4x2 + 7x – 5, when x = 3 and also when x = –3. 8. If p(x) = x 2 – 4x + 3, evaluate : p(2) – p(–1) + p ⎜⎟⎛ 1 ⎞ ⎝ 2 ⎠ 9. Find p(0),p(1),p(–2) for the following polynomials : (i) p(x) = 10x – 4x 2 – 3 (ii) p(y) = (y + 2) (y – 2) 10. Verify whether the following are True or False : (i) –3 is a zero of x – 3 (ii) – is a zero of 3x + 1 1 3–4 (iii) is a zero of 4 –5y5(iv) 0 and 2 are the zeroes of t2 – 2t (v) –3 is a zero of y2 + y – 6 11. Find the zeroes of the polynomial in each of the following : (i) p(x) = x – 4 (ii) g(x) = 3 – 6x (iii) q(x) = 2x –7 (iv) h(y) = 2y 12. Find the zeroes of the polynomial : p(x) = (x – 2)2 – (x + 2)2 13. By actual division, find the quotient and the remainder when the first polynomial is divided by the second polynomial : x 4 + 1; x –1 14. By Remainder Theorem find the remainder, when p(x) is divided by g(x), where (i) p(x) = x3 – 2x2 – 4x – 1, g(x) = x + 1 (ii) p(x) = x3 – 3x2 + 4x + 50, g(x) = x – 3 (iii) p(x) = 4x 3 – 12x 2 + 14x – 3, g(x) = 2x – 1 3 (iv) p(x) = x 3 – 6x 2 + 2x – 4, g(x) = 1 – x 2 15. Check whether p(x) is a multiple of g(x) or not : (i) p(x) = x3 – 5x2 + 4x – 3, g(x) = x – 2 (ii) p(x) = 2x3 – 11x2 – 4x + 5, g(x) = 2x + 1 16. Show that : (i) x + 3 is a factor of 69 + 11x – x 2 + x 3 . (ii) 2x – 3 is a factor of x + 2x 3 – 9x 2 + 12 . 17. Determine which of the following polynomials has x – 2 a factor : (i) 3x2 + 6x – 24 (ii) 4x2 + x – 2 18. Show that p – 1 is a factor of p 10 – 1 and also of p 11 – 1. 19. For what value of m is x 3 – 2mx 2 + 16 divisible by x + 2 ? 20. If x + 2a is a factor of x 5 – 4a 2 x 3 + 2x + 2a + 3, find a. 21. Find the value of m so that 2x – 1 be a factor of 8x4 + 4x3 – 16x2 + 10x + m. 22. If x + 1 is a factor of ax 3 + x 2 – 2x + 4a – 9, find the value of a. 23. Factorise : (i) x2 + 9x + 18 (ii) 6x2 + 7x – 3 (iii) 2x2 – 7x – 15 (iv) 84 – 2r – 2r2 24. Factorise : (i) 2x 3 – 3x 2 – 17x + 30 (ii) x 3 – 6x 2 + 11x – 6 (iii) x 3 + x 2 – 4x – 4 (iv) 3x 3 – x 2 – 3x + 1 25. Using suitable identity, evaluate the following: (i) 1033 (ii) 101 × 102 (iii) 9992 26. Factorise the following: (i) 4x 2 + 20x + 25 (ii) 9y 2 – 66yz + 121z 2 ⎛ 1 ⎞ 2 ⎛ 1 ⎞ 2 (iii) ⎜ 2x + ⎟ − ⎜ x − ⎟⎝ 3 ⎠⎝ 2 ⎠ 27. Factorise the following : (i) 9x 2 – 12x + 3 (ii) 9x 2 – 12x + 4 28. Expand the following : (i) (4a – b + 2c)2 (ii) (3a – 5b – c)2 (iii) (– x + 2y – 3z)2 29. Factorise the following : (i) 9x2 + 4y2 + 16z2 + 12xy – 16yz – 24xz (ii) 25x2 + 16y2 + 4z2 – 40xy + 16yz – 20xz (iii) 16x 2 + 4y 2 + 9z 2 – 16xy – 12yz + 24 xz 30. If a + b + c = 9 and ab + bc + ca = 26, find a 2 + b2 + c 2. 31. Expand the following :⎛ 1 y ⎞3 ⎛ 1 ⎞ 3 (i) (3a – 2b)3 (ii) ⎜ + ⎟ (iii) ⎜ 4– ⎟⎝ x 3 ⎠⎝ 3x ⎠ 32. Factorise the following : (i) 1 – 64a 3 – 12a + 48a 2 12 13 26 (ii) 8 p + p + p + 5 25 125 33. Find the following products : ⎛ x ⎞⎛ x 22 ⎞ y(i) ⎜ + 2 y⎟⎜ – xy + 4 ⎟ (ii) (x2 – 1) (x4 + x2 + 1)⎝ 2 ⎠⎝ 4 ⎠ 34. Factorise : 3(i) 1 + 64x 3 (ii) a –2 2b3 35. Find the following product : (2x – y + 3z) (4x 2 + y 2 + 9z 2 + 2xy + 3yz – 6xz) 36. Factorise : (i) a 3 – 8b3 – 64c 3 – 24abc (ii) 22 a 3 + 8b3 – 27c 3 + 18 2 abc. 37. Without actually calculating the cubes, find the value of : 33 3⎛ 1 ⎞⎛ 1 ⎞ ⎛ 5 ⎞ (i) ⎜⎟ + ⎜⎟ – ⎜⎟ (ii) (0.2)3 – (0.3)3 + (0.1)3 ⎝ 2 ⎠⎝ 3 ⎠ ⎝ 6 ⎠ 38. Without finding the cubes, factorise (x – 2y)3 + (2y – 3z)3 + (3z – x)3 39. Find the value of (i) x 3 + y 3 – 12xy + 64, when x + y = –4 (ii) x 3 – 8y 3 – 36xy – 216, when x = 2y + 6 40. Give possible expressions for the length and breadth of the rectangle whose area is given by 4a2 + 4a –3. (E) Long Answer Questions Sample Question 1 : If x + y = 12 and xy = 27, find the value of x 3 + y 3. Solution : x3 + y3 =(x + y) (x2 – xy + y2) =(x + y) [(x + y)2 – 3xy] = 12 [122 – 3 × 27] = 12 × 63 = 756 Alternative Solution : x 3 + y 3 =(x + y)3 – 3xy (x + y) = 123 – 3 × 27 × 12 = 12 [122 – 3 × 27] = 12 × 63 = 756 EXERCISE 2.4 1. If the polynomials az3 + 4z2 + 3z – 4 and z3 – 4z + a leave the same remainder when divided by z – 3, find the value of a. 2. The polynomial p(x) = x 4 – 2x 3 + 3x 2 – ax + 3a – 7 when divided by x + 1 leaves the remainder 19. Find the values of a. Also find the remainder when p(x) is divided by x + 2. 3. If both x – 2 and x – are factors of px 2 + 5x + r, show that p = r. 1 24. Without actual division, prove that 2x 4 – 5x 3 + 2x 2 – x + 2 is divisible by x 2 – 3x + 2. [Hint: Factorise x 2 – 3x + 2] 5. Simplify (2x – 5y)3 – (2x + 5y)3. 6. Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (– z + x – 2y). 22 2 ab c 7. If a, b, c are all non-zero and a + b + c = 0, prove that ++ = 3. bc ca ab 8. If a + b + c = 5 and ab + bc + ca = 10, then prove that a 3 + b3 + c 3 –3abc = – 25. 9. Prove that (a + b + c)3 – a 3– b3– c 3 = 3(a + b ) (b + c) (c + a).