Question 1: The graphs of y = p(x) are given in following figure, for some Polynomials p(x). Find the number of zeroes of p(x), in each case.

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1 Class X - NCERT Maths EXERCISE NO:.1 Question 1: The graphs of y = p(x) are given in following figure, for some Polynomials p(x). Find the number of zeroes of p(x), in each case. (i) (ii) (iii) (iv) (v) (vi) Solution 1: (i) The number of zeroes is 0 as the graph does not cut the x-axis at any point. (ii) The number of zeroes is 1 as the graph intersects the x-axis at only 1 point. (iii) The number of zeroes is 3 as the graph intersects the x-axis at 3 points. (iv) The number of zeroes is as the graph intersects the x-axis at points. (v) The number of zeroes is 4 as the graph intersects the x-axis at 4 points. (vi) The number of zeroes is 3 as the graph intersects the x-axis at 3 points. 0. Polynomials 1

2 EXERCISE NO:. Question 1: Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. (i) x x 8 (ii) 4s 4s 1 (iii) 6x 37x (iv) 4u 8u (v) t 15 (vi) 3x x 4 Solution 1: x x 8 x 4 x (i) The value of x = 4 or x = Therefore, the zeroes of x 8 is zero when x 4 = 0 or x + = 0, i.e., when x x x 8are 4 and. Coefficient of x Sum of zeroes = 4 1 Coefficient of x 8 Constant term Product of zeroes 4x 8 1 Coefficient of x 4s 4s 1 s 1 (ii) The value of 4s 4s + 1 is zero when s 1 = 0, i.e., Therefore, the zeroes of 4s 4s + 1 are 1 and 1. Sum of zeroes = 4 Coefficient of s Coefficient of s 1 s Product of zeroes Constant term 4 Coefficient of s (iii) 6x 3 7x 6x 7x 3 3x 1x 3 The value of 6x 3 7x is zero when 3x + 1 = 0 or x 3 = 0, i.e., 1 3 x or x 3 Therefore, the zeroes of 6x x are and 3 0. Polynomials

3 Coefficient of x Sum of zeroes = Coefficient of x Constant term Product of zeroes = = 3 6 Coefficient of x (iv) 4u 8u 4u 8u 0 4uu The value of 4u + 8u is zero when 4u = 0 or u + = 0, i.e., u = 0 or u = Therefore, the zeroes of 4u + 8u are 0 and. 8 Coefficient of u Sum of zeroes = 0 4 Coefficient of u 0 Constant term Product of zeroes = 0 0 = 4 Coefficient of u (v) t 15 t 0t 15 t 15t 15 The value of t 15 is zero when t 15 = 0 or t 15 = 0, i.e., when t 15 or t 15 Therefore, the zeroes of t 15 are and 15 and Sum of zeroes = Coefficient of t 1 Coefficient of t 15 Constant term Product of zeroes = Coefficient of x (vi) 3x x 4 The value of 3x x 4 is zero when 3x 4 = 0 or x + 1 = 0, i.e., 4 when x or x = 1 3 Therefore, the zeroes of 3x x 4 are 4 and Polynomials 3

4 4 1 1 Sum of zeroes = 1 Coefficient of x Product of zeroes Coefficient of x 4 4 Constant term Coefficient of x Question : Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. (i) 1, 1 4 (ii) 1 1 1, (iii) 0, 5 (iv) 1, 1 (v), (vi) 4, 1 Solution : (i) 1, 1 4 Let the polynomial be 1 b 4 a 4 c 1 ax bx c, and its zeroes be and. 4 a If a = 4, then b = 1, c = -4 Therefore, the quadratic polynomial is 4x x 4. (ii) 1, 3 Let the polynomial be ax bx c, and its zeroes be and. 3 b 3 a 1 c 3 a If a = 3, then b = 3, c = 1 Therefore, the quadratic polynomial is 3x 3 x + 1. (iii) 0, 5 Let the polynomial be ax bx c, and its zeroes be and. 0. Polynomials 4

5 0 b 0 1 a 5 c 5 1 a If a = 1, then b = 0, c = 5 Therefore, the quadratic polynomial is (iv) 1, 1 Let the polynomial be 1 b 1 1 a 1 c 1 1 a If a = 1, then b = -1, c = 1 Therefore, the quadratic polynomial is 1 1 (v), 4 4 Let the polynomial be 1 b 4 a 1 c 4 a If a = 4, then b = 1, c = 1 Therefore, the quadratic polynomial is (vi) 4, 1 Let the polynomial be ax bx c. 4 b 4 1 a 1 c 1 1 a If a = 1, then b = 4, c =1 Therefore, the quadratic polynomial is x 5. ax bx c, and its zeroes be and. x x 1. ax bx c, and its zeroes be and. 4x x 1. x 4x Polynomials 5

6 EXERCISE NO:.3 Question 1: Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following: g x x 3 (i) px x 3x 5x 3, 4 (ii) px x 3x 4x 5, 4 (iii) px x 5x 6, Solution 1: (i) 3 p x x 3x 5x 3, g x x x 3 3 x x 3x 5x 3 x 3 x 3x 7x 3 3x 6 7x 9 Quotient = x 3 Remainder = 7x 9 g x x 1 x g x x (ii) px x 3x 4x 5 x 0.x 3x 4x 5 gx x 1 x x x 1 0. Polynomials 6

7 Quotient = x x 3 Remainder = 8 (iii) p x x 5x 6x 0.x 5x q x x x x x 4 x 0.x 5x 6 x x 4 x 5x 6 x 4 5x 10 Quotient = x Remainder = 5x Polynomials 7

8 Question : Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial: 4 3 (i) t 3,t 3t t 9t l 4 3 (ii) x 3x 1,3x 5x 7x x (iii) 5 3 x 3x 1,x 4x x 3x 1 Solution : 4 3 (i) t 3,t 3t t 9t l t 3 t 0.t 3 t 3t t 0.t 3 t 3t t 9t 1 t 0.t 6t t 4t 9t 1 3 3t 0.t 9t 4t 0.t 1 4t 0.t 1 0 Since the remainder is 0, 4 3 Hence, t 3 is a factor of t 3t t 9t 1 (ii) 4 3 x 3x 1, 3x 5x 7x x 0. Polynomials 8

9 Since the remainder is 0, 4 3 Hence, x 3x 1 is a factor of3x 5x 7x x (iii) 5 3 x 3x 1,x 4x x 3x 1 Since the remainder 0, 5 3 x 3x 1,x 4x x 3x 1 Question 3: 4 3 Obtain all other zeroes of3x 6x x 10x 5, if two of its zeroes are 5 5 and Polynomials 9

10 Solution 3: p x 3x 6x x 10x Since the two zeroes are and x x x is a factor of Therefore, we divide the given polynomial by 3x 6x x 0.x 3x 6x x 10x 5 3 3x 4 0.x 3 5x 3 6x 3x 10x 5 3 6x 0x 10x 3x 0x 5 3x 0x 5 0 x x 6x x 10x 5 x 3x 6x x x x 1 3 We factorize x x 1 x1 Therefore, its zero is given by x + 1 = 0 x = x 6x x 10x Polynomials 10

11 As it has the termx 1, therefore, there will be zeroes at x = 1. Hence, the zeroes of the given polynomial are 5, 5, 1 and Question 4: 3 On dividing x 3x x by a polynomial g(x), the quotient and remainder were x and x + 4, respectively. Find g(x). Solution 4: 3 px x 3x x (Dividend) g(x) =? (Divisor) Quotient = (x ) Remainder = ( x + 4) Dividend = Divisor Quotient + Remainder 3 x 3x x g x x x x 4 3 x 3x x x 4 g x x 3 x 3x 3x g x x g(x) is the quotient when we divide x 3 3x 3x by x 0. Polynomials 11

12 x x 1 3 x x 3x 3x x x 3 x 3x x x x x 0 g x x x 1 Question 5: Give examples of polynomial p(x), g(x), q(x) and r(x), which satisfy the division algorithm and (i) deg p(x) = deg q(x) (ii) deg q(x) = deg r(x) Solution 5: According to the division algorithm, if p(x) and g(x) are two polynomials with g(x) 0, then we can find polynomials q(x) and r(x) such that p(x) = g(x) q(x) + r(x), where r(x) = 0 or degree of r(x) < degree of g(x) Degree of a polynomial is the highest power of the variable in the polynomial. (i) deg p(x) = deg q(x) Degree of quotient will be equal to degree of dividend when divisor is constant ( i.e., when any polynomial is divided by a constant). 0. Polynomials 1

13 Let us assume the division of 6x x by. Here, p(x) = 6x x g(x) = q(x) = 3x x 1and r(x) = 0 Degree of p(x) and q(x) is the same i.e.,. Checking for division algorithm, p(x) = g(x) q(x) + r(x) 6x x 3x x 1 6x x Thus, the division algorithm is satisfied. (ii) deg q(x) = deg r(x) Let us assume the division of x 3 + x by x, Here, p(x) = x 3 + x g(x) = x q(x) = x and r(x) = x Clearly, the degree of q(x) and r(x) is the same i.e., 1. Checking for division algorithm, p(x) = g(x) q(x) + r(x) x 3 + x = (x ) x + x x 3 + x = x 3 + x Thus, the division algorithm is satisfied. (iii)deg r(x) = 0 Degree of remainder will be 0 when remainder comes to a constant. Let us assume the division of x by x. Here, p(x) = x g(x) = x q(x) = x and r(x) = 1 Clearly, the degree of r(x) is 0. Checking for division algorithm, p(x) = g(x) q(x) + r(x) x = (x ) x + 1 x = x Thus, the division algorithm is satisfied. 0. Polynomials 13

14 EXERCISE NO:.4 Question 1: Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case: (i) x x 5x ;,1, x 3 + x 5x +; ½, 1, - 3 (ii) x 4x 5x ;,1,1 Solution 1: 3 (i)p(x) = x x 5x 3 1 Zeroes for this polynomial are 1,1, p p p Therefore, 1, 1, and are the zeroes of the given polynomial. 3 Comparing the given polynomial withax bx cx d, we obtain a =, b = 1, c = 5, d = 1 We can take, 1,y 1 1 b 1 a c 11 a 1 1 d 1 1 a Therefore, the relationship between the zeroes and the coefficients is verified. 0. Polynomials 14

15 3 (ii) px x 4x 5x Zeroes for this polynomial are, 1, 1 3 p p Therefore,, 1, 1 are the zeroes of the given polynomial. 3 Comparing the given polynomial with ax bx cx d, we obtain a = 1, b = 4, c = 5, d =. Verification of the relationship between zeroes and coefficient of the given polynomial 4 b Sum of zeroes a Multiplication of zeroes taking two at a time = ()(1) + (1)(1) + ()(1) 5 c = = 5 1 a d Multiplication of zeroes = 1 1 = 1 a Hence, the relationship between the zeroes and the coefficients is verified. Question : Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as, 7, 14 respectively. Solution : Let the polynomial be It is given that b 1 a 3 ax bx cx dand the zeroes be,, and. 0. Polynomials 15

16 7 c 1 a 14 d 1 a If a = 1, then b =, c = 7, d = 14 3 Hence, the polynomial is x x 7x 14. Question 3: If the zeroes of polynomial Solution 3: 3 px x 3x x 1 Zeroes are a b, a + a + b Comparing the given polynomial with p = 1, q = 3, r = 1, t = 1 Sum of zeroes = a b + a + a + b q 3a p a 3a a1 The zeroes are 1 b, 1 + b. Multiplication of zeroes = 1(1 b)(1 + b) t 1 b p 1 1b 1 1 b b b Hence, a = 1 and b = or. 3 x 3x x 1are a b,a,a b, find a and b. 3 px qx rx t, we obtain 0. Polynomials 16

17 Question 4: It two zeroes of the polynomial other zeroes. 4 3 x 6x 6x 138x 35 are 3 Solution 4: Given that + 3 and 3 are zeroes of the given polynomial. Therefore, x 3x 3= x + 4 4x 3 = x 4x + 1 is a factor of the given polynomial For finding the remaining zeroes of the given polynomial, we will find 4 3 the quotient by dividing x 6x 6x 138x 35 by x 4x + 1. x x x 4x 1 x 6x 6x 138x 35 x 4x x x 7x 138x 35 3 x 8x x 35x 140x 35 35x 140x 35 0, find Clearly, = x 4 6x 3 6x 138x 35 x 4x 1 x x 35 It can be observed that x x 35 is also a factor of the given polynomial. And =x x 35 (x 7)(x 5) Therefore, the value of the polynomial is also zero when or x 7 = 0 Or x + 5 = 0 Or x = 7 or 5 Hence, 7 and 5 are also zeroes of this polynomial. = 0. Polynomials 17

18 Question 5: 4 3 If the polynomial x 6x 16x 5x 10 is divided by another Polynomial x x k, the remainder comes out to be x + a, find k and a. Solution 5: By division algorithm, Dividend = Divisor Quotient + Remainder Dividend Remainder = Divisor Quotient x 6x 16x 5x 10 x a x 6x 16x 6x 10 a will be perfectly divisible by x x k. Let us divide by x 4 6x 3 16x 6x 10 a by x x k 4 3 x x k x 6x 16x 6x 10 a x x kx x 16 k x 6x 3 4x 8x 4kx 8 kx 16 kx 8k k 10 kx 10 a 8k k x 4x 8 k 8 k x 6 4k x 10 a x 4x 1 x x 35 (x 7)(x 5) It can be observed that 10 kx 10 a 8k k Therefore, 10 k = 0 and 10 a 8k k = 0 For 10 k = 0, k =10 will be Polynomials 18

19 And thus, k = 5 For 10 a 8k k = 0 10 a = 0 10 a = 0 5 a = 0 Therefore, a = 5 Hence, k = 5 and a = 5 0. Polynomials 19

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