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**Exercise 2.4****Ques 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) 2x ^{3} + x^{2} - 5x + 2; **

âˆ´

â‡’ 1/2 is a zero of p(x).

Again,

p (1) = 2 (1)^{3} + (1)^{2} - 5 (1) + 2

= 2 + 1 - 5 + 2

= (2 + 2 + 1) - 5

= 5 - 5 = 0

â‡’ 1 is a zero of p (x).

Also p (- 2) = 2 (- 2)^{3} + (- 2)^{2} - 5 (- 2) + 2

= 2(- 8) + (4) + 10 + 2

= - 16 + 4 + 10 + 2

= - 16 + 16 = 0

â‡’ -2 is a zero of p (x).

Relationship

âˆµ p(x) = 2x^{3} + x^{2} - 5x + 2

âˆ´ Comparing it with ax^{3} + bx^{2} + cx + d, we have :

a = 2, b = 1, c = - 5 and d = 2

Also 1/2 , 1 and - 2 are the zeroes of p(x)

Let Î± = 1/2 , Î² = 1 and Î³ = - 2

âˆ´ Î± + Î² +Î³ =

Sum of the zeroes =

â‡’ Î± + Î² + Î³ = -b/a

Sum of product of zeroes taken in pair:

Product of zeroes

Also

â‡’

Thus, the relationship between the co-efficients and the zeroes of p (x) is verified.

(ii) Here, p(x) = x^{3} - 4x^{2} + 5x - 2

âˆ´ p (2) = (2)^{3} - 4 (2)^{2} + 5 (2) - 2

= 8 - 16 + 10 - 2

= 18 - 18 = 0

â‡’ 2 is a zero of p(x)

Again p (1) = (1)^{3} - 4 (1)^{2} + 5 (1) - 2

= 1 - 4 + 5 - 2

= 6 - 6 = 0

â‡’ 1 is a zero of p(x).

âˆ´ 2, 1, 1 are zeroes of p(x).

Now,

Comparing p (x) = x^{3} - 4 (x^{2}) + 5x - 2 with ax^{3} + bx^{2} + cx + d = 0, we have

a = 1, b = - 4, c = 5 and d = - 2

âˆµ 2, 1, and 1 are the zeroes of p (x)

âˆ´ Let Î± = 2

Î² = 1

Î³ = 1

Relationship, Î± + Î² + Î³ = 2 + 1 + 1 = 4

Sum of the zeroes =

â‡’ Î± + Î² + Î³ = (-b/a)

Sum of product of zeroes taken in pair:

Î±Î² + Î²Î³ + Î³Î± = 2 (1) + 1 (1) + 1 (2)

= 2 + 1 + 2 = 5

and = c/a = 5/1 = 5

â‡’ Î±Î² + Î²Î³ + Î³Î± = c/a

Product of zeroes = Î±Î²Î³ = (2) (1) (1) = 2

â‡’

Thus, the relationship between the zeroes and the co-efficients of p(x) is verified.**Ques 2: Find the cubic polynomial with the sum, sum of the products of its zeroes taken two at a time and the product of its zeroes as 2, - 7, - 14 respectively.****Sol: **Let the required cubic polynomial be ax^{3} + bx^{2} + cx + d and its zeroes be a, b and g.

âˆ´

Since,

If a = 1, then -b/a = 2 â‡’ b = - 2

c/a = - 7 â‡’ c = - 7

-d/a = - 14 â‡’ d = 14

âˆ´ The required cubic polynomial

= 1x^{3} + (-2) x^{2} + (-7) x + 14

= x^{3} - 2x^{2} - 7x + 14**Ques 3: If the zeroes of the polynomial x ^{3} - 3x^{2} + x + 1 are a - b, a and a + b then find â€˜aâ€™ and â€˜bâ€™.**

p (x) = x

comparing it with Ax

We have

A = 1, B = - 3, C = 1 and D = 1

âˆµ It is given that (a - b), a and (a + b) are the zeroes of the polynomial.

âˆ´ Let, Î± = (a - b)

Î² = a

and Î³ = (a + b)

âˆ´ Î± + Î² + Î³ =

â‡’ (a - b) + a + (a + b) = 3

â‡’ 3a = 3

â‡’ a = 3/3 = 1

Again, Î±Î²Î³ = -D/A = -1

â‡’ (a - b) Ã— a Ã— (a + b) = - 1

â‡’ (1 - b) Ã— 1 Ã— (1 + b) = - 1 [âˆµ a = 1, proved above]

â‡’ 1 - b

â‡’ b

â‡’ b = âˆš2

Thus, a = 1 and b = .âˆš2

âˆµ Two of the zeroes of p (x) are :

or (x - 2)

or (x

or x

Now, dividing p (x) by x

âˆ´ (x

â‡’ (x

i.e., (x - 7) and (x + 5) are other factors of p(x).

âˆ´ 7 and - 5 are other zeroes of the given polynomial.

x

âˆ´ Remainder = (2k - 9) x - k (8 - k) + 10

But the remainder = x + Î±

Therefore, comparing them, we have :

2k - 9 = 1

â‡’ 2k = 1 + 9 = 10

â‡’ k = 10/2 = 5

and

Î± = - k (8 - k) + 10

= - 5 (8 - 5) + 10

= - 5 (3) + 10

= - 15 + 10

= - 5

Thus, k = 5 and Î± = - 5

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