Saturday, December 15, 2018

Polynomials Exercise 2.4

Polynomials Exercise 2.4

 

Page No: 36

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) 2x3 + x2 - 5+ 2; 1/2, 1, -2
(ii) x3 - 4x2 + 5x - 2; 2, 1, 1

Answer

(i) p(x) = 2x3 + x2 - 5+ 2
Now for zeroes, putting the given value in x.

p(1/2) = 2(1/2)3 + (1/2)2 - 5(1/2) + 2
= (2×1/8) + 1/4 - 5/2 + 2
= 1/4 + 1/4 - 5/2 + 2
= 1/2 - 5/2 + 2 = 0

p(1) = 2(1)3 + (1)2 - 5(1) + 2
= (2×1) + 1 - 5 + 2
= 2 + 1 - 5 + 2 = 0

p(-2) = 2(-2)3 + (-2)2 - 5(-2) + 2
= (2 × -8) + 4 + 10 + 2
= -16 + 16 = 0

Thus, 1/2, 1 and -2 are the zeroes of the given polynomial.

Comparing the given polynomial with ax3 + bx2 + c+ d, we get a=2, b=1, c=-5, d=2
Also, α=1/2, β=1 and γ=-2
Now,
-b/a = α+β+γ
⇒ 1/2 = 1/2 + 1 - 2
⇒ 1/2 = 1/2

c/a = αβ+βγ+γα
⇒ -5/2 = (1/2 × 1) + (1 × -2) + (-2 × 1/2)
⇒ -5/2 = 1/2 - 2 - 1
⇒ -5/2 = -5/2

-d/a = αβγ
⇒ -2/2 = (1/2 × 1 × -2)
⇒ -1 = 1

Thus, the relationship between zeroes and the coefficients are verified.

(ii)  p(x) = x3 - 4x2 + 5x - 2
Now for zeroes, putting the given value in x.

p(2) = 23 - 4(2)2 + 5(2) - 2
= 8 - 16 + 10 - 2
= 0

p(1) = 13 - 4(1)2 + 5(1) - 2
= 1 - 4 + 5 - 2
= 0

p(1) = 13 - 4(1)2 + 5(1) - 2
= 1 - 4 + 5 - 2
= 0

Thus, 2, 1 and 1 are the zeroes of the given polynomial.

Comparing the given polynomial with ax3 + bx2 + c+ d, we get a=1, b=-4, c=5, d=-2
Also, α=2, β=1 and γ=1
Now,
-b/a = α+β+γ
⇒ 4/1 = 2 + 1 + 1
⇒ 4 = 4

c/a = αβ+βγ+γα
⇒ 5/1 = (2 × 1) + (1 × 1) + (1 × 2)
⇒ 5 = 2 + 1 + 2
⇒ 5 = 5

-d/a = αβγ
⇒ 2/1 = (2 × 1 × 1)
⇒ 2 = 2

Thus, the relationship between zeroes and the coefficients are verified.

2. 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 2, –7, –14 respectively.

Answer

Let the polynomial be ax3 + bx+ cx + d and the zeroes be α, β and γ
Then, α + β + γ = -(-2)/1 = 2 = -b/a
αβ + βγ + γα = -7 = -7/1 = c/a
αβγ = -14 = -14/1 = -d/a

∴ a = 1, b = -2, c = -7 and d = 14
So, one cubic polynomial which satisfy the given conditions will be x3 - 2x2  - 7x + 14

3. If the zeroes of the polynomial x3 – 3x2 + x + 1 are a–b, a, a+b, find a and b.

Answer

Since, (a - b), a, (a + b) are the zeroes of the polynomial x3 – 3x2 + x + 1.
Therefore, sum of the zeroes = (a - b) + a + (a + b) = -(-3)/1 = 3

⇒ 3a = 3 ⇒ a =1

∴ Sum of the products of is zeroes taken two at a time = a(a - b) + a(a + b) + (a + b) (a - b) =1/1 = 1
a2 - ab + a2 + ab + a2 - b= 1
⇒ 3a2 - b2 =1

Putting the value of a,

⇒ 3(1)2 - b2 = 1
⇒ 3 - b2 = 1
⇒ b2 = 2
⇒ b = ±√2
Hence, a = 1 and b = ±√2

4. If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are  2±√3,  find other zeroes.

Answer

2+√3 and 2-√3 are two zeroes of the polynomial p(x) = x4 – 6x3 – 26x2 + 138x – 35.
Let x = 2±√3
So, x-2 = ±√3
On squaring, we get x2 - 4x + 4 = 3,
⇒ x2 - 4x + 1= 0

Now, dividing p(x) by x2 - 4x + 1

∴ p(x) = x4 - 6x3 - 26x2 + 138x - 35
= (x2 - 4x + 1) (x2 - 2x - 35)
= (x2 - 4x + 1) (x2 - 7x + 5x - 35)
= (x2 - 4x + 1) [x(x - 7) + 5 (x - 7)]
= (x2 - 4x + 1) (x + 5) (x - 7)

∴ (x + 5) and (x - 7) are other factors of p(x).
∴ - 5 and 7 are other zeroes of the given polynomial.

5. If the polynomial x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be x + a, find k and a.

Answer

On dividing x4 – 6x3 + 16x2 – 25x + 10 by x2 – 2x + k


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

But the remainder is given as x+ a. 

On comparing their coefficients,

2k - 9 = 1

⇒ k = 10 

⇒ k = 5 and,

-(8-k)k + 10 = a

⇒ a = -(8 - 5)5 + 10 =- 15 + 10 = -5 

Hence, k = 5 and a = -5 

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