Class10 NCRT Polynomials Exercise – 2.4 pdf || UP Board

 Polynomials
Exercise – 2.4*

 

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 -5x +2; 1/2 ,1, -2

Solution:

Let p(x) = 2x3 + x2 -5x +2

and the zeroes for p(x) are 1/2 , 1, -2, then putting the zeroes in the equation,

Similarly, putting 1, -2,

p(1) = 2(1)3 + 12 – 5(1) + 2 = 0

p(-2) = 2(-2)3 + (-2)2 – 5(-2) + 2 = 0

Therefore, all the zeroes of p(x).

Now, comparing with the general polynomial equation,

2x3 + x2 -5x +2 = ax3 + bx2 +cx +d

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

Let α=1/2, β=1, γ = -2

½ + 1 + (-2) = -1/2

½ - 1 = -1/2

-1/2 = -1/2

½×1 + 1×-2 + -2×1/2 = -5/2

½ + (-2) + (-1) = -5/2

½ -3 = -5/2

-5/2 = -5/2

½ × 1× (-2) = -2/2

 -2/2 = -2/2

Hence the relationship between the zeroes and the coefficients is verified.

 

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

Solution:

Let p(x) = x3 - 4x2 + 5x – 2

and the zeroes for p(x) are 2, 1, 1 then putting the zeroes in the equation,

p(2) = 23 -4(2)2 + 5×2 – 2 = 18 – 18 = 0

Similarly, putting 1, 1,

p(1) = (1)3 – 4(1)2 + 5(1) - 2 = 0

Therefore, all the zeroes of p(x).

Now, comparing with the general polynomial equation,,

x3 - 4x2 + 5x – 2 = ax3 + bx2 +cx +d

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

Let α=2, β=1, γ = 1

2 + 1 + 1 = -(-4/1)

4 = 4


2×1 + 1×1 + 1×2 = 5/1

5 = 5

2 × 1× 1 = -(-2/1)

 2 = 2

Hence the relationship between the zeroes and the coefficients is 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.

Solution:

Let p(x) = ax3 + bx2 +cx +d be a cubic polynomial and α, β, γ are zeroes

α + β + γ = 2

αβ + βγ + γα = -7

αβγ = -14

We know that,

α + β + γ = -b/a = 2/1

by comparison,

a = 1, b = -2

αβ + βγ + γα = c/a = -7/1

αβγ = -d/a = -14/1

इसी प्रकार c = -7, d = 14

Therefore, cubic polynomial is p(x) = 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.

Solution:

Let α = a-b, β= a, γ= a+b

Sum of zeroes α + β + γ = -b/a

(a-b) + a + (a+b) = -(-3/1)

3a = 3

a = 1

Product of zeroes αβγ = -d/a

(a-b) × a × (a+b) = -(1/1)  [a = 1]

1-b2 = -1

b2 = 2

b = ±2

Hence, a = 1 and b = ±√2

 

4. If two zeroes of the polynomial x4-6x3-26x2+13x-35 and 2±√3 find other zeroes.

Solution:

Let p(x) = x4-6x3-26x2+13x-35

Given that,

2+3 and 2-3 are zeroes of the polynomial p(x) so,

x = 2+3, x = 2-3

(x-2-3), (x-2+3) are factors of the polynomial p(x).

i.e. x2-4x+1 is a factor of the polynomial p(x)

p(x) = x4-6x3-26x2+13x-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-7)(x+5)

Hence x-7=0, x+5=0

The other zeroes of the polynomial are x = 7, –5.

 

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.

Solution:

Given thatolutioninder comes out to be alroduct of its zeroes taken two at a time and the product of its zeroes asrelationship between th,

Divisor = x2-2x+k

Dividend = x4-6x3+16x2-25x+10

Remainder = x+a

we know that,

   Dividend = Divisor × Quotient + Remainder

 x4-6x3+16x2-25x+10 = (x2-2x+k) × Quotient + (x + a)

Quotient

Thus, if the polynomial x4-6x3+16x2-26x+10-a is divided by the polynomial x2-2x+k, the remainder will be 0.

On comparing,

-10+2k = 0

2k = 10

k = 5

and

10 – a – 8k + k2 = 0

10 – a – 8×5 + 52 = 0  [k = 5]

-a-5 = 0

a = -5                  

Hence, k = 5, a = -5.

a ) = x

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