NCERT Solutions For Class 10 Math Chapter – 2 Polynomial Exercise – 2.4
Q1. Verify that the numbers given alongside of the cubic polynomials below as their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
- 2x3+ x2 – 5x + 2 ; 1/2 , 1 , -2
- X3– 4x2 + 5x – 2 ; 2 , 1 , 1
Solutions:-
- P(1/2) = 2 x (1/2)3+ (1/2)2 – 5(1/2) + 2
=0
P(1) = 2 x (1)3 + (1)2 – 5(1) + 2
= 3 – 3
= 0
P(-2) = 2 x (-2)3 + (-2)2 – 5(-2) + 2
= -16 + 4 + 10 + 2
= 0
Hence verified.
- P(2) = (2)3– 4(2)2 + 5(2) – 2
= 0
P(1) = (1)3 – 4(1)2 + 5 x 1 – 2
= 0
Hence verified.
Q2. 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:-
= x3 – (a + B + y) x2 + (aB + By + ya) x – aBy
= x3 – 2x2 – 1x + 14
Q3. If the zeroes of the polynomial is x3 – 2x2 – 7x + 14.
Solution:-
Sum of the zeroes = a + B + y
3 = (a – b) + a + (a + b) = 3
3a = 3
A = 1
Product of zeroes = aBy
-1 = (a – b) a (a + b)
= a3 – ab2 = -1
(1)3 – (1)b2 = -1
B = √2
Q4. If tow zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2 + √3 find other zeroes.
Solution:-
[x -(2 + √3)] [ x – (2 – √3)]
= (x – 2)2 – (√3)2
X4 – 6x3 – 26x2 + 138x – 35
= (x2 – 4x + 1) (x2 – 2x – 35)
= (x2 – 4x + 1) (x – 7) (x + 5)
Hence , the other zeroes of the given polynomial are 7 and -5.
Q5. 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:-
P(x) = x4 – 6x3 + 16x2 – 25x + 10
Remainder = x + a
Using equation (I) , we get:
(-9 + 2k)x + 10 – 8 k + k2 = x + a
On comparing the like coefficients , we have:
-9 + 2k = 1
K = 5
10 – 8k + k2 – a
A = -5
Hence k = 5 an a = -5