2. Polynomials
Exercise 2.2
Q1. Find the value of the polynomial 5x – 4x2 + 3 at
(i) x = 0 (ii) x = –1 (iii) x = 2
Solution:
(i) p(x) = 5x - 4x2 + 3
The value of the polynomial p(x) at x = 0 is given by
P(0) = 5(0) - 4(0)2 + 3
= 0 - 0 + 3
= 3
(ii) p(x) = 5x - 4x2 + 3
The value of the polynomial p(x) at x = 1 is given by
P(1) = 5(1) - 4(1)2 + 3
= 5 - 4 + 3
= 4
(iii) p(x) = 5x - 4x2 + 3
The value of the polynomial p(x) at x = 2 is given by
P(2) = 5(2) - 4(2)2 + 3
= 10 -16 + 3
= - 9
Q2.Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y) = y2 – y + 1
(ii) p(t) = 2 + t + 2t 2 – t 3
(iii) p(x) = x3
(iv) p(x) = (x – 1) (x + 1)
Solution:
(i) p(y) = y2 - y + 1
P(0) = (0)2- 0 + 1
=1
P(1) = (1)2- 1 + 1
= 1 - 1 + 1
= 1
P(2) = (2)2- 2 + 1 = 4 - 2 + 1 = 3
(ii) p(t) = 2 + t + 2t2- t3
P(0) = 2 + 0 + 2(0)2- (0)3
=2
P(1) = 2 + 1 + 2(1)2 - (1)3
= 4
P(2) = 2 + 2 + 2(2)2- (2)3
= 4 + 8 - 8
= 4
(iii) p(x) = x3
P(0)=(0)3 =0
P(1)=(1)3 =1
P(2)=(2)3=8
(iv)p(x) = (x – 1) (x + 1)
P(0)= (0-1) (0+1)=(-1) (1) =-1
P(1)= (1-1) (1+1) =0(1) =0
P(2)= (2-1) (2+1)=1(3) =3
Q3. Verify whether the following are zeroes of the polynomial, indicated against them.
Solution:
Q4. Find the zero of the polynomial in each of the following cases:
(i) P(x) = x + 5
(ii) P(x) = x – 5
(iii) Px) = 2x + 5
(iv) P(x) = 3x – 2
(v) P(x) = 3x
(vi) P(x) = ax, a ≠ 0
Solution (i) :
(i) P(x) = x + 5
⇒ x + 5 = 0
⇒ x = - 5
The zero of the polynomial is - 5.
Solution (ii) :
(ii) P(x) = x – 5
⇒ x – 5 = 0
⇒ x = 5
The zero of the polynomial is 5.
The zero of the polynomial is - 5/2.
(iv) P(x) = 3x - 2
3x - 2 = 0
The zero of the polynomial is 2/3.