Exercise 2.2 Page: 31
1. Find the value of the polynomial (x)=5x−4x2+3.
(i) x = 0
(ii) x = – 1
(iii) x = 2
Solution:
Let f(x) = 5x−4x2+3
(i) When x = 0
f(0) = 5(0)-4(0)2+3
= 3
(ii) When x = -1
f(x) = 5x−4x2+3
f(−1) = 5(−1)−4(−1)2+3
= −5–4+3
= −6
(iii) When x = 2
f(x) = 5x−4x2+3
f(2) = 5(2)−4(2)2+3
= 10–16+3
= −3
2. Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y)=y2−y+1
Solution:
p(y) = y2–y+1
∴ p(0) = (0)2−(0)+1 = 1
p(1) = (1)2–(1)+1 = 1
p(2) = (2)2–(2)+1 = 3
(ii) p(t)=2+t+2t2−t3
Solution:
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=2+1+2–1 = 4
p(2) = 2+2+2(2)2–(2)3=2+2+8–8 = 4
(iii) p(x)=x3
Solution:
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)
Solution:
p(x) = (x–1)(x+1)
∴ p(0) = (0–1)(0+1) = (−1)(1) = –1
p(1) = (1–1)(1+1) = 0(2) = 0
p(2) = (2–1)(2+1) = 1(3) = 3
3. Verify whether the following are zeroes of the polynomial indicated against them.
(i) p(x)=3x+1, x = −1/3
Solution:
For, x = -1/3, p(x) = 3x+1
∴ p(−1/3) = 3(-1/3)+1 = −1+1 = 0
∴ -1/3 is a zero of p(x).
(ii) p(x) = 5x–π, x = 4/5
Solution:
For, x = 4/5, p(x) = 5x–π
∴ p(4/5) = 5(4/5)- π = 4-π
∴ 4/5 is not a zero of p(x).
(iii) p(x) = x2−1, x = 1, −1
Solution:
For, x = 1, −1;
p(x) = x2−1
∴ p(1)=12−1=1−1 = 0
p(−1)=(-1)2−1 = 1−1 = 0
∴ 1, −1 are zeros of p(x).
(iv) p(x) = (x+1)(x–2), x =−1, 2
Solution:
For, x = −1,2;
p(x) = (x+1)(x–2)
∴ p(−1) = (−1+1)(−1–2)
= (0)(−3) = 0
p(2) = (2+1)(2–2) = (3)(0) = 0
∴ −1, 2 are zeros of p(x).
(v) p(x) = x2, x = 0
Solution:
For, x = 0 p(x) = x2
p(0) = 02 = 0
∴ 0 is a zero of p(x).
(vi) p(x) = lx+m, x = −m/l
Solution:
For, x = -m/l ; p(x) = lx+m
∴ p(-m/l)= l(-m/l)+m = −m+m = 0
∴ -m/l is a zero of p(x).
(vii) p(x) = 3x2−1, x = -1/√3 , 2/√3
Solution:
For, x = -1/√3 , 2/√3 ; p(x) = 3x2−1
∴ p(-1/√3) = 3(-1/√3)2-1 = 3(1/3)-1 = 1-1 = 0
∴ p(2/√3 ) = 3(2/√3)2-1 = 3(4/3)-1 = 4−1 = 3 ≠ 0
∴ -1/√3 is a zero of p(x), but 2/√3 is not a zero of p(x).
(viii) p(x) =2x+1, x = 1/2
Solution:
For, x = 1/2 p(x) = 2x+1
∴ p(1/2) = 2(1/2)+1 = 1+1 = 2≠0
∴ 1/2 is not a zero of p(x).
4. Find the zero of the polynomials in each of the following cases:
(i) p(x) = x+5
Solution:
p(x) = x+5
⇒ x+5 = 0
⇒ x = −5
∴ -5 is a zero polynomial of the polynomial p(x).
(ii) p(x) = x–5
Solution:
p(x) = x−5
⇒ x−5 = 0
⇒ x = 5
∴ 5 is a zero polynomial of the polynomial p(x).
(iii) p(x) = 2x+5
Solution:
p(x) = 2x+5
⇒ 2x+5 = 0
⇒ 2x = −5
⇒ x = -5/2
∴x = -5/2 is a zero polynomial of the polynomial p(x).
(iv) p(x) = 3x–2
Solution:
p(x) = 3x–2
⇒ 3x−2 = 0
⇒ 3x = 2
⇒x = 2/3
∴ x = 2/3 is a zero polynomial of the polynomial p(x).
(v) p(x) = 3x
Solution:
p(x) = 3x
⇒ 3x = 0
⇒ x = 0
∴ 0 is a zero polynomial of the polynomial p(x).
(vi) p(x) = ax, a≠0
Solution:
p(x) = ax
⇒ ax = 0
⇒ x = 0
∴ x = 0 is a zero polynomial of the polynomial p(x).
(vii) p(x) = cx+d, c ≠ 0, c, d are real numbers.
Solution:
p(x) = cx + d
⇒ cx+d =0
⇒ x = -d/c
∴ x = -d/c is a zero polynomial of the polynomial p(x).