Chapter 2.2 Polynomials in One Variable Solutions

Q1. Find the value of the polynomial 5x – 4x² + 3 at 
(i) x = 0 
(ii) x = –1 
(iii) x = 2
Ans: 
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

Q2. Find p(0), p(1) and p(2) for each of the following polynomials: 
(i) p(y) = y² − y + 1
Ans: 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
Ans:  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
Ans: ∴ p(0) = (0)³ = 0
p(1) = (1)³ = 1
p(2) = (2)³ = 8

(iv) P(x) = (x − 1) (x + 1)
Ans: ∴ 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

Q3. Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1, x = −1/3
Ans: 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
Ans: 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
Ans: 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
Ans: 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
Ans: 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
Ans: 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
Ans: 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
Ans: 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).

Q4. Find the zero of the polynomial in each of the following cases: 
(i) p(x) = x + 5 
Ans: p(x) = x + 5
⇒ x + 5 = 0
⇒ x = −5
∴ -5 is a zero polynomial of the polynomial p(x).

(ii) p(x) = x – 5
Ans: p(x) = x − 5
⇒ x − 5 = 0
⇒ x = 5
∴ 5 is a zero polynomial of the polynomial p(x).

(iii) p(x) = 2x + 5
Ans: 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 
Ans: 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 
Ans: p(x) = 3x
⇒ 3x = 0
⇒ x = 0
∴ 0 is a zero polynomial of the polynomial p(x).

(vi) p(x) = ax, a ≠ 0
Ans: 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.
Ans: p(x) = cx + d
⇒ cx + d =0
⇒ x = -d/c
∴ x = -d/c is a zero polynomial of the polynomial p(x).